few_shots: 
Example:
Concept/skill: 
Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.

Problem/activity:
Problem 1
The table below contains single equations in a single variable, single equations in two variables, and pairs of simultaneous equations in two variables. Check the appropriate box to show whether there is no solution, exactly 1 solution, or if there is an infinite number of solutions.
{"table": [{"_attributes": {"align": "center", "class": ["table", "table-bordered"], "style": "table-layout:fixed; width:70%"}, "tbody": [{"tr": [{"td": [{"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "No solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Exactly 1 solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Infinite number of solutions"}]}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9=4x+7}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y=3y+5}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y={x\\over2}}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${3x+6y-9=0}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${x=y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${1.25x=1.25y}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9.9=6x+8y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9x=2.5y-8.8}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=1.5}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=2.5}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {}]}]}]}]}
Problem 2
Kim has a small container and a large container as shown.
[IMG]
It takes 16 of the small containers to fill the large container. Three small containers leave 1.95 gallons of space in the big container.
What is the size of each of the two containers?


Answer: 
No. Completing neither of these problems requires replacing one equation by the sum of that equation and a multiple of the other equation.

Now, assess whether the following problem or activity is 'aligned' with the given Common Core standard in Mathematics.

shot_info: fl_problem_001392_neg_0

8
standard_description: Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
problem_activity: Problem 1
Flares are used to warn cars that there is an obstacle on the road. Flares are often part of cars’ emergency kits and burn for a long time.
Fred’s Flare Company used to make flares that burned for 100 minutes on average. Fred changed the chemical formula in his flares, and he thinks that his new formula will burn for slightly over 120 minutes. Fred wants to make the claim on his packaging that his flares “Now burn for over 2 hours on average!!”
To make this claim, Fred needs to back it up with data. He decides that performing some random sampling is the best way to do this. Since he runs a 24-hour production operation, he decides that taking a sample each hour is the easiest way to do this.
Fred has tried two different ways of sampling, as shown below.
{"table": [{"_attributes": {"class": ["table"], "style": "table-layout:fixed; width:100%"}, "tbody": [{"tr": [{"td": [{"p": [{"_value": "Sample size: 20 flares"}, {"_value": "24 total samples (one per hour of production)"}, {"_attributes": {"style": "text-align:center"}, "img": [{"_attributes": {"alt": "", "src": "https://d37su8u1z9ynie.cloudfront.net/uploads/2018/06/05/algiiu8_l12_at1.png", "style": "height:76px; width:374px"}}]}]}, {"p": [{"_value": "Sample size: 100 flares"}, {"_value": "24 total samples (one per hour of production)"}, {"img": [{"_attributes": {"alt": "", "src": "https://d37su8u1z9ynie.cloudfront.net/uploads/2018/06/05/algiiu8_l12_at1a.png", "style": "height:101px; width:414px"}}]}]}]}]}]}]}
If we further extrapolate that Fred continues to take samples for 30 days in the same manner as above, he charts his results in histograms shown below.
{"table": [{"_attributes": {"class": ["table"], "style": "table-layout:Fixed; width:100%"}, "tbody": [{"tr": [{"td": [{"_attributes": {"style": "text-align:center"}, "img": [{"_attributes": {"alt": "", "src": "https://d37su8u1z9ynie.cloudfront.net/uploads/2018/06/05/algiiu8_l12_at1b.png", "style": "height:244px; width:358px"}}]}, {"img": [{"_attributes": {"alt": "", "src": "https://d37su8u1z9ynie.cloudfront.net/uploads/2018/06/05/algiiu8_l12_at1c.png", "style": "height:231px; width:350px"}}]}]}]}]}]}
The margin of error is the largest expected difference between the estimate of a population parameter and the actual population parameter.
What would you guess is the mean and the margin of error of burn time of Fred’s flares (with the new and improved formula)?
Problem 2
The margin of error can also be calculated in by
$${ME=2\sqrt{{\hat{p}}(1-{\hat{p}})\over n}}$$
, where
$${\hat{p}}$$
is the proportion of a sample of an event occurring,
$$1-{\hat{p}}$$
is the proportion of a sample of an event not occurring,
$$n$$
is the sample size, and
$${\sigma}$$
is the sample standard deviation.
Suppose you draw a random sample of 36 chips from a mystery bag and find 20 red chips. Find:
The sample proportion of red chips
The standard error (standard deviation from the mean)
The margin of error


aligns: no