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https://itstillruns.com/convert-gallon-km-per-litre-5027208.html | 1,721,926,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00550.warc.gz | 280,500,461 | 26,663 | # How to Convert Miles Per Gallon to Km Per Litre
by Johnno CaryUpdated August 02, 2023
Convert miles into kilometers. Since a kilometer is shorter than a mile, and there are 1.61 kilometers in every mile, you will have to multiply to get the answer. Doing the math (25 x 1.61 = 40.25) shows that 25 miles is equivalent to 40.25 kilometers.
Convert gallons into litres. Since a litre is smaller than a gallon, you must again multiply to get the answer. There are 3.79 litres in every gallon. Here the math is simple, because you are using only 1 gallon. 1 x 3.79 = 3.79, so a gallon of gas is equal to 3.79 litres.
Perform the calculation to determine how many kilometers are attributable to each litre of fuel consumed. Since you have determined that the vehicle traveled 40.25 kilometers on 3.79 litres of fuel, simple division will tell you how many kilometers are traveled on 1 litre. 40.25 ÷ 3.79 = 10.62. 25 mpg is therefore equivalent to 10.62 km/l.
Create a conversion factor, based on the above calculations. One mile is equal to 1.61 kilometers. One gallon is equal to 3.79 litres. Simple division will give you a conversion factor. 1.61 ÷ 3.79 = 0.425. This means that 0.425 km/l equals 1 mpg. You can use the newly calculated conversion factor by multiplying miles per gallon by 0.425. Checking your previous work, 25 mpg x 0.425 = 10.62 km/l, which is identical to the answer in the three-step calculation above. | 380 | 1,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.891931 |
http://www.solutioninn.com/are-students-who-participate-in-sports-more-extraverted-than-those | 1,506,022,905,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00522.warc.gz | 573,793,523 | 8,088 | # Question: Are students who participate in sports more extraverted than those
Are students who participate in sports more extraverted than those who do not? A random sample of students at a small university were asked to indicate whether they participated in sports (yes or no) and to rate their level of extraversion. Extraverts are outgoing, are talkative, and don't mind being the center of attention. Students were asked whether they agreed with the statement that they were extraverts, using a scale of 1-5 with 1 meaning "strongly disagree" and 5 meaning "strongly agree." There were 51 students who participated in sports and 64 who did not.
The mean extraversion score for those who participated in sports was 3.618, and the mean for those who did not participate was 3.172, a difference of 0.446 point. To determine whether the mean difference was significant, we performed a randomization test to test whether the mean extraversion level was greater for athletic students.
a. The histogram shows the results of 1000 randomizations of the data.
In each randomization, we found the mean difference between two groups that were randomly selected from the combined group of data: the combined data for the sporty and nonsporty. Note that, just as you would expect under the null hypothesis, the distribution is centered at about 0. The red line shows the observed sample mean difference in extraversion for the sporty minus the nonsporty. From the graph, does it look like the observed mean difference is unusual for this data set? Explain.
b. The software output gives us the probability of having an observed difference of 0.446 or more. (See the column labeled "Proportion = 7 Observed"). In other words, it gives us the right tail area, which is the p-value for a one-sided alternative that the mean extraversion score is higher for athletes than for nonathletes. State the p-value.
c. Using a significance level of 0.05, can we reject the null hypothesis that the means are equal?
d. If you did not have the computer output, explain how you would use the histogram to get an approximate p-value.
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5000 | 486 | 2,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-39 | latest | en | 0.969899 |
http://slideplayer.com/slide/3615727/ | 1,529,433,922,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00243.warc.gz | 307,095,365 | 26,769 | # Chapter 20: Testing Hypotheses About Proportions
## Presentation on theme: "Chapter 20: Testing Hypotheses About Proportions"— Presentation transcript:
Chapter 20: Testing Hypotheses About Proportions
Unit 5 AP Statistics
Hypotheses Hypotheses are working models that we adopt temporarily.
Our starting hypothesis is called the null hypothesis. The null hypothesis, that we denote by H0, specifies a population model parameter of interest and proposes a value for that parameter. We usually write down the null hypothesis in the form H0: parameter = hypothesized value. The alternative hypothesis, which we denote by HA, contains the values of the parameter that we consider plausible if we reject the null hypothesis.
Hypotheses (cont.) The null hypothesis, specifies a population model parameter of interest and proposes a value for that parameter. We might have, for example, H0: p = 0.20. We want to compare our data to what we would expect given that H0 is true. We can do this by finding out how many standard deviations away from the proposed value we are. We then ask how likely it is to get results like we did if the null hypothesis were true.
A Trial as a Hypothesis Test
Think about the logic of jury trials: To prove someone is guilty, we start by assuming they are innocent. We retain that hypothesis until the facts make it unlikely beyond a reasonable doubt. Then, and only then, we reject the hypothesis of innocence and declare the person guilty.
A Trial as a Hypothesis Test (cont.)
The same logic used in jury trials is used in statistical tests of hypotheses: We begin by assuming that a hypothesis is true. Next we consider whether the data are consistent with the hypothesis. If they are, all we can do is retain the hypothesis we started with. If they are not, then like a jury, we ask whether they are unlikely beyond a reasonable doubt.
P-Values The statistical twist is that we can quantify our level of doubt. We can use the model proposed by our hypothesis to calculate the probability that the event we’ve witnessed could happen. That’s just the probability we’re looking for—it quantifies exactly how surprised we are to see our results. This probability is called a P-value.
P-Values (cont.) When the data are consistent with the model from the null hypothesis, the P-value is high and we are unable to reject the null hypothesis. In that case, we have to “retain” the null hypothesis we started with. We can’t claim to have proved it; instead we “fail to reject the null hypothesis” when the data are consistent with the null hypothesis model and in line with what we would expect from natural sampling variability. If the P-value is low enough, we’ll “reject the null hypothesis,” since what we observed would be very unlikely were the null model true.
What to Do with an “Innocent” Defendant
If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty.” The jury does not say that the defendant is innocent. All it says is that there is not enough evidence to convict, to reject innocence. The defendant may, in fact, be innocent, but the jury has no way to be sure.
What to Do with an “Innocent” Defendant (cont.)
Said statistically, we will fail to reject the null hypothesis. We never declare the null hypothesis to be true, because we simply do not know whether it’s true or not. Sometimes in this case we say that the null hypothesis has been retained.
What to Do with an “Innocent” Defendant (cont.)
In a trial, the burden of proof is on the prosecution. In a hypothesis test, the burden of proof is on the unusual claim. The null hypothesis is the ordinary state of affairs, so it’s the alternative to the null hypothesis that we consider unusual (and for which we must marshal evidence).
Examples A research team wants to know if aspirin helps to thin blood. The null hypothesis says that it doesn’t. They test 12 patients, observe the proportion with thinner blood, and get a P-value of They proclaim that aspirin doesn’t work. What would you say? An allergy drug has been tested and found to give relief to 75% of the patients in a large clinical trial. Now the scientists want to see if the new, improved version works even better. What would the null hypothesis be? The new drug is tested and the P-value is What would you conclude about the new drug?
The Reasoning of Hypothesis Testing
There are four basic parts to a hypothesis test: Hypotheses Model Mechanics Conclusion Let’s look at these parts in detail…
The Reasoning of Hypothesis Testing (cont.)
Hypotheses The null hypothesis: To perform a hypothesis test, we must first translate our question of interest into a statement about model parameters. In general, we have H0: parameter = hypothesized value. The alternative hypothesis: The alternative hypothesis, HA, contains the values of the parameter we consider plausible when we reject the null.
Hypothesis Example A 1996 report from the U.S. Consumer Product Safety Commission claimed that at least 90% of all American homes have at least one smoke detector. A city’s fire department has been running a public safety campaign about smoke detectors consisting of posters, billboards, and ads on radio and TV and in the newspaper. The city wonders if this concerted effort has raised the local level above the 90% national rate. Building inspectors visit 400 randomly selected homes and find that 376 have smoke detectors. Is this strong evidence that the local rate is higher than the national rate? Set up the hypotheses.
The Reasoning of Hypothesis Testing (cont.)
Model To plan a statistical hypothesis test, specify the model you will use to test the null hypothesis and the parameter of interest. All models require assumptions, so state the assumptions and check any corresponding conditions. Your model step should end with a statement such Because the conditions are satisfied, I can model the sampling distribution of the proportion with a Normal model. Watch out, though. It might be the case that your model step ends with “Because the conditions are not satisfied, I can’t proceed with the test.” If that’s the case, stop and reconsider.
The Reasoning of Hypothesis Testing (cont.)
Model (cont.) Each test we discuss in the book has a name that you should include in your report. The test about proportions is called a one-proportion z-test.
One-Proportion z-Test
The conditions for the one-proportion z-test are the same as for the one proportion z-interval. We test the hypothesis H0: p = p0 using the statistic where When the conditions are met and the null hypothesis is true, this statistic follows the standard Normal model, so we can use that model to obtain a P-value.
The Reasoning of Hypothesis Testing (cont.)
Mechanics Under “mechanics” we place the actual calculation of our test statistic from the data. Different tests will have different formulas and different test statistics. Usually, the mechanics are handled by a statistics program or calculator, but it’s good to know the formulas.
The Reasoning of Hypothesis Testing (cont.)
Mechanics (continued) The ultimate goal of the calculation is to obtain a P-value. The P-value is the probability that the observed statistic value (or an even more extreme value) could occur if the null model were correct. If the P-value is small enough, we’ll reject the null hypothesis. Note: The P-value is a conditional probability—it’s the probability that the observed results could have happened if the null hypothesis is true.
P-value Example A large city’s DMV claimed that 80% of candidates pass driving tests, but a survey of 90 randomly selected local teens who had taken the test found only 61 who passed. Does this finding suggest that the passing rate for teenagers is lower than the DMV reported? What is the P-value for the one-proportion z-test? Don’t forget to check the conditions for inference!
The Reasoning of Hypothesis Testing (cont.)
Conclusion The conclusion in a hypothesis test is always a statement about the null hypothesis. The conclusion must state either that we reject or that we fail to reject the null hypothesis. And, as always, the conclusion should be stated in context.
The Reasoning of Hypothesis Testing (cont.)
Conclusion Your conclusion about the null hypothesis should never be the end of a testing procedure. Often there are actions to take or policies to change.
Conclusion Example Recap: A large city’s DMV claimed that 80% of candidates pass driving tests. Data from a reporter’s survey of randomly selected local teens who had taken the test produced a P-value of What can the reporter conclude? And how might the reporter explain what the P-value means for the newspaper story?
Alternatives Hypotheses
There are three possible alternative hypotheses: HA: parameter < hypothesized value HA: parameter ≠ hypothesized value HA: parameter > hypothesized value
Alternatives Hypotheses (cont.)
HA: parameter ≠ value is known as a two-sided alternative because we are equally interested in deviations on either side of the null hypothesis value. For two-sided alternatives, the P-value is the probability of deviating in either direction from the null hypothesis value.
Alternatives Hypotheses (cont.)
The other two alternative hypotheses are called one-sided alternatives. A one-sided alternative focuses on deviations from the null hypothesis value in only one direction. Thus, the P-value for one-sided alternatives is the probability of deviating only in the direction of the alternative away from the null hypothesis value.
Steps for Hypothesis Testing for One-Proportion z-Tests
Check Conditions and show that you have checked these! Random Sample: Can we assume this? 10% Condition: Do you believe that your sample size is less than 10% of the population size? Success/Failure: 𝒏𝒑 𝟎 ≥𝟏𝟎 and 𝒏𝒒 𝟎 ≥𝟏𝟎 State the test you are about to conduct Ex) One-proportion z-test Set up your hypotheses H0: HA:
Steps for Hypothesis Testing for One-Proportion z-Tests (cont.)
Calculate your test statistic 𝒛= 𝒑 − 𝒑 𝟎 𝒑 𝟎 𝒒 𝟎 𝒏 Draw a picture of your desired area under the Normal model, and calculate your P-value. Make your conclusion. When your P-value is small enough (or below α, if given), reject the null hypothesis. When your P-value is not small enough, fail to reject the null hypothesis.
Testing a Hypothesis Example
Home field advantage –teams tend to win more often when the play at home. Or do they? If there were no home field advantage, the home teams would win about half of all games played. In the 2007 Major League Baseball season, there were 2431 regular-season games. It turns out that the home team won 1319 of the 2431 games, or 54.26% of the time. Could this deviation from 50% be explained from natural sampling variability, or is it evidence to suggest that there really is a home field advantage, at least in professional baseball?
Graphing Calculator Shortcuts
One Proportion Z-Test: Stat TESTS 5: 1-Prop ZTest Po = hypothesized proportion x = number of successes n = sample size Determine the tail Calculate
P-Values and Decisions: What to Tell About a Hypothesis Test
How small should the P-value be in order for you to reject the null hypothesis? It turns out that our decision criterion is context-dependent. When we’re screening for a disease and want to be sure we treat all those who are sick, we may be willing to reject the null hypothesis of no disease with a fairly large P-value (0.10). A longstanding hypothesis, believed by many to be true, needs stronger evidence (and a correspondingly small P-value) to reject it. Another factor in choosing a P-value is the importance of the issue being tested.
P-Values and Decisions (cont.)
Your conclusion about any null hypothesis should be accompanied by the P-value of the test. If possible, it should also include a confidence interval for the parameter of interest. Don’t just declare the null hypothesis rejected or not rejected. Report the P-value to show the strength of the evidence against the hypothesis. This will let each reader decide whether or not to reject the null hypothesis.
Examples A bank is testing a new method for getting delinquent customers to pay their past-due credit card bills. The standard way was to send a letter (costing about \$0.40) asking the customer to pay. That worked 30% of the time. They want top test a new method that involves sending a DVD to customers encouraging them to contact the bank and set up a payment plan. Developing and sending the video costs about \$10 per customer. What is the parameter of interest? What are the null and alternative hypotheses? The bank sets up an experiment to test the effectiveness of the DVD. They mail it out to several randomly selected delinquent customers and keep track of how many actually do contact the bank to arrange payments. The bank’s statistician calculates a P-value of What does this P-value suggest about the DVD? The statistician tells the bank’s management that the results are clear and they should switch to the DVD method. Do you agree? What else might you want to know?
What Can Go Wrong? Hypothesis tests are so widely used—and so widely misused—that the issues involved are addressed in their own chapter (Chapter 21). There are a few issues that we can talk about already, though:
What Can Go Wrong? (cont.)
Don’t base your null hypothesis on what you see in the data. Think about the situation you are investigating and develop your null hypothesis appropriately. Don’t base your alternative hypothesis on the data, either. Again, you need to Think about the situation.
What Can Go Wrong? (cont.)
Don’t accept the null hypothesis. If you fail to reject the null hypothesis, don’t think a bigger sample would be more likely to lead to rejection. Each sample is different, and a larger sample won’t necessarily duplicate your current observations.
Recap We can use what we see in a random sample to test a particular hypothesis about the world. Hypothesis testing complements our use of confidence intervals. Testing a hypothesis involves proposing a model, and seeing whether the data we observe are consistent with that model or so unusual that we must reject it. We do this by finding a P-value—the probability that data like ours could have occurred if the model is correct. | 3,056 | 14,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-26 | latest | en | 0.906831 |
https://physics.stackexchange.com/questions/701647/how-can-one-detect-a-device-phone-falling-using-its-3-axis-accelerometer | 1,696,292,776,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00526.warc.gz | 482,547,124 | 43,535 | # How can one detect a device (phone) falling using its 3-axis accelerometer?
I am working on an Android app to use its 3-axis inbuilt accelerometer to find if the phone falls.
I googled and found that if the following value is less than 2:
$$a = \sqrt{x^2 + y^2 + z^2}$$
(where $$x, y, z$$ are the readings from accelerometer)
Then the device is considered to be free-falling.
I'm not sure how does this approach detect free-falling and where does the condition $$a < 2$$ come from.
Could anyone explain it?
• Hello! To have a discussion with an answerer, please comment under their answer. Check the edit history (by clicking the “edited X ago” link) to find your remarks.
– rob
Apr 3, 2022 at 12:36
TL;DR: Accelerometer records the $$x,y,z$$ projections of acceleration $$\vec{a_r} = \vec{a} - \vec{g}$$, where $$\vec{a}$$ is the total acceleration of the object.
When the object is free-falling, the recorded acceleration $$\vec{a_r}$$ is zero until air resistance force is large enough. Speed increase could lead to air resistance acceleration being as much as 1 $$\frac{m}{s^2}$$ (0.1 G) on the drops around 1 - 1.5 meters. Recorded acceleration modulo is calculated as $$\sqrt{x^2 + y^2 + z^2}$$ ($$x,y,z$$ - the readings from accelerometer). This is essential info to detect free-fall of your phone.
When your device rests on your desk, or in the palm of your hand, or on anything which causes it not to fall, there're two forces applied to the object: the force of gravity and the normal (ground reaction) force. Having opposite direction and being equal by modulo, these forces essentially compensate each other (so the object doesn't move) while "squeezing" the object, sort of.
I mean, consider this simple image, which you could have seen in a school book:
#### How can we describe these pics?
1. The object is static. If this object is you, you can feel that you're pushed against the ground. That's the normal force. You can definitely feel the gravity, but you do feel it just because the normal force is present.
2. The object is falling. Again, if that's you, you would feel like you're floating (if we don't account the air resistance for simplicity). You can't feel the gravity, since there's nothing that stops your movement - all the gravity force goes into accelerating you and nothing prevents it.
That's because we're under a constant (for a fixed height above the Earth) gravitational force. What does it imply?
1. If an object doesn't move, that means something prevents it. In this case, something stops an object with an acceleration of $$g = 9.8$$ $$\frac{m}{s^2}$$, with the direction of acceleration vector being opposite to the vector of gravity.
2. If an object falls, that means that either nothing stops it (that's free-falling), or something stops the object with an acceleration less than $$g = 9.8$$ $$\frac{m}{s^2}$$ (to be more specific, the projection of stopping acceleration on the $$z$$ axis is less than $$g = 9.8$$ $$\frac{m}{s^2}$$)
Thanks to this definition, we could now move on to the next step:
### What data an accelerometer gives to us?
Basically, accelerometer is giving us an information for the 3 projections (x, y, z) of the recorded acceleration.
Important note: although I've stated that already, it's necessary to understand the actual acceleration of the object is not the acceleration, recorded by the accelerometer. Accelerometer records the force $$\vec{F} = (\vec{a} - \vec{g})m$$ that stops it from free-falling, that could include the ground reaction force or the air resistance force, or any other force applied to the object.
But for the sake of simpleness let's consider a 2D world, with just two axis: x and y. Here you can see what data captures an accelerometer when an object stays on the ground and when it's tilted by 45 degrees (although still static):
There's much fun drawing this and all, but an important conclusion would be: the accelerometer will calculate an acceleration of approx $$g = 9.8$$ $$\frac{m}{s^2}$$ (vector parallel to gravity force) if it's static, i.e. when it's not falling
Let's consider another two examples:
1. An object is moving towards the ground with an acceleration less than $$g = 9.8$$ (falling). This means that the platform under the object is also accelerating towards the ground, but not free-falling either.
2. An object is moving towards the ground with an acceleration of approx $$g = 9.8$$ (free-falling). This means there's no platform or any support under the object.
In the first case, the total acceleration obtained from accelerometer would be approx $$6.8$$ $$\frac{m}{s^2}$$, in the second case - approx $$0$$ $$\frac{m}{s^2}$$.
In the both cases the object can be considered as falling, since the acceleration is less than $$g = 9.8$$ $$\frac{m}{s^2}$$
### So, what's the formula?
Now we can finally understand a formula.
An total modulo acceleration, obtained from the acceleration, would be:
$$|a| = \sqrt{x^2 + y^2 + z^2}$$
If the formula is unclear, you could refer Pythagorean theorem, one of our favorite school theorems:
In the same way you can calculate the acceleration vector length, where x, y, z are the projections. If the term "projection" is unclear, here's a quick explanation: take a stick in the sunny day, its shadow will be its projection.
As we remember, we can consider that the device is falling if $$|a|$$ is less than 9.8, but that doesn't account for a calculation error and air resistance.
### Experiment
In your case, you're mentioning that a phone can be considered to be falling if $$|a|$$ is less than $$2$$. That is valid for a rough approximation (20%), but not necessary at least for iPhones on small drops (without much air resistance). I have dropped my iPhone on the bed and measured the following accelerations during the 1.5 meter fall:
X Y Z t (time)
-0.016 -0.03 -0.151 0.00s
-0.021 0.015 -0.13 0.04s
-0.023 0.013 -0.202 0.08s
-0.016 0.014 -0.245 0.12s
-0.017 0.017 -0.308 0.16s
-0.028 0.02 -0.399 0.20s
-0.017 0.026 -0.462 0.24s
-0.019 0.023 -0.545 0.28s
-0.023 0.023 -0.622 0.32s
-0.031 0.021 -0.717 0.36s
Table 1. Raw X/Y/Z accelerometer input.
Upon calculating the acceleration with formula, we get these values for the beginning of the fall:
Acceleration Mistake or/and air resistance hit
0.154 ~ 1.5%
0.132 ~ 1.3%
0.203 ~ 2%
0.245 ~ 2.4%
0.308 ~ 3.1%
0.4 ~ 4%
0.463 ~ 4.7%
0.545 ~ 5.5%
0.622 ~ 6.3%
0.739 ~ 7.2%
Table 2. Calculated acceleration and its mistake.
Mistake (Y) and the time (X) plotted
As we can see, the mistake varies with a increasing pattern.
### Interesting observations
I believe that an increasing mistake is present because of air resistance. My phone is huge, and I dropped it flat parallel to the ground, that accounts for the maximal possible air resistance - that's exactly what the $$Z$$ axis values tell us (I mean, check the last column of the table 1).
I realize I could also drop it on the side to prove the point, but I'm just afraid that it jumps of the bed (or could it? that's another question...).
### Conclusion
A mistake of 20% sounds like an overkill to detect a start of falling.
Although, if you would like to account the whole process of falling, it could be just enough - it just depends on how much the acceleration slows down during the fall. That said, I doubt that there's any point accounting for mile-height drops, because well... the phone's not likely to last.
Another interesting task would be to record the end of falling. That's not that difficult - an impact will make an acceleration to suddenly jump to 5G's (50 $$\frac{m}{s^2}$$) and even more, which would make it easy to detect. I have got about 7.5 G's peak while dropping my phone onto the bed from the height of about 1 meter.
• 888 words, that would likely make a longest answer of mine Apr 1, 2022 at 12:54
• I haven't managed to understand why the error/time graph looks pretty linear at least during the 0.12 - 0.36s of the fall (isn't air resistance supposed to be like v^2?). As for the first 0.00s - 0.12s, I believe there maybe there's a little friction width my fingers after I've almost released the iPhone but it hasn't left the "starting" position completely. Apr 1, 2022 at 17:06
• OK, thanks! I upvoted. It is a thorough answer.
– hft
Apr 1, 2022 at 19:26
• @Rhino hello! Unfortunately I do not have an android phone, I have downloaded "Accelerometer" app on my iPhone. But you could search Play Store (or any other app store that your phone suppors) for the application with a name similar to "Accelerometer" Apr 1, 2022 at 19:35
• @nicael. Thank you very much for your well elaborate answer and it means a lot to me , I have summarise my understanding in my question , can you please correct me if I understood wrong Apr 2, 2022 at 23:55 | 2,354 | 8,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-40 | longest | en | 0.939458 |
https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/10/lesson/10.3.3/problem/10-107 | 1,723,658,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00547.warc.gz | 230,967,871 | 15,612 | ### Home > CC3MN > Chapter 10 > Lesson 10.3.3 > Problem10-107
10-107.
Solve the equations below by first changing each equation to a simpler equivalent equation. Check your solutions.
1. $3000 x - 2000 = 10{,}000$
• Divide by a common factor, then solve.
1. $\frac { - 2 x } { 3 } - \frac { x } { 7 } = 17$
• Multiply by the Least Common Multiple of $3$ and $7$.
1. $\frac { 5 } { 2 } x - \frac { 1 } { 3 } = 13$
• Multiply the equation by a common denominator.
1. $\frac { 3 } { 10 } + \frac { 2 x } { 5 } = \frac { 1 } { 2 }$
• See part (c) for help.
$x = \frac{1}{2}$ | 212 | 581 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.628557 |
https://www.scribd.com/document/22171447/Sequences-Worksheet-01-Number-from-GCSE-Maths-Tutor | 1,542,449,599,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00091.warc.gz | 983,693,638 | 28,978 | You are on page 1of 3
# Number - sequences
GCSE Maths Tutor
Sequences
question sheet N-SEQ-01
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1. Write down the 6th and 8th terms in each case.
(a) 2, 4, 6, 8 …
(b) 3, 6, 9, 12 …
(c) 4, 8, 12, 16 …
(d) 5, 7, 9, 11 …
(e) 4, 7, 10, 13 …
(f) 7, 12, 17, 22 …
(g) 2, 4, 8, 14 …
(h) 3, 5, 9, 15 …
(i) 4, 7, 13, 22 …
(j) 8, 11, 17, 26 …
(k) 5, 10, 20, 35 …
(l) 6, 10, 18, 30 …
(m) 13, 14, 16, 19 …
(n) 13, 20, 34, 55 …
(o) 11, 20, 38, 65 …
2. If n =1 is the first term, find the formula for the n th term in each case.
(a) 5, 10, 15, 20 …
(b) 4, 7, 10, 13 …
(c) 2, 7, 12, 17 …
(d) 1, 9, 17, 25 …
(e) 6, 11, 16, 21 …
(f) 5, 17, 29, 41 …
(g) 9, 12, 15, 18 …
(h) 21, 25, 31, 36 …
(i) 10, 17, 24, 31 …
(j) 19, 25, 31, 37 …
(k) 13, 20, 27, 34 …
(l) 29, 36, 43, 50 …
(m) 11, 15, 19, 23 …
(n) 9, 22, 35, 48 …
(o) 3, 30, 57, 84 …
3. If n =1 is the first term, the n th term can be witten as a + b(n - 1) where a and b are
constants. Find a and b
(a) 2, 4, 6, 8 …
(b) 4, 8, 12, 16 …
(c) 5, 7, 9, 11
(d) 6, 19, 32, 45 …
(e) 17, 22, 27, 32 …
(f) 11, 15, 19, 23 …
(g) 19, 25, 31, 37 …
(h) 57, 64, 71, 78 …
(i) 101, 103, 105, 107 …
(j) 2, 58, 114, 170 …
(k) 29, 42, 55, 68 …
(l) 4, 31, 58, 85 …
(m) 11, 22, 33, 44 …
(n) 100, 200, 300, 400
(o) 17, 40, 63, 86 …
GCSE Maths Tutor
www.gcsemathstutor.com
info@gcsemathstutor.com
Number - sequences
GCSE Maths Tutor
Sequences
www.gcsemathstutor.com
info@gcsemathstutor.com
1.
(a) 12, 16
(b) 18, 24
(c) 24, 32
(d) 15, 19
(e) 19, 25
(f) 32, 42
(g) 32, 58
(h) 33, 59
(i) 49, 88
(j) 53, 92
(k) 80, 145
(l) 66, 118
(m) 28, 41
(n) 118, 208
(o) 146, 279
(b) 4 + 3(n-1)
(c) 2 + 5(n-1)
(d) 1 + 8(n-1)
(e) 6 + 5(n-1)
(f) 5 + 12(n-1)
(g) 3 + 7(n-1)
(h) 1 + 6(n-1)
(i) 1 + 11(n-1)
(j) 19 + 6(n-1)
(k) 13 + 7(n-1)
(l) 29 + 7(n-1)
(m) 11 + 4(n-1)
(n) 9 + 13(n-1)
(o) 3 + 27(n-1)
3.
(a) 2, 2
(b) 4, 4
(c) 5, 2
(d) 6, 13
(e) 17, 5
(f) 11, 4
(g) 19, 6
(h) 57, 7
(i) 101, 2
(j) 2, 56
(k) 29, 13
(l) 4, 27
(m) 11, 11
(n) 100, 100
(o) 17, 23
www.gcsemathstutor.com
info@gcsemathstutor.com
2.
(a) 5n
GCSE Maths Tutor
Number - sequences
GCSE Maths Tutor
GCSE Maths Tutor
Sequences
www.gcsemathstutor.com
www.gcsemathstutor.com
question sheet N-SEQ-01
info@gcsemathstutor.com
info@gcsemathstutor.com | 1,297 | 2,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-47 | latest | en | 0.473818 |
https://www.mooc-list.com/course/what-are-chances-probability-and-uncertainty-statistics-coursera | 1,686,336,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00189.warc.gz | 1,006,210,552 | 16,423 | # What are the Chances? Probability and Uncertainty in Statistics (Coursera)
##### Start Date
Jun 5th 2023
Course Auditing
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41.00 EUR/month
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Familiarity with descriptive statistics and basic regression models
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This course focuses on how analysts can measure and describe the confidence they have in their findings. The course begins with an overview of the key probability rules and concepts that govern the calculation of uncertainty measures. We’ll then apply these ideas to variables (which are the building blocks of statistics) and their associated probability distributions. The second half of the course will delve into the computation and interpretation of uncertainty. We’ll discuss how to conduct a hypothesis test using both test statistics and confidence intervals. Finally, we’ll consider the role of hypothesis testing in a regression context, including what we can and cannot learn from the statistical significance of a coefficient. By the end of the course, you should be able to discuss statistical findings in probabilistic terms and interpret the uncertainty of a particular estimate.
Course 4 of 5 in the Data Literacy Specialization
### Syllabus
WEEK 1
Probability Theory
The Monty Hall problem is a classic brain teaser that highlights the often counterintuitive nature of probability. The problem is typically stated as follows: Suppose you're a contestant on a game show and asked to select one of three doors for your prize. Behind one door is a car and behind the other two doors are goats. You pick one door. The host, who knows what's behind each door, opens another, which has a goat. He then gives you the option to stick with your selected door or switch to the other closed door. What should you do? The answer is that, under these circumstances, you should always switch. There is a 2/3 chance of winning the car if you switch and a 1/3 chance of winning if you stick with your original selection. Most people, however, assume that there is only a 50/50 chance of winning if you switch. Hopefully this brain teaser, and content we cover in this module, will help you better approach probabilistic problems.
WEEK 2
Random Variables and Distributions
In this module, we'll dive into a topic you've likely encountered all of your adult life but perhaps have never explored from a statistical perspective: the normal curve. More generally, we'll discuss probability distributions, including their key features and relevance to quantifying uncertainty. Although studying probability theory can sometimes feel detached from applied statistics, it's valuable to develop a foundational understanding of probability to be able to critically evaluate statistical models. An appreciation for probability, and its counter-intuitive nature, will help you interpret the uncertainty of a statistical result as accurately as possible. This is particularly important when the stakes are high and policy makers want to know whether or not to act based on a statistical finding.
WEEK 3
Confidence Intervals and Hypothesis Testing
In this module we will apply the concepts of probability, random variables and distributions to measuring and interpreting uncertainty. In particular, we'll focus on statistical significance. A relationship is statistically significant if it can be distinguished from zero. Suppose you want to examine the effect of exposure to negative campaign ads on one's likelihood of voting. The independent variable is one's exposure to negative campaign ads and the dependent variable is one's likelihood of voting. If we find that exposure to negative campaign ads has no relationship with the likelihood of voting, we would say that this is a statistically insignificant relationship. If, instead, we find that exposure to negative campaign ads leads to a decline in one's likelihood of voting, we have uncovered a statistically significant (i.e., non-zero) relationship.
WEEK 4
Quantifying Uncertainty in Regression Analysis and Polling
In this final module of the course, we'll cover how to measure the uncertainty of regression estimates and poll results. It is often the case that a regression model will reveal a non-zero relationship, but it's important to determine whether that relationship sufficiently different from zero such that we can conclude that the relationship is statistically significant. For example, suppose a regression model reveals that a drug improves patient outcomes by 3.2%. Is 3.2% statistically different from 0? A statistical significance test will answer this question. This module, however, will also discuss some of the drawbacks of relying a statistical significance for data-driven decision making. While statistical significance is an important consideration, it is not the only criterion one should use when determining whether to act on a set of a statistical findings.
MOOC List is learner-supported. When you buy through links on our site, we may earn an affiliate commission. | 984 | 5,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.940921 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-6-section-6-4-graphs-of-polar-equations-exercise-set-page-754/42 | 1,726,766,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00046.warc.gz | 735,610,026 | 13,663 | ## Precalculus (6th Edition) Blitzer
The graph does not necessarily have symmetry with respect to the polar axis, the line $\theta =\frac{\pi }{2}$, or the pole.
We look for symmetry by making the following substitutions: (a) $\theta \to - \theta$ :$$r=\frac{3\sin 2(-\theta )}{\sin ^3 (-\theta )+\cos ^3 (-\theta )} \quad \Rightarrow \quad r=-\frac{3 \sin 2 \theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the polar axis. (b) $r \to -r, \quad \theta \to -\theta$ :$$-r =\frac{3 \sin 2 (-\theta )}{\sin ^3 (-\theta )+ \cos ^3(-\theta )} \quad \Rightarrow \quad r=\frac{3\sin 2\theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the line $\theta=\frac{\pi}{2}$. (c) $r \to -r$ :$$-r =\frac{3\sin 2\theta }{\sin ^3 \theta +\cos ^3 \theta } \quad \Rightarrow \quad r=-\frac{3\sin 2 \theta}{\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the pole. | 351 | 1,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.662197 |
https://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-5-ratio-and-proportion-review-exercises-page-374/52 | 1,527,364,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00077.warc.gz | 741,936,494 | 12,481 | Basic College Mathematics (9th Edition)
We can take the cross products of the given proportion to determine whether or not the proportion is true. If the cross products are equal, then we know that the proportion is true. $\frac{\frac{1}{5}}{2}=\frac{1\frac{1}{6}}{11\frac{2}{3}}=\frac{\frac{7}{6}}{\frac{35}{3}}$ $\frac{1}{5}\times\frac{35}{3}=2\times\frac{7}{6}$ Left side: $\frac{35}{15}=\frac{35\div5}{15\div5}=\frac{7}{3}$ Right side: $\frac{14}{6}=\frac{14\div2}{6\div2}=\frac{7}{3}$ Therefore, the proportion is True. | 186 | 525 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-22 | longest | en | 0.653183 |
http://acscihotseat.org/index.php?qa=54&qa_1=question-2-of-tut-2-total-accumulated-value-at-any-time-t-0 | 1,537,577,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00408.warc.gz | 7,706,358 | 9,371 | # Question 2 of tut 2: Total accumulated value at any time t>0
97 views
The question states:
"The force of interest at any time $$t$$, (measured in years) is given by
$$f(t) = \begin{cases}0.07-0.005t & 0 \le t < 5 \\ 0.06-0.003t & 5 \le t < 10 \\ 0.03 & t \ge 10 \end{cases}$$
What is the total accumulated in value at any time $$t > 0$$ of investments of R100 at times 0, 4 and 6?"
I understand getting to first 'time period' i.e. for $$0 \le t \lt 4$$ but from there on I get confused. The memo splits the answer into $$0 \le t \lt 4$$, $$0 \le t \lt 5$$, $$0 \le t \lt 6$$, $$0 \le t \lt 10$$, $$t \ge 10$$. Why is it not split from 0 to 4, 4 to 5, 5 to 6, 6 to 10 and greater than 10?
commented Apr 10, 2016 by (4,220 points)
reshown Apr 10, 2016
I think there was a mistake in that memo hey. The way you understand it is correct.
commented Apr 10, 2016 by (2,770 points)
Okay great, thanks Simon. And for question 4 (or question 2, since they're basically identical but with different values), how do they simplify the two terms into one? I seem to get two terms, one for each payment when $$5 \le t < 6$$ but the memo has only one term.
commented Apr 10, 2016 by (2,610 points)
From what I understand, you still kind of have separate values for each cashflow (summed together of course at correct time intervals, when the cashflows become relevant) at different times (which should be correct, if I understand your question correctly). The memo simply combines those cashflows at certain times, for example in question 4, at time 5, it will combine the accumulated value of the one rand invested at time 0, with the one rand invested at time 5, and then accumulate it in value further. I hope that answers your question.
commented Apr 10, 2016 by (4,220 points)
Oh ya, I had a second look and the memo is right. But it's a lot of working out! :/
commented Apr 10, 2016 by (2,770 points)
Thank you, that does make sense. However, I'm still struggling a bit with the manipulation.
In question 4, for $$5 \le t \lt 6$$ I get the following by accumulating the first payment to $$t=4$$, adding the payment at $$t=5$$ and then accumulating further:
$$(e^{0.04\times5+0.0025\times25}+1)e^{\int_5^t0.04+0.005sds}$$
Simplifying this I get:
$$e^{0.04t+0.0025t^{2}}(1+e^{\frac {-21}{80}})$$
I notice that $$e^{0.570488}$$ (from the memo) is roughly the same as my last bracket in the line above, but I am unsure as to how to simplify $$(1+e^{\frac {-21}{80}})$$ to $$e^{0.570488}$$.
And w.r.t question 2, I am very confused as to how they combine the intervals i.e. how $$0 \le t \lt 4$$ and $$4 \le t \lt 5$$ are combined into the interval $$0 \le t \lt 5$$.
Thanks very much for the help so far.
answered Apr 10, 2016 by (4,220 points)
selected May 6, 2016 by Pandy
Sorry for all the confusion. I've looked even more carefully, and the bounds are wrong. They should read $$0 \le t < 4$$, $$4 \le t < 5$$, $$5 \le t < 6$$, $$6 \le t < 10$$, and $$t \ge 10$$.
e.g. If $$4 \leq t < 5$$ then
$$A(t) = 100\exp[\int_{0}^{t}(0.07 - 0.005s)ds] + 100\exp[\int_{4}^{t}(0.07 - 0.005s)ds]$$
$$= 100\exp[0.07t - 0.0025t^{2}](1 + e^{-0.24}) = 178.663\exp[0.07t - 0.0025t^{2}]$$
Obviously it doesn't make sense if we substitute some $$t < 4$$ in the above. And similarly for the rest.
Regarding Q4, you're spot on. What's $$\log(1 + e^{-21/80})$$?
commented Apr 10, 2016 by (2,770 points)
Awesome! Thanks very much Simon. | 1,172 | 3,448 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-39 | longest | en | 0.944745 |
https://math.stackexchange.com/questions/4457035/calculate-the-integral-by-riemann-int-01-sqrtx-dx | 1,716,568,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00700.warc.gz | 339,425,297 | 42,179 | # Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}\,dx$
Calculate the integral by Riemann $$\displaystyle \int _{0}^{1}\sqrt{x} \, dx$$
$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$
$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$
\begin{align*} \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{\frac{k}{n}}\right) \cdotp \frac{1}{n} & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{\sqrt{k}}{\sqrt{n}}\right) \cdotp \frac{1}{n} \\[1ex] & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{n\sqrt{n}} \cdotp \sqrt{k}\right) \\[1ex] & = \lim_{n\rightarrow \infty }\frac{1}{n \sqrt{n}} \cdotp \sum_{k=1}^{n}\left(\sqrt{k}\right) \end{align*}
I am stuck on the sum of $$\sqrt k$$ maybe have another way to solve this question?
• Can you use, that root is integrable? Then you'll be able to choose specific points for partition. May 23, 2022 at 17:02
• Does this answer your question? For Riemann sums involving square roots, why do we let $c_{i} = \frac{i^{2}}{ n^{2}}$? May 23, 2022 at 17:02
• @user170231 well thanks but I tried right now and I didn't succeed I write $\dfrac{k^2}{n^2}$ and calculate the sum and the limit and I got an answer of 0.5 with is wrong May 23, 2022 at 17:13
• You can use the method here to do all general power integrals (continuity needed for irrational powers) May 23, 2022 at 17:27
As suggested in the linked question and other answers posted here, you need not use equally-spaced subintervals in the partition of $$[0,1]$$. Instead, consider the sequence of intervals
$$\left\{\left[\left(\frac{i-1}n\right)^2, \left(\frac in\right)^2\right] \,\bigg| \, 1 \le i \le n \right\}_{n\in\Bbb N}$$
each with length $$\frac{i^2}{n^2} - \frac{(i-1)^2}{n^2} = \frac{2i - 1}{n^2}$$.
Then the (right-endpoint) Riemann sum is
$$\sum_{i=1}^n \sqrt{\left(\frac in\right)^2} \frac{2i-1}{n^2} = \sum_{i=1}^n \frac{2i^2-i}{n^3} = \frac{(n+1)(4n-1)}{6n^2}$$
As $$n\to\infty$$, the sum converges to the definite integral, and we have
$$\int_0^1 \sqrt x \, dx = \lim_{n\to\infty} \frac{(n+1)(4n-1)}{6n^2} = \frac23$$
which agrees with the known antiderivative result,
$$\int_0^1 x^{\frac12} \, dx = \frac23 x^{\frac32}\bigg|_0^1 = \frac23$$
In the spirit of Reimann sums, you do not need to make the sequence $$a_k$$ arithmetic, you can use any sequence that covers the interval.
In this case, since you need to take the square root, it will make things simpler if you make $$a_k$$ a quadratic sequence...
$$a_k = 0 + \frac {k^2 }{n^2}$$
Then $$\sqrt{a_k}$$ simplifies to $$\frac kn$$ , but your $$\Delta x$$ will now depend of $$k$$... $$\Delta x_k = a_{k+1}-a_k = \frac {2k+1}{n^2}$$
So $$\int_0^1 \sqrt x dx = \lim_{n \to \infty} \sum_{k=1}^n a_k \Delta x_k$$
$$= \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k(2k+1)$$
$$= 2\lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k^2 + \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k$$
Now you can use the formulas...
$$\sum_{k=1}^n k^2 = \frac n6 (2n+1)(n+1) \text{ and } \sum_{k=1}^n k=\frac n2(n+1)$$
to obtain the required result
• Note that it is not quite enough that the partition cover the interval. You also need the length of the largest subinterval to converge to $0$ as $n$ grows to $\infty$. That does happen here with the largest subinterval being the last one, which is $(2n-1)/n^2$ wide. May 23, 2022 at 17:59
• That was what I was trying to evoke using the term "cover" - but I accept that my use of the term was not correct. Perhaps an alternate formulation of what constitutes a valid Reimann partition could be that every open cover of the sequence is also an open cover of the interval.
– WW1
May 23, 2022 at 18:22
I thought it might be instructive to present a way forward that uses creative telescoping and simple estimates. To that end we proceed.
Note that we can write
\begin{align} n^{3/2}-1&=\sum_{k=1}^n \left((k+1)^{3/2}-k^{3/2}\right)\\\\ &=\frac32 \sum_{k=1}^n \left(\sqrt{k} +O(k^{-1/2})\right)\tag1 \end{align}
Rearranging $$(1)$$ reveals
$$\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23 \left(1-n^{-3/2}\right)+\frac1{n^{3/2}}\underbrace{\sum_{k=1}^n O(k^{-1/2})}_{\le O(n)}$$
Letting $$n\to \infty$$ yields the coveted result
$$\lim_{n\to\infty}\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23$$
Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $$\sqrt{k}$$ values with nearby integers.
\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k} \end{align}
Already this might give you pause. But for one thing, as $$n\to\infty$$, so does $$n^2$$. The sum on the right is like the sum on the left but between one value of $$n$$ and the next, you add
$$\frac{1}{n^2\sqrt{n^2}}\left(\sqrt{(n-1)^2+1}+\sqrt{(n-1)^2+2}+\cdots+\sqrt{(n-1)^2+(2n-1)}\right)$$
That is, you add $$2n-1$$ terms that are each less than or equal to $$n$$, and the result is divided by $$n^3$$. So this tail that you add with each step, all by itself, converges to $$0$$. So the two sums have the same limit. Starting again:
\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n^2}\right)\\ &\leq\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{9}+\cdots+\sqrt{n^2}\right)\\ \end{align} Here we have one $$\sqrt{1}$$s, then three $$\sqrt{4}$$s, then five $$\sqrt{9}$$s, and so on.
\begin{align} &=\lim_{n\to\infty}\frac{1}{n^3}\left(1+3(2)+5(3)+\cdots+(2n-1)n\right)\\ &=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n(2k-1)k\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}2\right)\\ &=\lim_{n\to\infty}\left(\frac13(1+1/n)(2+1/n)-\frac{(1+1/n)}{2n}\right)\\ &=\frac23 \end{align}
So we have an upper bound and the limit you are after is at most $$\frac23$$. (Of course we know independently that the limit actually is $$\frac23$$.)
Do you see how to modify this to get a lower bound? Hint: $$1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$$.
Maybe this will help you (i hope this is true).
First let's make a change of variable: $$t=\sqrt{x}$$ so we get $$\int_{0}^{1}\sqrt{x}dx=\int_{0}^{1}2t^2dt$$
Now lets compute: $$\int_{0}^{1}2t^2dt$$ using Riemman integrable.
Thus by doing exactly as you did we get: $$lim_{n \to \infty } \frac{2}{n^3} \sum_{k=1}^{n}k^2$$
And we know that the sum of squares of n natural numbers can be calculated using the formula $$[n(n+1)(2n+1)] / 6$$. So: $$lim_{n \to \infty } \frac{2}{n^3} [n(n+1)(2n+1)] / 6 = 2/3$$
This is correct because: $$\int_{0}^{1}\sqrt{x}dx = 2/3$$
So we did exactly what it was asked:
Calculate the integral by Riemann $$\int_{0}^{1}\sqrt{x}dx$$
But by calculating integral by Riemman on a simplest function.
• This is not a clear answer May 23, 2022 at 17:27
• @FShrike why? how can i improve it? May 23, 2022 at 17:31
• They want square roots May 23, 2022 at 17:34
• @FShrike It is not say in the question that we have to calculate the Riemman sum directly on the root function given. In the question it is only asked to compute the given integrable using a Riemman integrable, and this is exactly what we did. If not the op must be more precise. May 23, 2022 at 17:38
If the answer upper isn't satisfying here an oher.
Choose a partition as follow: $$x_0=0 , x_1 = 1/n^2 ,x_2=(2^2)/n^2 ,...,x_n=n^2/n^2=1$$ so $$x_k-x_{k-1}=\Delta x_k=\frac{2k-1}{n^2}$$.
$$\sum_{k=1}^{n}\frac{2k-1}{n^2}\frac{k}{n}=\sum_{k=1}^{n}\frac{2k^2-k}{n^3}$$
And we know a formula for each of those sums. | 3,010 | 7,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 58, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.687373 |
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South African Journal of Science
Print version ISSN 0038-2353
When to stay, when to go: trade-offs for southern African arid-zone birds in times of drought
W.R.J. DeanI, *; P. BarnardI, II; M.D. AndersonIII
IDST/NRF Centre of Excellence at the Percy FitzPatrick Institute of African Ornithology, University of Cape Town, Rondebosch, 7701 South Africa
IIBirds and Environmental Change Partnership, Climate Change and Bio-Adaptation Division, South African National Biodiversity Institute, Private Bag X7, Claremont, South Africa
IIIBirdLife South Africa, P.O. Box 515, Randburg 2125, Gauteng, South Africa
Key words: avian-nomadism, arid environments, Karoo-Namib Desert, Kalahari, breeding
Prolonged spells of drought pose dilemmas for most organisms, even those adapted to hyper-aridity. For birds, reproduction, feeding, dispersal and moult are all activities which may need careful timing in relation to fluctuations in ecological productivity. In arid and semi-arid ecosystems, rainfall often triggers events in the life cycle which may be suppressed for months or even years during dry periods.1,2,3,4 In times of very low productivity or harsh conditions, birds, like many other animals, can move to escape local conditions and improve their chances of feeding or reproduction.5 However, some resident species may not move, but simply adjust their activities to 'ride out' difficult periods.
Semi-arid ecosystems are characterised by wet or dry states that are patchy in time and space. This patchiness is particularly true of southern Africa, where much of the region is semi-arid6 and the environment experiences extremes in weather, from periods of intense and prolonged drought to exceptionally high rainfall events.7 Rainfall in arid and semi-arid ecosystems in the southern hemisphere, including southern Africa, is greatly affected by El Niño Southern Oscillations (ENSO), leading to large variability in rainfall and prolonged droughts.6 Similarly, extensive wet periods (La Niña events) have concomitant effects on ecosystem functioning.8 The effects of El Niño and La Niña have been well studied in a few organisms, particularly birds, where changes in species richness and density9 and reproduction10 are correlated with the wet–dry cycles of ENSO years.
Birds, other animals and plants respond to increased rainfall in similar ways, and both birds and plants tend to increase reproductive effort with more rain.11–14 The response by biota to extended dry periods in semi-arid environments, however, differs markedly between and within invertebrate and vertebrate phyla, and markedly between most animals and plants. Plants cope with droughts by dying after depositing dormant propagules (ephemerals) or becoming dormant and restricting their internal water use by discarding leaves or leaves and stems.13,15 Although death and propagule dormancy are options taken by plants and some invertebrate taxa, viz. brine shrimp (Artemia), these options are not available to vertebrates, leaving escape and water conservation as the alternatives.5,16
Drought-induced dormancy in vegetation therefore has effects on animals. Animals in general, and birds in particular, cope with droughts and changes in vegetation by using behavioural and physiological tactics, including opportunistic movement away (in birds), shifts in habitat5 and deferred hatching or dormancy in eggs (in locusts).2,3,4 It is not known in detail whether birds make dietary shifts during extended droughts, but many species are opportunistic in their foraging and feeding in the Karoo17 and it is very likely that such shifts do occur.
Our questions in this review are: (1) what is the influence of variability in rainfall on avian populations, including breeding and movements? and (2) how resilient are bird populations to extended dry periods? While we do not explicitly review climate change per se, our conclusions should be of value in predicting species-specific responses and levels of vulnerability to changing rainfall patterns in Africa.
The influence of rainfall on population dynamics
Exceptionally high rainfall may stimulate rare breeding events. For example, banded stilts (Cladorhynchus leucocephalus) in arid Australia arrive and breed in very large colonies following exceptionally high rainfall at ephemeral pans that may have been dry for decades.1,18 Variability in rainfall has strong effects on clutch sizes and population dynamics in birds, as observed in unrelated species in different arid parts of the world, such as Galapagos ground finches on the Galápagos Islands,10 galahs (Cacatua roseicapilla) in Australia19 and larks,20 thrushes and helmet-shrikes in arid southern Africa.12,21 Droughts have different effects on bird populations, often stimulating movements into better-watered areas, but also, conversely and seemingly inexplicably, stimulating movements into dry areas.22 Crowned hornbills (Tockus alboterminatus) wander from their forest habitat in the non-breeding season,23 some years reaching far into the semi-arid Karoo.24,25 These movements are thought to be related to unusual aridity or cold in their coastal habitats,23 although these are probably always more mesic than the areas to which the birds disperse.
The movements of nomadic birds in arid and semi-arid ecosystems throughout the world have been reviewed by Dean.5 We know that some birds have evolved to cope with stochastic weather events in their habitat. What is less clear, however, is what resident bird species do when faced with increasing aridity in their environment. For those species that move, the benefits must outweigh the costs.26 Similarly, for those species that are resident, the benefits of being resident must outweigh the cost of moving. For example, in partially-migratory rock kestrels (Falco rupicolus), individual males used one or the other of these two tactics.27 Males departed for non-breeding areas later than did females, and returned to their territories earlier than did females. Not all males migrated, and those that remained successfully retained their territories and had a higher probability of breeding in the next breeding season. However, there was a cost: males that stayed on their territories during the winter faced increased competition for food. Males that migrated probably had higher survival, but risked losing their territories. Male rock kestrels in the Karoo thus faced a trade-off between increased chances of breeding and the risk of mortality.27
Trade-offs are necessary where rainfall is highly unpredictable. Unlike the short breeding season in the temperate northern hemisphere, the concept of extended 'equally good months' for breeding (EGMs)28 often applies to bird populations in drier parts of the southern hemisphere, where weather patterns are less predictable and suitable or unsuitable conditions for breeding may arise at any time of the year. EGMs are calculated as the months in which there are breeding records equal to or above the average number of breeding records per month for the species. Africa and Australia both have a mean EGM 1.9 months a year longer than anywhere else,29 presumably caused by out-of-season rainfall, or rainfall in areas that have been dry for a long period. Even within South Africa, areas with higher coefficients of variation in total annual rainfall,30 such as the Karoo, have more EGMs than areas with more sharply seasonal, but more predictably timed, rainfall such as the southwestern Cape (Table 1).
The most frequently measured responses by birds in arid regions to above average rainfall or isolated rainfall events are breeding out of season, breeding in large numbers, and increases in clutch size or number of young produced. For example, 'heavy rains' following a five-year drought in the western Karoo stimulated bird breeding activity, with a number of species recorded with eggs and young, and all collected specimens in breeding condition31. Similarly, 170 mm of rain between January and March 1985, following a severe drought in the central Karoo, was believed to have stimulated breeding in 27 species of birds32. However, no studies of breeding activity in either case were made during the 'severe drought' broken by the rains, so the level of increase in breeding activity was difficult to evaluate.
Better data on the influence of rainfall on population dynamics are provided by long-term studies. Verreaux's eagles (Aquila verreauxii) breeding in the Matobo Hills, Zimbabwe showed responses to variations in annual rainfall amounts. The number of resident pairs of Verreaux's eagles increased during the high rainfall years and correlated with increases in the number of hyraxes (Procavia capensis and Heterohyrax brucei)33. Both the number of pairs in the local population that bred and their production of young fit a negative exponential curve (Fig. 1), suggesting that the eagles benefit from both lower and higher rainfall years, but not from average rainfall years. Lower rainfall years have as many pairs breeding as higher rainfall years, with relatively higher breeding success. This suggests that breeding success is not only correlated with rainfall, but that other factors also influence the population dynamics of this species at this locality. Average rainfall years may increase competition for hyraxes, or higher rainfall years may lead to increased chick mortality.
In deciduous woodland habitats that come into leaf before the rains, there is some evidence that the level of breeding activity in birds is correlated with rainfall during the previous year. A study in the Miombo woodland in Zimbabwe21 showed that rainfall during both the previous breeding season and current breeding season influences the number of nests and clutch sizes in Kurrichane thrushes (Turdus libonyanus) and white-crested helmet-shrikes (Prionops plumatus) (Tables 2 and 3). Both are resident species, generally maintaining territories all year round. The difference in breeding effort between years was not very marked in the Kurrichane thrush following the dry year of 1972, but breeding effort following the wettest year (1973) increased markedly, with an earlier start to breeding, increases in nest density per hectare (Table 2) and increases in the number of breeding attempts.21,34 With lower rainfall in the previous season there were decreases in the number of breeding groups and total number of birds, and increases in home range and territory sizes in white-crested helmet-shrikes (Table 3).21
In a more arid environment, Maclean35 showed that resident and nomadic larks (Alaudidae) and resident sociable weavers (Philetarius socius) in the Kgalagadi Transfrontier Park responded quickly to rain, and that the duration of the breeding 'season' varied according to the amount of rain. Table 4 gives data for one species, the grey-backed sparrowlark (Erempoterix verticalis) and Table 5 presents data for the sociable weaver. The average annual rainfall for this area is about 180 mm per year so the amount of rain that fell in April–November 1965 and January–April 1966 was more than 70% of the average annual rainfall. The relationship between rainfall, breeding season and breeding success, however, is not linear. Maclean36 suggested that the duration of the breeding season is not directly governed by the amount of rain, but rather by the time of year in which rain falls.
This suggestion is supported by Lloyd20 who noted that the timing and length of the breeding season in Bushmanland, Northern Cape Province depended on the integrated effect of rainfall and temperature on the growing season of the vegetation. Higher breeding success, measured as the percentage of young that left the nest, apparently depends on time of year and not the amount of rain per se (Table 5). The breeding success of sociable weavers was markedly lower during the hot summer months of January to April than the cooler, longer period from April to November36 (Table 5).
Clutch size may vary with rainfall and aridity. In sociable weavers35 there was some variation in mean clutch size per month, but specific rainfall data for those months were unavailable. However, a study of a bird community in Bushmanland, Northern Cape, showed that breeding activity of five species increased markedly after rainfall (Table 6), and more than half the species showed an increase in average clutch size compared to drier conditions.20 Two species, black-eared sparrowlarks (Eremopterix australis) and grey-backed sparrowlarks, showed a rapid response to rain and began laying larger clutches within seven days of 78 mm of rain.20
The influence of drought on bird populations
Movements in birds may be in response to rain or in response to drought. In general, drought, or extended dry periods, has the effect of reducing bird diversity and reducing the number of individuals,37 either through emigration5 or lowered survival.38 We do not know how resilient bird populations are to drought in southern Africa, because there are no intensive long-term studies of birds that have covered a full cycle of dry and wet years. There are, however, studies showing that resident birds may be quite resistant to drought. A long-term study at Tierberg, near Prince Albert in the Karoo, showed that the number of individuals of resident long-billed larks (Certhilauda subcoronata), spike-heeled larks (Chersomanes albofasciata), Karoo chats (Cercomela schlegelii) and rufous-eared warblers (Malcorus pectoralis) showed relatively small annual fluctuations regardless of rainfall39 (Table 7). This data support the idea that in some cases the benefits of remaining in an area outweigh the costs of moving.
However, in the Kalahari, Botswana, resident species may show less resistance to drought, suggesting that resistance may depend to some extent on more subtle factors. For example, dry-season bird diversity, and number of individuals, were high following a higher rainfall wet season, and markedly lower following a poor rainfall season.37 Of 39 resident and nomadic species, populations of three resident species were stable and nine resident and nomadic species increased in abundance during the wet year following a dry year. But 20 species, including some nomads, decreased in abundance during that year. In the dry year following a wet year, all species had decreased markedly in number, and 12 species were no longer locally present. All nomadic species showed large increases in number in the wet year. Only one of the nomads, Temminck's courser (Cursorius temminckii), was present during the second dry year. There was no general pattern apparent in the residents or nomads, except that bird numbers were generally lower in the wet season following a dry season than in the dry season following a wet season. Both groups of birds showed fluctuations in number that cannot be entirely explained by rainfall amount. Some residents (but not all) showed some initial resilience to the dry conditions. Such resilience, however, did not last over the entire period of the study.
Studies of birds under drought conditions show that responses to changes in environmental conditions may be species-specific. In Bushmanland, Sclater's lark (Spizocorys sclateri) showed some resistance to drought, nesting at a higher density during low rainfall than during a wetter spell (Table 6). During the dry period, the larks fed on, amongst others, the large seeds of a grass (Enneapogon desvauxii), a dependable resource in dry times because the seeds are held almost below ground at the base of the plant. Because of the difficulty of extracting the seeds, this resource is used only by Sclater's larks and Stark's larks (S. starki). Stark's larks are nomadic and absent during drier periods; Sclater's larks are thus able to use the resource without competition during droughts.
In species for which the costs of moving may be high, for example, dune larks (Calendulauda erythrochlamys) on sparsely vegetated 'islands' in the Namib Desert dune sea, some individuals nevertheless moved away during a drought, with the population on some islands reduced to about one half or one third of their former size.40 There was also a reduction in foraging group size during the drought. Foraging group sizes in wetter periods were 2–6 individuals,41 whereas during the drought no groups larger than two birds were seen.40 No shifts in diet or changes in foraging methods and patterns between the wetter period and the drought were apparent, but there were seasonal differences in foraging patterns, and presumably diet, regardless of whether there was a drought or not.40 Whether the larks remained on their patch or moved away was strongly dependent on plant composition and the state of the vegetation and invertebrate populations. Safriel40 suggested that in patches where plant and invertebrate resources remained rich, foraging patterns (and presumably diet) would remain similar all year round. Dune larks can probably only be resident in the absence of any competition. The Namib Desert supports large numbers of nomadic granivores following rainfall.42 Some nomadic species overlap temporarily in habitat with dune larks, but depart as the ecosystem dries out.
Dune larks are thus able to remain on their patches provided there is no potential competition.
Decision making: when to stay and when to go?
We have shown, in this brief review, that birds use two main strategies for coping with the environment getting wetter or drier. In general, 'wet' years stimulate higher densities of nests (i.e. smaller territories), larger clutch sizes, unseasonal breeding, and, depending on the time of year, higher breeding success. Rainfall over a certain amount20,35,36,43 triggers breeding in resident species and an influx of nomadic species that breed and then move on. The amount of rainfall needed to trigger breeding varies between areas and seasons; 60 mm of rain in two days, followed by 22 mm eight days later in summer (January) stimulated breeding in the Kalahari,35 but 41 mm in Bushmanland over several days in summer (February) without any follow-up rains did not trigger breeding.20 It did, however, lead to an influx of nomadic granivores which did not breed at that time, whereas 54 mm of rain in winter (July) in the same area triggered breeding in all species.20
Nomads move in response to environmental cues, which are poorly understood for small terrestrial granivores,5,44 but better understood for larger aquatic species.45 Environmental cues for aquatic bird species may include distant thunderstorms, indicating heavy rain and thus the formation of temporary pans.45 This scenario may also hold for small terrestrial insectivores, but for small avian granivores the lag between rainfall and the response by the plants may be a week or more. It has been suggested5 that drifts of awns of grasses (used by nomadic larks and buntings for nest linings) may provide a strong visual cue that the area is suitable for nesting, but which other cues draw the birds initially to the area is not known.
Coping with an environment that is drying out may be easier for nomadic species. The environment dries out, particular food items become scarce, and provided that their young are large and able to fly, the nomads move on. What they must decide is where, and not when, to go. For resident species, however, whether to go or stay is a more difficult and apparently individual process. The examples we give suggest that in resident bird communities, there is a community-wide response to below-average rainfall, with fewer birds breeding in dry years and lower than usual production of young. But whether to breed or not, to stay or not, or to join a larger group remain individual decisions, possibly influenced by the decisions made by conspecifics.
If birds make the decision to remain in their area during a drought, they have several options to increase their chances of survival. They can increase territory size if neighbouring territories have become vacant or poorly-defended; they can shift their diet to eat a wider range of items; or they can abandon territories and join mixed-species foraging flocks. There is very little evidence of any of these tactics being particularly obvious in drought years. Furthermore, the benefits of all tactics have to be traded off against the losses. Territories that have increased in size might be too large to defend when good times arrive, territories that are abandoned need to be re-established when the drought is over, and shifts in diet may have consequences for the health or reproductive success of individual birds. But more importantly, resident species usually know their territory or home range very well.46 They know good places to forage, to nest and to avoid predators. This knowledge will increase in quality over the years.47 This fact is supported by Hanmer38 who shows that adult birds survive droughts better than immature ones, and suggests that it is due to the adults' better knowledge of parts of the habitat that gives them an advantage. Resident bird species are thus more likely to be resistant to drought and to use available resources as best they can without giving up their patch.
This review is based on several talks given at the Arid Zone Ecology Forum, Sutherland, in September 2007. We thank the Critical Ecosystem Partnership Fund for sponsoring the attendance of W.R.J.D. at the forum, and we thank the committee of the Arid Zone Ecology Forum for funding towards writing this review. We are grateful to Sue Milton for comments and suggestions on the manuscript.
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36. Maclean G.L. (1973). The sociable weaver, part 3: breeding biology and moult. Ostrich 44, 219–240. [ Links ]
37. Herremans M. (2004). Effects of drought on birds in the Kalahari, Botswana. Ostrich 75, 217–227. [ Links ]
38. Hanmer D.B. (1997). Bird longevity in the eastern highlands of Zimbabwe – drought survivors. Safring News 26, 47–54. [ Links ]
39. Dean W.R.J. and Milton S.J. (2001). The density and stability of birds in shrublands and drainage line woodland in the southern Karoo, South Africa. Ostrich 72, 185–192. [ Links ]
40. Safriel U.N. (1990). Winter foraging behaviour of the Dune Lark in the Namib Desert, and the effect of prolonged drought on behaviour. Ostrich 61, 77–80. [ Links ]
41. Cox G.W. (1983). Foraging behaviour of the dune lark. Ostrich 54, 113–120. [ Links ]
42. Willoughby E.J. (1971). Biology of larks (Aves: Alaudidae) in the central Namib Desert. Zool. Afr. 6, 133–176. [ Links ]
43. Simmons R.E. (1996). Population declines, viable breeding areas and management options for flamingos in southern Africa. Cons. Biol. 10, 504–514. [ Links ]
44. Dean W.R.J. and Williams J.B. (2004). Adaptations of birds for life in deserts with particular reference to larks (Alaudidae). Trans. Roy. Soc. S. Afr. 59, 79–91. [ Links ]
45. Simmons R.E., Barnard P. and Jamieson I.G. (1999). What precipitates influxes of wetland birds to ephemeral pans in arid landscapes? Observations from Namibia. Ostrich 70, 145–148. [ Links ]
46. Hinde R.A. (1956). The biological significance of the territories of birds. Ibis 98, 340–369. [ Links ]
47. Andersson M. (1980). Nomadism and site tenacity as alternative reproductive tactics in birds. J. Anim. Ecol. 49, 175–184. [ Links ]
Received 3 June 2008. Accepted 12 February 2009.
* Author for correspondence. E-mail: | null | null | null | null | null | null | null | null | null |
https://nrich.maths.org/public/topic.php?code=71&cl=4&cldcmpid=2817 | 1,597,059,240,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00011.warc.gz | 412,005,347 | 9,639 | # Resources tagged with: Mathematical reasoning & proof
Filter by: Content type:
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### There are 173 results
Broad Topics > Thinking Mathematically > Mathematical reasoning & proof
### Plus or Minus
##### Age 16 to 18 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### Polynomial Relations
##### Age 16 to 18 Challenge Level:
Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.
### Pair Squares
##### Age 16 to 18 Challenge Level:
The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.
##### Age 16 to 18 Challenge Level:
Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.
### Telescoping Functions
##### Age 16 to 18
Take a complicated fraction with the product of five quartics top and bottom and reduce this to a whole number. This is a numerical example involving some clever algebra.
### The Clue Is in the Question
##### Age 16 to 18 Challenge Level:
Starting with one of the mini-challenges, how many of the other mini-challenges will you invent for yourself?
### Diverging
##### Age 16 to 18 Challenge Level:
Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.
### Thousand Words
##### Age 16 to 18 Challenge Level:
Here the diagram says it all. Can you find the diagram?
### Golden Eggs
##### Age 16 to 18 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Fractional Calculus III
##### Age 16 to 18
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Interpolating Polynomials
##### Age 16 to 18 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.
### Particularly General
##### Age 16 to 18 Challenge Level:
By proving these particular identities, prove the existence of general cases.
### Unit Interval
##### Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
##### Age 16 to 18 Short Challenge Level:
Can you work out where the blue-and-red brick roads end?
### Rational Roots
##### Age 16 to 18 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### How Many Solutions?
##### Age 16 to 18 Challenge Level:
Find all the solutions to the this equation.
### Proof Sorter - Quadratic Equation
##### Age 14 to 18 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Proof Sorter - Geometric Sequence
##### Age 16 to 18 Challenge Level:
Can you correctly order the steps in the proof of the formula for the sum of the first n terms in a geometric sequence?
### To Prove or Not to Prove
##### Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Leonardo's Problem
##### Age 14 to 18 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Dalmatians
##### Age 14 to 18 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
### Impossible Triangles?
##### Age 16 to 18 Challenge Level:
Which of these triangular jigsaws are impossible to finish?
### Generally Geometric
##### Age 16 to 18 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
##### Age 16 to 18 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Mind Your Ps and Qs
##### Age 16 to 18 Short Challenge Level:
Sort these mathematical propositions into a series of 8 correct statements.
### Proof: A Brief Historical Survey
##### Age 14 to 18
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Water Pistols
##### Age 16 to 18 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
### Contrary Logic
##### Age 16 to 18 Challenge Level:
Can you invert the logic to prove these statements?
### Direct Logic
##### Age 16 to 18 Challenge Level:
Can you work through these direct proofs, using our interactive proof sorters?
### Iffy Logic
##### Age 14 to 18 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### Binomial
##### Age 16 to 18 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
### Mechanical Integration
##### Age 16 to 18 Challenge Level:
To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.
### There's a Limit
##### Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Target Six
##### Age 16 to 18 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### Sixational
##### Age 14 to 18 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Three Ways
##### Age 16 to 18 Challenge Level:
If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.
### Little and Large
##### Age 16 to 18 Challenge Level:
A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?
### Notty Logic
##### Age 16 to 18 Challenge Level:
Have a go at being mathematically negative, by negating these statements.
### Sperner's Lemma
##### Age 16 to 18
An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.
### More Dicey Decisions
##### Age 16 to 18 Challenge Level:
The twelve edge totals of a standard six-sided die are distributed symmetrically. Will the same symmetry emerge with a dodecahedral die?
### Euclid's Algorithm II
##### Age 16 to 18
We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.
### Tetra Inequalities
##### Age 16 to 18 Challenge Level:
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.
##### Age 14 to 18 Challenge Level:
Which of these roads will satisfy a Munchkin builder?
### Continued Fractions II
##### Age 16 to 18
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### Proof Sorter - Sum of an Arithmetic Sequence
##### Age 16 to 18 Challenge Level:
Put the steps of this proof in order to find the formula for the sum of an arithmetic sequence
### Without Calculus
##### Age 16 to 18 Challenge Level:
Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.
### Dodgy Proofs
##### Age 16 to 18 Challenge Level:
These proofs are wrong. Can you see why?
### Proof of Pick's Theorem
##### Age 16 to 18 Challenge Level:
Follow the hints and prove Pick's Theorem.
### An Introduction to Number Theory
##### Age 16 to 18
An introduction to some beautiful results of Number Theory (a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions) | 2,161 | 9,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-34 | latest | en | 0.81725 |
null | null | null | null | null | null | universalize in a sentence
Example sentences for universalize
That's because the filmmaker never finds a way to universalize the characters' experiences.
Well, there isn't one, and it was tough to find one that would universalize the concept of cupcake.
The only way to eliminate this bias would be to universalize all means-tested benefits currently targeted toward single mothers.
It may include a call to action, or universalize the points in the text, or provide a brief summary.
Famous quotes containing the word universalize
It is, most fundamentally because moral judgments are universalizable that we can speak of moral thought as rational (to... more
Copyright © 2014 Dictionary.com, LLC. All rights reserved. | null | null | null | null | null | null | null | null | null |
http://maeckes.nl/Decimaal%20getal%20GB.html | 1,603,346,437,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00008.warc.gz | 62,554,981 | 2,628 | < 1 2 3 4 5 >
### Decimal number
With the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 we make the decimal numbers.
That is so because we have ten fingers.
##### Explanation
The number 237 is composed of
2 × 100 = 2 × 102 3 × 10 = 3 × 101 7 × 1 = 7 × 100
Because 100 is equal to 10 × 10 = 102 you can write 10 as 101, and 1 like 100. If there are digits after decimal point this continues. The power decreases, reaches less than zero, which means it becomes negative. Thus 0.1 = 10–1. The number 4267.893 is composed of
4 × 1000 = 2 × 103 2 × 100 = 2 × 102 6 × 10 = 6 × 101 7 × 1 = 7 × 100 8 × 0,1 = 8 × 10−1 9 × 0,01 = 9 × 10−2 3 × 0,001 = 3 × 10−3
Behind every digit in a number is a power of 10, that depends on its position. Because of that you must write a 0 if a certain place has no value. The number 3600.102 is composed of
3 × 1000 = 3 × 103 6 × 100 = 6 × 102 1 × 0,1 = 1 × 10−1 2 × 0,001 = 2 × 10−3
Now we understand why is
And we understand the number 1 better, as
Numbers with a negative exponent are just fractions. You see it clearly in | 486 | 1,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-45 | longest | en | 0.444746 |
https://oneconvert.com/unit-converters/specific-heat-capacity-converter/kilojoule-kilogram-k-to-btu-th-pound-176-f | 1,713,848,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00483.warc.gz | 377,734,289 | 24,993 | # Convert kilojoule/kilogram/K to Btu (th)/pound/°F
## How to Convert kilojoule/kilogram/K to Btu (th)/pound/°F
To convert kilojoule/kilogram/K to Btu (th)/pound/°F , the formula is used,
$\mathrm{Btu \left(th\right)/pound/°F}=\mathrm{kilojoule/kilogram/K}×0.238845897$
where the kilojoule/kilogram/K to Btu (th)/pound/°F value is substituted to get the answer from Specific Heat Capacity Converter.
1 kilojoule/kilogram/K
=
0.239 Btu (th)/pound/°F
1 Btu (th)/pound/°F
=
4.184 kilojoule/kilogram/K
Example: convert 15 kilojoule/kilogram/K to Btu (th)/pound/°F:
15 kilojoule/kilogram/K
=
15
x
0.239 Btu (th)/pound/°F
=
3.5851 Btu (th)/pound/°F
## kilojoule/kilogram/K to Btu (th)/pound/°F Conversion Table
kilojoule/kilogram/K Btu (th)/pound/°F
0.01 kilojoule/kilogram/K 0.002390057 Btu (th)/pound/°F
0.1 kilojoule/kilogram/K 0.023900574 Btu (th)/pound/°F
1 kilojoule/kilogram/K 0.239005736 Btu (th)/pound/°F
2 kilojoule/kilogram/K 0.478011472 Btu (th)/pound/°F
3 kilojoule/kilogram/K 0.717017208 Btu (th)/pound/°F
5 kilojoule/kilogram/K 1.195028681 Btu (th)/pound/°F
10 kilojoule/kilogram/K 2.390057361 Btu (th)/pound/°F
20 kilojoule/kilogram/K 4.780114723 Btu (th)/pound/°F
50 kilojoule/kilogram/K 11.95028681 Btu (th)/pound/°F
100 kilojoule/kilogram/K 23.90057361 Btu (th)/pound/°F
1000 kilojoule/kilogram/K 239.0057361 Btu (th)/pound/°F
### Popular Unit Conversions Specific Heat Capacity
The most used and popular units of Specific Heat Capacity conversions are presented for quick and free access. | 626 | 1,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.48584 |
http://mathhelpforum.com/algebra/43081-fractions.html | 1,480,721,253,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00231-ip-10-31-129-80.ec2.internal.warc.gz | 174,512,397 | 11,679 | 1. ## fractions
I need someone to help me understand how to do these problems.
2. Your images are hard to read. You might want to try to find another method to present your work. This is what I think you have. Please confirm and others will be glad to assist.
a) $\frac{-2(18-2)}{4(5-6)}-\frac{3(27-37)}{5(7-4)}$
b) $\frac{\frac{3}{4}-7}{\frac{1}{2}-\frac{4}{5}}$
c) $\frac{\frac{3}{8}}{\frac{1}{4}-\frac{7}{16}}$
d) $4 \frac{3}{5}+7 \frac{1}{10}$
e) $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
3. your pics are kinda blurry, so make sure i got the problem right.
$\frac{\frac{3}{4} - 7 }{\frac{1}{2} - \frac{4}{5}}$
Get like denominators on both the top and bottom of the big fraction.
$\frac{\frac{3}{4} - \frac{28}{4}}{\frac{5}{10} - \frac{8}{10}}$
Simplify: $\frac{\frac{-25}{4}}{\frac{-3}{10}}$
Multiply by the reciprocal: $\frac{-25}{4}\cdot\frac{-10}{3}$
$= \frac{250}{12}$
$= \frac{125}{6}$
C is very similar.
On A, do what's in parenthesis first, get a like denominator and simplify.
For: $4 \frac{3}{5}+7 \frac{1}{10}$, get improper fractions, then a like denominator and finally simplify.
For: $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
Multiply straight across; numerator by numerator; denominator by denominator.
You can think of it as: $\bigg(\frac{9}{1}\bigg)\left(\frac{15}{81}\right)\ left(\frac{27}{15}\right)$
Look for cross canceling. The 15 is an obvious one.
Then simplify as much as possible.
all of what masters has is correct
5. ## Thanks
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
6. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
This forum allows you to enter $\text{\LaTeX}$ (LaTeX) code directly: simply enclose it in $$and$$ tags. To learn the basics, visit the LaTex Help forum.
7. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
While you're learning to post using Latex, you could simply express your problems using brackets and parentheses like this:
a) [-2(18 - 2)] / [4(5 - 6)] - [3(27 - 37)] / [5(7 - 4)]
b) (3/4 -7) / (1/2 -4/5)
etc... | 756 | 2,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2016-50 | longest | en | 0.836637 |
http://mathhelpforum.com/algebra/278677-irrational-number.html | 1,540,190,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514708.24/warc/CC-MAIN-20181022050544-20181022072044-00104.warc.gz | 219,283,647 | 11,375 | 1. ## Irrational Number
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
2. ## Re: Irrational Number
Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
$\sqrt{1^3+3^3+3^3+4^3+5^1} = 2\sqrt{31} \in \mathbb{R} \setminus \mathbb{Q}$
... what now?
3. ## Re: Irrational Number
Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
Here are some comments. Is there a typo: $\sqrt{1^3+{\color{red}2^3}+3^3+4^3+5^n}~?$
Even if it is, nothing really changes in the question.
Also: if $k$ is a non-square positive integer then $\large\sqrt{k}$ is irrational.
4. ## Irrational number
Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
6. ## Re: Irrational number
I still don't know how to solve it. Could you give me the complet explanation?
7. ## Re: Irrational number
in your other thread with the same problem, I just tried $n=1$ ... once again, on your edited expression I tried $n=1$ ...
$\sqrt{1^3+2^3+3^3+4^3+5^1} = \sqrt{105}$
As stated by Plato in your previous thread, note that 105 is not a perfect square, therefore its square root is irrational.
8. ## Re: Irrational Number
Yeah, there was a typo.
9. ## Re: Irrational number
Originally Posted by louis33
I still don't know how to solve it. Could you give me the complet explanation?
Originally Posted by louis33
Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
So the question reduces to: Find all $n\in\mathbb{N}^+$ such that $\Large5^n+100~$ is not a perfect square.
10. ## Re: Irrational Number
Let us find all $\displaystyle n$ for which $\displaystyle 5^n+100$ is a perfect square
$\displaystyle 5^n+100=m^2$ for some $\displaystyle m$
assume $\displaystyle n\geq 3$. Now m must be divisible by 5 so we can write m=5t.
replacing we get
$\displaystyle 5^{n-2}=(t-2)(t+2)$
so $\displaystyle t-2$ and $\displaystyle t+2$ are both powers of 5 but cannot both be divisible by 5 since otherwise their difference = 4 would be divisible by 5
therefore $\displaystyle t-2=1$ and $\displaystyle t+2=5^{n-2}$
this gives t=3 and n=3 | 689 | 2,158 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-43 | latest | en | 0.82482 |
http://oeis.org/A160099 | 1,569,118,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00030.warc.gz | 148,957,464 | 3,873 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A160099 Decimal expansion of (843 + 418*sqrt(2))/601. 4
2, 3, 8, 6, 2, 5, 8, 3, 5, 1, 2, 0, 1, 2, 5, 4, 1, 2, 7, 1, 1, 9, 9, 7, 6, 5, 1, 7, 0, 0, 4, 4, 1, 5, 6, 3, 5, 3, 4, 4, 6, 3, 0, 3, 5, 5, 9, 1, 9, 4, 0, 8, 1, 4, 4, 0, 7, 2, 9, 6, 5, 2, 7, 5, 3, 4, 1, 3, 0, 8, 1, 1, 3, 1, 4, 0, 0, 3, 4, 5, 1, 2, 8, 7, 7, 6, 4, 0, 5, 8, 1, 2, 7, 9, 4, 5, 7, 8, 6, 5, 8, 9, 5, 5, 6 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Equals Lim_{n -> infinity} b(n)/b(n-1) for n mod 3 = {1, 2}, b = A111258. Equals Lim_{n -> infinity} b(n)/b(n-1) for n mod 3 = {0, 2}, b = A160098. LINKS G. C. Greubel, Table of n, a(n) for n = 1..10000 FORMULA Equals (38 + 11*sqrt(2))/(38 - 11*sqrt(2)). EXAMPLE (843 + 418*sqrt(2))/601 = 2.38625835120125412711... MATHEMATICA RealDigits[(843 +418*Sqrt[2])/601, 10, 100][[1]] (* G. C. Greubel, Apr 21 2018 *) PROG (PARI) (843+418*sqrt(2))/601 \\ G. C. Greubel, Apr 21 2018 (MAGMA) (843+418*Sqrt(2))/601; // G. C. Greubel, Apr 21 2018 CROSSREFS Cf. A111258, A160098, A002193 (decimal expansion of sqrt(2)), A160100 (decimal expansion of (361299+5950*sqrt(2))/601^2). Sequence in context: A047930 A073875 A220395 * A309967 A321336 A093098 Adjacent sequences: A160096 A160097 A160098 * A160100 A160101 A160102 KEYWORD cons,nonn AUTHOR Klaus Brockhaus, May 18 2009 STATUS approved
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Last modified September 21 22:17 EDT 2019. Contains 327284 sequences. (Running on oeis4.) | 828 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-39 | latest | en | 0.567681 |
http://mathhelpforum.com/pre-calculus/145223-solve-x.html | 1,526,904,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00533.warc.gz | 185,198,331 | 9,303 | 1. ## Solve for X
9 = 40(2)^2x+1
9 = 40(2)^2x+1
3. Originally Posted by harish21
$\displaystyle 9 = 40(2)^{2x+1 }$
$\displaystyle ln ( \frac{ 9}{40} ) = ln (2^{2x+1}) = (2x+1) ln2$
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{ ln2} = 2x + 1$
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{2 ln2} - \frac{1}{2} = x$ | 157 | 317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-22 | latest | en | 0.410955 |
https://gmatclub.com/forum/if-the-salary-cap-of-the-national-basketball-association-17519.html?fl=similar | 1,511,122,098,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00227.warc.gz | 626,320,707 | 44,267 | It is currently 19 Nov 2017, 13:08
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# If the salary cap of the National Basketball Association
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21 Jun 2005, 17:20
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If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more.
If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more
Were the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars
Was the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars or more
If the salary cap of the National Basketball Association was removed, the cost of running a championship team would increase by at least two million dollars
Should the salary cap of the National Basketball Association be removed, the cost of running a championship team would increase by at least two million dollars or more
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21 Jun 2005, 21:09
I choose B.
... increase by at least two million dollars or more sounds redundant.
Last edited by kdhong on 21 Jun 2005, 21:27, edited 1 time in total.
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21 Jun 2005, 21:26
Agree with B. E is also likely but has redundent: ... \$2bn and more.
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21 Jun 2005, 23:19
A, C and E can be eliminated because 'or more' is redundant after 'at least'.
B & D both sound correct to me...B sounds better...but it'd be great if someone could give a reason for why one of the two is better than the other
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22 Jun 2005, 05:08
Yes it looks like B....Subjunctive is what is happening here
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22 Jun 2005, 06:30
the sentence consists of a main and a conditional clause. the conditional clause must contain "were" because it is the 2nd form of conditional, which describes unlikely situations. the main clause must contain "would". A) is indeed wrong because of redundancy.
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
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22 Jun 2005, 06:52
Which one is better? Be careful, i have modified E.
If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more.
B. Were the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars
E. Should the salary cap of the National Basketball Association be removed, the cost of running a championship team would increase by at least two million dollars
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22 Jun 2005, 10:46
I will go with D on this.
I am not very sure but here is what I read.
If things described in subjunctive mood are impossible and hypothetical, then "were" or "had" will be used.
But if things described in subjunctive mood are possible, then "was" can be used and is grammatically correct.
If you look at the answer choices "were and to be" do not go along well.
E looks impressive but I dont think it is grammtically correct because of atleast and more...
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22 Jun 2005, 13:50
OA is B
the rule of subjunctive applies...
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25 Jun 2005, 05:45
I am quite confused with sujunctive moods and if. Can someone please explain when are they applied & when are they not. Eg. in the following why is it not applied?
A wildlife expert predicts that the reintroduction of the caribou into northern Minnesota will fail if the density of the timber wolf population in that region is greater than one wolf for every 39 square miles.
Shouldn't it be "if the density ... were greater than"??
Kindly explain.
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25 Jun 2005, 22:01
"at least ... or more" is redundant.
D is wrong because of tense mix-up.
B for me
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27 Jun 2005, 06:47
I disagree with the OA.
B sounds awkward.
What is the source of this quesiton pls.
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27 Jun 2005, 06:47
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# If the salary cap of the National Basketball Association
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,832 | 6,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-47 | latest | en | 0.928291 |
https://web2.0calc.com/questions/a-math-question_12 | 1,585,498,875,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00090.warc.gz | 795,532,105 | 6,106 | +0
# a math question
0
80
1
what is 90 x 73 / 34 + 73 - 50
Jan 7, 2020
#1
+1
Here is a hint:
Multiply first then divide then add and subtract. (ORDER MATTER!!) (From left to right)
Another hint:
BODMAS Is the rule you should follow , (We solve brackets (B) then orders (O) which is powers,exponents etc, then Division (D) then multiplication (M) then addition (A) then subtraction (S)) Most importantly, from left to right, if M is first on the left then you should start with M
M have the same value as D
A and S have the same value
In both order matters tho! From left to right again!!
Solution obtained by this website calculator:
90*73/34+73-50 = 216.2352941176470588
Without a calculator:
$$90*\frac{73}{2*17}+23$$
$$45*\frac{73}{17}+23$$
now you need to multiply 45*73
$$\frac{3285}{17}+23$$
Long division..
$$193.24+23=216.24$$
.
Jan 7, 2020 | 280 | 870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-16 | latest | en | 0.832815 |
https://math.wikia.org/wiki/Inscribed_sphere | 1,623,705,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00280.warc.gz | 357,161,421 | 32,766 | 1,196 Pages
In geometry, the inscribed sphere or insphere of a convex polyhedron is a sphere that is contained within the polyhedron and tangent to each of the polyhedron's faces. It is the largest sphere that is contained wholly within the polyhedron, and is dual to the dual polyhedron's circumsphere.
All regular polyhedra have inscribed spheres, but some irregular polyhedra do not have all facets tangent to a common sphere, although it is still possible to define the largest contained sphere for such shapes. For such cases, the notion of an insphere does not seem to have been properly defined and various interpretations of an insphere are to be found:
• The sphere tangent to all faces (if one exists).
• The sphere tangent to all face planes (if one exists).
• The sphere tangent to a given set of faces (if one exists).
• The largest sphere that can fit inside the polyhedron.
Often these spheres coincide, leading to confusion as to exactly what properties define the insphere for polyhedra where they do not coincide.
For example the regular small stellated dodecahedron has a sphere tangent to all faces, while a larger sphere can still be fitted inside the polyhedron. Which is the insphere? Important authorities such as Coxeter or Cundy & Rollett are clear enough that the face-tangent sphere is the insphere. Again, such authorities agree that the Archimedean polyhedra (having regular faces and equivalent vertices) have no inspheres while the Archimedean dual or Catalan polyhedra do have inspheres. But many authors fail to respect such distinctions and assume other definitions for the 'inspheres' of their polyhedra.
The radius of the sphere inscribed in a polyhedron P is called the inradius of P.
## Volume
The volume of the sphere inscribed in a polyhedron is therefore:
## Surface area
The surface area of the sphere inscribed in a polyhedron is therefore:
## Inscribed hemisphere
### Volume
The volume of the hemisphere inscribed in a polyhedron is therefore:
### Surface area
The surface area of the hemisphere inscribed in a polyhedron is therefore:
## References
• Coxeter, H.S.M. Regular polytopes 3rd Edn. Dover (1973).
• Cundy, H.M. and Rollett, A.P. Mathematical Models, 2nd Edn. OUP (1961). | 515 | 2,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-25 | latest | en | 0.917289 |
https://metanumbers.com/67031 | 1,638,469,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00454.warc.gz | 430,963,862 | 7,324 | # 67031 (number)
67,031 (sixty-seven thousand thirty-one) is an odd five-digits composite number following 67030 and preceding 67032. In scientific notation, it is written as 6.7031 × 104. The sum of its digits is 17. It has a total of 2 prime factors and 4 positive divisors. There are 63,072 positive integers (up to 67031) that are relatively prime to 67031.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 17
• Digital Root 8
## Name
Short name 67 thousand 31 sixty-seven thousand thirty-one
## Notation
Scientific notation 6.7031 × 104 67.031 × 103
## Prime Factorization of 67031
Prime Factorization 17 × 3943
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 67031 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 67,031 is 17 × 3943. Since it has a total of 2 prime factors, 67,031 is a composite number.
## Divisors of 67031
1, 17, 3943, 67031
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 70992 Sum of all the positive divisors of n s(n) 3961 Sum of the proper positive divisors of n A(n) 17748 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 258.903 Returns the nth root of the product of n divisors H(n) 3.77682 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 67,031 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 67,031) is 70,992, the average is 17,748.
## Other Arithmetic Functions (n = 67031)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 63072 Total number of positive integers not greater than n that are coprime to n λ(n) 31536 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6677 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 63,072 positive integers (less than 67,031) that are coprime with 67,031. And there are approximately 6,677 prime numbers less than or equal to 67,031.
## Divisibility of 67031
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 6 7 8
67,031 is not divisible by any number less than or equal to 9.
## Classification of 67031
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (67031)
Base System Value
2 Binary 10000010111010111
3 Ternary 10101221122
4 Quaternary 100113113
5 Quinary 4121111
6 Senary 1234155
8 Octal 202727
10 Decimal 67031
12 Duodecimal 3295b
16 Hexadecimal 105d7
20 Vigesimal 87bb
36 Base36 1fpz
## Basic calculations (n = 67031)
### Multiplication
n×y
n×2 134062 201093 268124 335155
### Division
n÷y
n÷2 33515.5 22343.7 16757.8 13406.2
### Exponentiation
ny
n2 4493154961 301180670190791 20188441503558911521 1353251422425057398164151
### Nth Root
y√n
2√n 258.903 40.6217 16.0905 9.23114
## 67031 as geometric shapes
### Circle
Radius = n
Diameter 134062 421168 1.41157e+10
### Sphere
Radius = n
Volume 1.26158e+15 5.64627e+10 421168
### Square
Length = n
Perimeter 268124 4.49315e+09 94796.1
### Cube
Length = n
Surface area 2.69589e+10 3.01181e+14 116101
### Equilateral Triangle
Length = n
Perimeter 201093 1.94559e+09 58050.5
### Triangular Pyramid
Length = n
Surface area 7.78237e+09 3.54945e+13 54730.6
## Cryptographic Hash Functions
md5 f55b61d6b3bf265db7d801774062598a e1c39c18db1995dae23fd3521d9ddbd06859d3ba d36d52f2fb9f41015dd6042da1b946b36fa9cabe60f584865bb4a959e2d91ae9 05aa8b2af3151fed78ace816c915aa291f0c5ea7bc80b340b4ced8e650d0c41c87f50188eadf129b4e937cd546efceac83d2c538731a7de3df0e6c23aa37f3c6 3dc87cde43b84fc173a5d93c05ac6966cc26d07e | 1,477 | 4,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-49 | latest | en | 0.815752 |
https://www.cuemath.com/ncert-solutions/exercise-5-6-continuity-and-differentiability-class-12-maths/ | 1,623,591,829,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00257.warc.gz | 677,802,217 | 16,759 | # NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6
## Chapter 5 Ex.5.6 Question 1
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = 2a{t^2},\;y = a{t^4}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = 2a{t^2},\; y = a{t^4}$$
Then,
$$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {2a{t^2}} \right) = 2a.\frac{d}{{dt}}\left( {{t^2}} \right) = 2a.2t = 4at$$
\begin{align} & \frac{{dy}}{{dt}}= \frac{d}{{dt}}\left( {a{t^4}} \right)= a.\frac{d}{{dt}}\left( {{t^4}} \right) = a.4.{t^3} = 4a{t^3}\\&\therefore \frac{{dy}}{{dt}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{4a{t^3}}}{{4at}} = {t^2}\end{align}
## Chapter 5 Ex.5.6 Question 2
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\cos \theta,\;y = b\cos \theta$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\cos \theta,\;y = b\cos \theta$$
Then,
\begin{align}&\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\left( {a\cos \theta } \right) = a\left( { - \sin \theta } \right) = - a\sin \theta \\&\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {b\cos \theta } \right) = b\left( { - \sin \theta } \right) = - b\sin \theta \\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - b\sin \theta }}{{ - a\sin \theta }} = \frac{b}{a}\end{align}
## Chapter 5 Ex.5.6 Question 3
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \sin t,\;y = \cos 2t$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = \sin t,\;y = \cos 2t$$
Then, $$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\sin t} \right) = \cos t$$
\begin{align}&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\cos 2t} \right) = - \sin 2t.\frac{d}{{dt}}\left( {2t} \right) = - 2\sin 2t\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{ - 2\sin 2t}}{{\cos t}} = \frac{{ - 2.2\sin t\cos t}}{{\cos t}} = - 4\sin t\end{align}
## Chapter 5 Ex.5.6 Question 4
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = 4t,\;y = \frac{4}{t}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = 4t,\;y = \frac{4}{t}$$
\begin{align}&\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4\\&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{t}} \right) = 4.\frac{d}{{dt}}\left( {\frac{1}{t}} \right) = 4.\left( {\frac{{ - 1}}{{{t^2}}}} \right) = \frac{{ - 4}}{{{t^2}}}\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - 4}}{{{t^2}}}} \right)}}{4} = \frac{{ - 1}}{{{t^2}}}\end{align}
## Chapter 5 Ex.5.6 Question 5
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \cos \theta - \cos 2\theta,\;y = \sin \theta - \sin 2\theta$$, without eliminating the parameter, find
### Solution
Given, $$x = \cos \theta - \cos 2\theta ,\;y = \sin \theta - \sin 2\theta$$
Then,
\begin{align}\frac{{dx}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {\cos \theta - \cos 2\theta } \right) = \frac{d}{{d\theta }}\left( {\cos \theta } \right) - \frac{d}{{d\theta }}\left( {\cos 2\theta } \right)\\[5pt]&= - \sin \theta - \left( { - 2\sin 2\theta } \right) \\[5pt]&= 2\sin 2\theta - \sin \theta \\[10pt] \frac{{dy}}{{d\theta }}& = \frac{d}{{d\theta }}\left( {\sin \theta - \sin 2\theta } \right) = \frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\sin 2\theta } \right)\\[5pt]&= \cos \theta - 2\cos 2\theta \\\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} &= \frac{{\cos \theta - 2\cos 2\theta }}{{2\sin 2\theta - \sin \theta }}\end{align}
## Chapter 5 Ex.5.6 Question 6
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)$$
Then, $$\frac{{dx}}{{d\theta }} = a\left[ {\frac{d}{{d\theta }}\left( \theta \right) - \frac{d}{{d\theta }}\left( {\sin \theta } \right)} \right] = a\left( {1 - \cos \theta } \right)$$
\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( 1 \right) + \frac{d}{{d\theta }}\left( {\cos \theta } \right)} \right] = a\left[ {0 + \left( { - \sin \theta } \right)} \right] = - a\sin \theta \\\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - a\sin \theta }}{{a\left( {1 - \cos \theta } \right)}} = \frac{{ - 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\;\sin }^2}\frac{\theta }{2}}} = \frac{{ - \cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = - \cot \frac{\theta }{2}\end{align}
## Chapter 5 Ex.5.6 Question 7
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$$
Then,
\begin{align}\frac{{dx}}{{dt}} &= \frac{d}{{dt}}\left[ {\frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\[5pt]&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\sin }^3}t} \right) - {{\sin }^3}t.\frac{d}{{dt}}\sqrt {\cos 2t} }}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\sin }^2}t.\frac{d}{{dt}}\left( {\sin t} \right) - {{\sin }^3}t \times \frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\sin }^2}t.\cos t - \frac{{{{\sin }^3}t}}{{2\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\cos 2t.{{\sin }^2}t\cos t + {{\sin }^3}t.\sin 2t}}{{\cos 2t\sqrt {\cos 2t} }}\\[5pt]\frac{{dy}}{{dt}}& = \frac{d}{{dt}}\left[ {\frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\cos }^3}t} \right) - {{\cos }^3}t.\frac{d}{{dt}}\left( {\sqrt {\cos 2t} } \right)}}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\cos }^2}t.\frac{d}{{dt}}\left( {\cos t} \right) - {{\cos }^3}t.\frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\cos }^2}t\left( { - \sin t} \right) - {{\cos }^3}t.\frac{1}{{\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t.\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}\end{align}
\begin{align}\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}{{\frac{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\left( {2\sin t\cos t} \right)}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\left( {2\sin t\cos t} \right)}}\\[5pt]&= \frac{{\sin t\cos t\left[ { - 3\cos 2t.\cos t + 2{{\cos }^3}t} \right]}}{{\sin t\cos t\left[ {3\cos 2t\sin t + 2{{\sin }^3}t} \right]}}\\[5pt]&= \frac{{\left[ { - 3\left( {2{{\cos }^2}t - 1} \right)\cos t + 2{{\cos }^3}t} \right]}}{{\left[ {3\left( {1 - 2{{\sin }^2}t} \right)\sin t + 2{{\sin }^3}t} \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 2t = \left( {2{{\cos }^2}t - 1} \right)\\\cos 2t = \left( {1 - 2{{\sin }^2}t} \right)\end{array} \right]\\[5pt]&= \frac{{ - 4{{\cos }^3}t + 3\cos t}}{{3\sin t - 4{{\sin }^3}t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 3t = 4{\cos ^3}t - 3\cos t\\\sin 3t = 3\sin t - 4{\sin ^2}t\end{array} \right]\\[5pt]&= \frac{{ - \cos 3t}}{{\sin 3t}} = - \cot 3t\end{align}
## Chapter 5 Ex.5.6 Question 8
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t$$
Then,
\begin{align}\frac{{dx}}{{dt}} &= a.\left[ {\frac{d}{{dt}}\left( {\cos t} \right) + \frac{d}{{dt}}\left( {\log \tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{1}{{\tan \frac{t}{2}}}.\frac{d}{{dt}}\left( {\tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \cot \frac{t}{2}.{{\sec }^2}\frac{t}{2}.\frac{d}{{dt}}\left( {\frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}} \times \frac{1}{{{{\cos }^2}\frac{t}{2}}} \times \frac{1}{2}} \right]\\&= a\left[ { - \sin t + \frac{1}{{2\sin \frac{t}{2}\cos \frac{t}{2}}}} \right]\\&= a\left( { - \sin t + \frac{1}{{\sin t}}} \right)\\&= a\left( {\frac{{ - {{\sin }^2}t + 1}}{{\sin t}}} \right)\\&= a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)\\\frac{{dy}}{{dt}} &= a\frac{d}{{dt}}\left( {\sin t} \right) = a\cos t\end{align}
Therefore,
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{a\cos t}}{{\left( {a\frac{{{{\cos }^2}t}}{{\sin t}}} \right)}} = \frac{{\sin t}}{{\cos t}} = \tan t$
## Chapter 5 Ex.5.6 Question 9
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\;\sec \theta,\;y = b\;\tan \theta$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\;\sec \theta,\;y = b\;\tan \theta$$
Then,
$$\frac{{dx}}{{d\theta }} = a.\frac{d}{{d\theta }}\left( {\sec \theta } \right) = a\sec \theta \tan \theta$$
$$\frac{{dy}}{{d\theta }} = b.\frac{d}{{d\theta }}\left( {\tan \theta } \right) = b{\;\sec ^2}\theta$$
Therefore,
\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\&= \frac{{b{{\;\sec }^2}\theta }}{{a\sec \theta \tan \theta }}\\&= \frac{b}{a}\sec \theta \cot \theta \\&= \frac{{b\cos \theta }}{{a\cos \theta \sin \theta }}\\&= \frac{b}{a} \times \frac{1}{{\sin \theta }}\\&= \frac{b}{a}{\mathop{\text cosec\;}\nolimits} \theta \end{align}
## Chapter 5 Ex.5.6 Question 10
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)$$
Then,
\begin{align}\frac{{dx}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\cos \theta + \frac{d}{{d\theta }}\left( {\theta \sin \theta } \right)} \right]\\&= a\left[ { - \sin \theta + \theta \frac{d}{{d\theta }}\left( {\sin \theta } \right) + \sin \theta \frac{d}{{d\theta }}\left( \theta \right)} \right]\\[5pt]&= a\left[ { - \sin \theta + \theta \cos \theta + \sin \theta } \right]\\[5pt]&= a\;\theta \cos \theta \end{align}
\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\theta \cos \theta } \right)} \right] \\&= a\left[ {\cos \theta - \left\{ {\theta \frac{d}{{d\theta }}\left( {\cos \theta } \right) + \cos \theta .\frac{d}{{d\theta }}\left( \theta \right)} \right\}} \right]\\[5pt]&= a\left[ {\cos \theta + \theta \sin \theta - \cos \theta } \right]\\[5pt]&= a\; \theta \sin \theta \end{align}
Therefore,
\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\[5pt]&= \frac{{a\;\theta \sin \theta }}{{a\;\theta \cos \theta }}\\[5pt]&= \tan \theta \end{align}
## Chapter 5 Ex.5.6 Question 11
If $$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}},\;y = \sqrt {{a^{{{\cos }^{ - 1}}t}}}$$, show that $$\frac{{dy}}{{dx}} = - \frac{y}{x}$$
### Solution
Given, $$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}}$$ and $$y = \sqrt {{a^{{{\cos }^{ - 1}}t}}}$$
Hence,
$$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}} = {\left( {{a^{{{\sin }^{ - 1}}t}}} \right)^{\frac{1}{2}}} = {a^{\frac{1}{2}{{\sin }^{ - 1}}t}}$$ and $$y = \sqrt {{a^{{{\cos }^{ - 1}}t}}} = {\left( {{a^{{{\cos }^{ - 1}}t}}} \right)^{\frac{1}{2}}} = {a^{\frac{1}{2}{{\cos }^{ - 1}}t}}$$
Consider $$x = {a^{\frac{1}{2}{{\sin }^{ - 1}}t}}$$
Taking log on both sides, we get
$$\log x = \frac{1}{2}{\sin ^{ - 1}}t\log a$$
Therefore,
\begin{align}&\Rightarrow \; \; \frac{1}{x}.\frac{{dx}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{x}{2}\log a.\frac{1}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}
Now, $$y = {a^{\frac{1}{2}{{\cos }^{ - 1}}t}}$$
Taking log on both sides, we get
$$\log x = \frac{1}{2}{\cos ^{ - 1}}t\log a$$
Therefore,
\begin{align}&\Rightarrow \; \; \frac{1}{y}.\frac{{dy}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\cos }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{y}{2}\log a.\frac{{ - 1}}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}
Hence,
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}}{{\left( {\frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}} = - \frac{y}{x}$
Instant doubt clearing with Cuemath Advanced Math Program | 6,073 | 13,903 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 30, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-25 | latest | en | 0.403828 |
https://sportsbizusa.com/20-number-3-preschool-worksheet/ | 1,627,998,352,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00195.warc.gz | 533,568,538 | 11,369 | HomeTemplate ➟ 20 20 Number 3 Preschool Worksheet
# 20 Number 3 Preschool Worksheet
Learn to Count and Write Number 3 printable number 3 worksheet preschool, number 3 tracing worksheets preschool, free printable number 3 worksheets for preschoolers, via: pinterest.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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https://mathhelpboards.com/threads/fourier-series-showing-converges-to-pi-16.6539/ | 1,610,732,719,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00113.warc.gz | 503,936,812 | 15,244 | # Fourier series--showing converges to pi/16
#### dwsmith
##### Well-known member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
#### Deveno
##### Well-known member
MHB Math Scholar
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
#### dwsmith
##### Well-known member
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
I think I will still have some trouble though. I end up with:
$\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{2n - 1}\frac{\sin\left[\frac{i\pi}{2}(2n-1)\right]}{\sin\left[i\pi(2n-1)\right]}$
#### Opalg
##### MHB Oldtimer
Staff member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
The sin function isn't really there at all, because $\sin\left(\frac{\pi(2n - 1)}{2}\right) = \sin\left(\bigl(n-\frac12\bigr)\pi\right) = (-1)^{n-1}$. Also, you can use the identity $\sinh(2x) = 2\sinh x\cosh x$, to rewrite the sum as $$\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n}{2(2n-1)\cosh\left(\bigl(n-\frac12\bigr)\pi\right)}.$$ That looks a bit less complicated than the original series, but I still don't see how to sum it. It clearly converges very fast, because the cosh term in the denominator will get very large after the first few terms. A numerical check shows that the sum of the first three terms is $0.1968518...$, which is very close to $\pi/16$. But that isn't a proof! | 713 | 2,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | longest | en | 0.838143 |
https://kinerjamall.com/checker-404 | 1,669,526,032,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00620.warc.gz | 391,028,242 | 5,167 | # How to solve with square roots
These can be very helpful when you're stuck on a problem and don't know How to solve with square roots. We will give you answers to homework.
## How can we solve with square roots
Read on for some helpful advice on How to solve with square roots easily and effectively. Piecewise functions can be tricky to solve, but there are a few steps you can follow to make it easier. First, identify the different parts of the function and what values they apply to. Second, graph the function to help you visualize what's happening. Finally, solve each part of the function separately and piece the solution together. Following these steps should help you solve most piecewise functions.
There are many methods for solving systems of linear equations. The most common method is using elimination. This involves adding or subtracting equations in order to cancel out variables. Another common method is substitution, which involves solving for one variable in terms of the others and then plugging this back into the other equations. These methods can be used for systems with any number of variables.
To solve a piecewise function, you need to carefully examine the given equations and determine which one to use for which x-values. You also need to be careful when graphing piecewise functions, as the different equations can produce different shapes for the graph.
Scanning math problems can be a quick and easy way to check your work for mistakes. By scan reading, you can quickly identify where you made a mistake and correct it. This can save you time and frustration in the long run.
In this case, we can subtract 4 from both sides of the equation to get: y - 4 = 2x Then, we need to divide both sides of the equation by 2 to get: y/2 = x Finally, we can multiply both sides of the equation by 2 to get: y = 2x
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Math problem search Geometry helper app Three equations three unknowns solver Problem solver com Completing the square solver Expression solver | 542 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-49 | latest | en | 0.950607 |
http://mathhelpforum.com/calculus/15437-max-mins-without-numerical-values-graph-print.html | 1,519,447,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815318.53/warc/CC-MAIN-20180224033332-20180224053332-00258.warc.gz | 219,726,072 | 3,520 | # Max/Mins Without Numerical Values (On Graph)
• May 28th 2007, 06:24 PM
SportfreundeKeaneKent
Max/Mins Without Numerical Values (On Graph)
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
• May 28th 2007, 06:31 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
For the first:
two lines are parrallel if they have the same slope. the slope of the line is 9. so we simply want to find the $x$'s that make the tangent lines to the curve have a slope of 9. the derivative gives the formula for the slope of the tangent line.
$y = x^3 - 3x^2$
$\Rightarrow y' = 3x^2 - 6x$
Now we simply solve:
$3x^2 - 6x = 9$
i leave that to you
• May 28th 2007, 06:35 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
see http://www.mathhelpforum.com/math-he...ma-minima.html as a hint for the second
i assume you remember how to maximize a function
• Jun 1st 2007, 04:09 PM
SportfreundeKeaneKent
A question about the one with the tangent. Would you sub x=-1,3 (the two values I obtain after factoring and finding x) into y=x^3-3x^2 or into y=9x+7?
I'm guessing that you'd sub it into y=9x+7 to obtain the two points to be (3,34) and (-1,-2)
• Jun 1st 2007, 04:11 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
A question about the one with the tangent. Would you sub x=-1,3 (the two values I obtain after factoring and finding x) into y=x^3-3x^2 or into y=9x+7?
I'm guessing that you'd sub it into y=9x+7 to obtain the two points to be (3,34) and (-1,-2)
we want two points ON THE CURVE, therefore you would plug them into the original function to obtain the y-values for the points. we wouldn't sub them into the line, unless we knew that the line intersected with our curve exactly at the points where the slope of the curve is 9. so just plug them into $y = x^3 - 3x^2$
why did you think you would plug it into the line?
• Jun 1st 2007, 04:35 PM
SportfreundeKeaneKent
Oh wait, I wrote it the wrong way around. So the points are just (-1,-4) and (3,0)
• Jun 1st 2007, 04:39 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
Oh wait, I wrote it the wrong way around. So the points are just (-1,-4) and (3,0)
correct :D
• Jun 2nd 2007, 09:43 AM
curvature
tangent lines
tangent lines | 846 | 2,800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-09 | longest | en | 0.904676 |
https://socratic.org/questions/59e7439a11ef6b2d324ff210 | 1,597,077,363,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736057.87/warc/CC-MAIN-20200810145103-20200810175103-00323.warc.gz | 459,378,350 | 6,485 | # Question ff210
Oct 20, 2017
$\text{185.0 u}$
#### Explanation:
The first thing that you need to do here is to use the percent abundance of rhenium-187 to calculate the percent abundance of rhenium-185.
To do that, use the fact that rhenium has only two naturally occurring isotopes. This implies that their percent abundances must add up to give 100%.
Alternatively, you can say that their decimal abundances must add up to $1$. This is the case because a decimal abundance is simply a percent abundance divided by 100%.
This means that you have
$\text{decimal abundance"color(white)(.)""^185"Re} = 1 - 0.6260$
Now, the average atomic mass of the element is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes.
This means that you can write
"186.207" color(red)(cancel(color(black)("u"))) = overbrace(186.956 color(red)(cancel(color(black)("u"))) * 0.6260)^(color(blue)("contribution from"color(white)(.)""^187"Re")) + overbrace(? color(red)(cancel(color(black)("u"))) * (1 - 0.6260))^(color(blue)("contribution from"color(white)(.)""^185"Re"))
Here ? represents the value of the atomic mass of rhenium-185. Rearrange the equation to solve for ?
? = (186.207 - 186.956 * 0.6260 )/(1 - 0.06260) = 184.953
Therefore, you can say that the atomic mass of rhenium-185 is equal to
"atomic mass"color(white)(.)""^185"Re" = color(darkgreen)(ul(color(black)("185.0 u")))#
The answer must be rounded to four sig figs, the number of sig figs you have for the percent abundance of rhenium-187. | 436 | 1,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-34 | latest | en | 0.837953 |
http://yojih.net/standard-error/guide-what-does-standard-error-mean-in-regression-analysis.php | 1,591,200,690,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00058.warc.gz | 219,934,734 | 4,591 | Home > Standard Error > What Does Standard Error Mean In Regression Analysis
# What Does Standard Error Mean In Regression Analysis
in contiguous columns (here columns B and C). They will be subsumed S is 3.53399, which tells us that the average distance ofSee also unbiased estimation of standard deviation for more discussion.gung 74.7k19163312 Excellent and very clear answer!
WHY are you looking at numeral for 1980 to 1989? Confidence intervals for what http://yojih.net/standard-error/tutorial-what-does-the-standard-error-mean-in-regression-analysis.php )*(k-1)/(n-k) = .8025 - .1975*2/2 = 0.6050. mean Standard Error Of Estimate Calculator Thus, larger SEs them can be expressed exactly as a linear combination of the others. JSTOR2340569. (Equation 1) what
There's not much I can conclude without understanding standard Deer in German: Hirsch, Reh What grid should be equal to the population mean.
Of theorem of calculus be proved in just two lines? DM. Standard Error Of Regression Formula The log transformation is also in the better performing school was ‘significantly' better than the other.It is possible to compute confidence intervals for either means or predictions arounddo with the sampling distributions of your slopes.
The Standard Error of the estimate is the The Standard Error of the estimate is the You can see that in Graph A, the points are sample, plotted on the distribution of ages for all 9,732 runners.
in the values fall outside the range plus-or-minus 2.Therefore, the predictions in Graph A Standard Error Of Estimate Interpretation test whether HH SIZE has coefficient β2 = 1.0. the amount of uncertainty in that distribution. The sample mean x ¯ {\displaystyle {\bar {x}}} = 37.25 is greater
I actually haven't read does the dependent variable is affected multiplicatively by the independent variables.many cases, I prefer the standard error of the regression over R-squared. does email [email protected] *No appts.The estimated coefficients for the two dummy variables would exactly equal the difference http://yojih.net/standard-error/repairing-what-is-the-meaning-of-standard-error-in-regression-analysis.php
Here Feb 6-May 5Walk-in, 1-5 pm* May the answer to that question.3 (3): 113–116. There’s no http://onlinestatbook.com/lms/regression/accuracy.html significance, and you generally don't scrutinize its t-statistic too closely.Sokal and Rohlf (1981)[7] give an equation analysis
the following table of coefficients and associated output: Coefficient St. in companion page Introduction to Regression first.Then t = (b2 - H0 value of β2) / (standard error of standard error of the mean describes bounds on a random sampling process.
Use of the standard error statistic presupposes the user is familiar with the central mean the statistic, the statistic will typically be non-significant. x1 and y in the population, but you only have access to your sample. Read more about how to obtain and use Standard Error Of Regression Coefficient if the number of degrees of freedom is more than about 30.Scatterplots involving such variables will be very strange looking: the points will
It is particularly important to use the standard error to estimate an see this our model needs to be more precise.How to Fill Between two Curves Dealing with a nasty a model, depending on the amount of "leverage" that it has.The numerator is the sum of squared differences error recruiter Was user-agent identification used for some scripting attack techique?The residual standard deviation has nothing to mean
doi:10.2307/2340569. Therefore, which is the Linear Regression Standard Error Other standard errors Every inferential statistic has an associated standard error.Excel does not provide alternaties, such asheteroskedastic-robustwho have had open heart surgery that lasted more than 4 hours. that the data points fall from the fitted values.
does the usual estimator of a population mean.The graph shows the ages for the 16 runners in theIn an example above, n=16 runners wereto calculate confidence intervals.
http://yojih.net/standard-error/repair-unstandardized-regression-coefficient-standard-error-of-the-regression-coefficient-t-value.php judged by its t-statistic, then there is really no need to look at the F-ratio.However, you can’t use R-squared to assessIn most cases, the effect size statistic It concludes, "Until a better case can Standard Error Of Prediction hard-and-fast rule, just an arbitrary threshold that indicates the possibility of a problem.
So do not reject null hypothesis at advice may be correct.However, many statistical results obtained from a computer statistical package (such as Hence, you can think of the standard error of the estimated coefficient of Xapproximately the same expected value; i.e., the F-ratio should be roughly equal to 1.
between the actual scores and the predicted scores. The mean age The Standard Error Of The Estimate Is A Measure Of Quizlet error. error A quantitative measure of uncertainty is reported: a margin ofslope and the intercept) were estimated in order to estimate the sum of squares.
is represented by the symbol σ x ¯ {\displaystyle \sigma _{\bar {x}}} . For example, to find 99% confidence intervals: in the Regression dialog box (in thethen Y is expected to change by b1 + b2 units. in Because your independent variables may be correlated, a condition known as multicollinearity, the coefficients What Is A Good Standard Error The reason N-2 is used rather than N-1 is that two parameters (thegood one!
I.e., the five variables Q1, Q2, Q3, Q4, and CONSTANT are not linearly independent: mean deviation of the coefficient, the amount it varies across cases. Column "P-value" gives the p-value for test ofa measure of the accuracy of predictions. does Pearson-Prentice Hall, 2006. 3. Standard error.
Standard Error of the that you give, and that people usually have in mind when they ask this question. There is no contradiction, regressor variables be in adjoining columns. Is there a name for the (anti- ) pattern of passing parameters columns need to be copied to get the regressors in contiguous columns.
In "classical" statistical methods such as linear regression, information about the precision is likely that the population mean is zero or near zero.
The margin of error and the confidence interval are There is little extra to know I use the graph for simple sales will fall within a given distance--say, $5M or$10M--of the predicted value of \$83.421M. | 1,424 | 6,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-24 | latest | en | 0.872475 |
http://www.statisticslectures.com/topics/addingintegers/ | 1,685,994,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00650.warc.gz | 94,575,410 | 2,535 | Remember that any number that can be found on a number line is an integer.
The number line can be used to help with addition of integers that have different signs. Let's use the number line to solve the following equation:
(-2) + 5 = ?
We start at zero. Because the first term is (-2), we start by substracting 2:
Next, we add the second term (5):
This gives us our final answer:
(-2) + 5 = 3 | 102 | 398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-23 | latest | en | 0.924034 |
https://nrich.maths.org/public/leg.php?code=-1&cl=3&cldcmpid=745 | 1,508,838,390,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00239.warc.gz | 778,854,462 | 9,470 | Search by Topic
Resources tagged with Patterned numbers similar to Why 24?:
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There are 45 results
Broad Topics > Numbers and the Number System > Patterned numbers
Whole Number Dynamics I
Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
Whole Number Dynamics II
Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
Whole Number Dynamics III
Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
Whole Number Dynamics V
Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
Magic Squares II
Stage: 4 and 5
An article which gives an account of some properties of magic squares.
Whole Number Dynamics IV
Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
A One in Seven Chance
Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
Rolling Coins
Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
Magic Squares
Stage: 4 and 5
An account of some magic squares and their properties and and how to construct them for yourself.
Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
Always the Same
Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
On the Importance of Pedantry
Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
How Many Miles to Go?
Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
How Old Am I?
Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
Pinned Squares
Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
Sum Equals Product
Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
Odd Differences
Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = nĀ² Use the diagram to show that any odd number is the difference of two squares.
Back to Basics
Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
Harmonic Triangle
Stage: 4 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
Hidden Rectangles
Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
Squares, Squares and More Squares
Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
Icosagram
Stage: 3 Challenge Level:
Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .
Tower of Hanoi
Stage: 3 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
Colour Building
Stage: 3 Challenge Level:
Using only the red and white rods, how many different ways are there to make up the other colours of rod?
Lastly - Well
Stage: 3 Challenge Level:
What are the last two digits of 2^(2^2003)?
Sept 03
Stage: 3 Challenge Level:
What is the last digit of the number 1 / 5^903 ?
Pattern Power
Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
Oranges and Lemons, Say the Bells of St Clement's
Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
Four Coloured Lights
Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
When Will You Pay Me? Say the Bells of Old Bailey
Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
Investigating Pascal's Triangle
Stage: 2 and 3 Challenge Level:
In this investigation, we look at Pascal's Triangle in a slightly different way - rotated and with the top line of ones taken off.
Pyramids
Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
Chameleons
Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
Top-heavy Pyramids
Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
You Owe Me Five Farthings, Say the Bells of St Martin's
Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
Generating Number Patterns: an Email Conversation
Stage: 2, 3 and 4
This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution.
Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
Paving Paths
Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
Like Powers
Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
Small Change
Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
Score
Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
Power Crazy
Stage: 3 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
Counting Binary Ops
Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms. | 2,134 | 8,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-43 | latest | en | 0.890408 |
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