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https://lokobar.pl/?p=30023 | [
"Request for Quotation\n\nYou can get the price list and a A&C representative will contact you within one business day.(",
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"[email protected])\n\nmotor power calculation for conveyor xls Description\n\n•",
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"Conveyor Horsepower Calculator - Superior …\n\nConveyor Horsepower Calculator Sara Hoidahl 2017-08-28T21:17:11+00:00. Conveyor Length (center-to-center) Belt Width. Vertical Lift . Belt Capacity. Calculated Minimum HP. 0.0 HP. Minimum HP + 10%. 0.0 HP. Backstop. Not needed. This required horsepower calculator is provided for reference only. It provides a reasonable estimation of required horsepower given user requirements. Superior ...\n\n•",
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"Motor Torque Calculations - NEPSI - Northeast …\n\nMOTOR TORQUE. The following calculators compute the various torque aspects of motors. These equations are for estimation only, friction, windage, and other factors are not taken into consideration. Calculator-1. Known variables: Horse Power and Speed in RPM Torque is the action of a force producing or tending to produce rotation. Torque = force x distance Torque Input Horse Power, hp : …\n\n•",
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"Belt Conveyors for Bulk Materials Practical Calculations\n\nBELT CONVEYORS - BASIC CALCULATIONS: 1. Mass of the Load per Unit Length: Load per unit length. Given the production capacity Qt = tph, the weight of the load per unit length (kg/m) – (lbs per ft) is calculated by: Wm = 2000. Qt or Wm = 33.333.Qt = (lb/ft) 60 x v v Q = 0.278.Qt or Q = Qt = (Kg/m) v 3.600 x v 2. Belt Tensions: In order to find the maximum tension is necessary to calculate the ...\n\n•",
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"Conveyor Power and Torque Calculator - EICAC\n\nCONVEYOR POWER CALCULATOR. Use this calculator to calculate the force, torque and power required from a conveyor to move a load at an angle. If your conveyor is horizontal enter an angle of 0. Enter your values for the Mass, Diameter, Beltspeed, Friction and Angle; select your units as required. MASS TO MOVE (M) DIAMETER OF DRIVE DRUM (D): BELTSPEED (S): COEFFICIENT OF …\n\n•",
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"Calculations for Screw conveyors - Bechtel\n\nCalculations for screw conveyors Power in Kw (P) Q x L x K 3600 x 102 P = power in Kw Q = capacity in 1000 kg per hour L = conveyor screw length (m) K = friction coeffi cient P = v = speed in m per sec v = estring 395 T +49 (0)212 64 50 94-0 [email protected] Wuppertal F +49 (0)212 64 50 94-10 K 102 Calculations for screw conveyors Capacity in m2 per hour (Q) Q = 47 ...\n\n•",
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"(DOC) erhitungan Daya Motor Conveyor …\n\nerhitungan Daya Motor Conveyor (Calculation of Conveyor Power Equipment\n\n•",
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"Conveyors - Load & Power Consumption\n\nLevel Ground Conveyors. Horsepower required for conveyors transporting material on level ground: 1 hp (English horse power) = 745.7 W = 0.746 kW; 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd; Lifting Conveyors. With lifting conveyors - add lifting power from the chart below to the level ground power from the chart above.\n\n•",
null,
"How to Calculate 3 Phase Motor Power …\n\nCalculate three-phase motor power consumption by multiplying amps by volts by the square root of three (W = AV(sqrt 3). For example, if the motor is drawing 30 amps at 250 volts, you have 30 x 250 x sqrt 3 (about 1.73) = 12,975 watts). Convert watts to kilowatts by dividing the number of watts by 1,000. Thus, a three-phase electric motor drawing 12,975 watts is consuming 12.975 kilowatts. For ...\n\n•",
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"Conveyor Belt Calculations - Bright Hub Engineering\n\nEvery calculation should contain a contingency factor to allow for occasional temporary overloads. It easy enough, given the low cost of low and fractional horsepower drives, to simply overpower your system. But your electrical controls contain a thermal overload which will trip the motor in the event of a jam or stall. This device not only protects the motor, it also protects from harm your ...\n\n•",
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"Electric Motor Calculator - Engineering ToolBox\n\nRLA - \"Running Load Amps\" - current drawn during normal operation of electric motor. FLA - \"Full Load Amps\" - amount of current drawn when full-load torque and horsepower is reached for the motor.FLA is usually determined in laboratory tests.Note! - in the calculator above FLA is RLA + 25%. 1 hp = 0.745 kW; Related Mobile Apps from The Engineering ToolBox ...\n\n•",
null,
"Motor Sizing Calculations\n\nCalculation for the Effective Load Torque ( Trms ) for Servo Motors and BX Series Brushless Motors. When the required torque for the motor varies over time, determine if the motor can be used by calculating the effective load torque. The effective load torque becomes particularly important for operating patterns such as fast-cycle operations ...\n\n•",
null,
"Electric Motor Calculator - Engineering ToolBox\n\nRLA - \"Running Load Amps\" - current drawn during normal operation of electric motor. FLA - \"Full Load Amps\" - amount of current drawn when full-load torque and horsepower is reached for the motor.FLA is usually determined in laboratory tests.Note! - in the calculator above FLA is RLA + 25%. 1 hp = 0.745 kW; Related Mobile Apps from The Engineering ToolBox ...\n\n•",
null,
"Conveyor Power and Torque Calculator - EICAC\n\nCONVEYOR POWER CALCULATOR. Use this calculator to calculate the force, torque and power required from a conveyor to move a load at an angle. If your conveyor is horizontal enter an angle of 0. Enter your values for the Mass, Diameter, Beltspeed, Friction and Angle; select your units as required. MASS TO MOVE (M) DIAMETER OF DRIVE DRUM (D): BELTSPEED (S): COEFFICIENT OF …\n\n•",
null,
"Conveyors - Load & Power Consumption\n\nLevel Ground Conveyors. Horsepower required for conveyors transporting material on level ground: 1 hp (English horse power) = 745.7 W = 0.746 kW; 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd; Lifting Conveyors. With lifting conveyors - add lifting power from the chart below to the level ground power from the chart above.\n\n•",
null,
"Screw Conveyor Interactive Calculators | …\n\nEng. Guide Index Download Guide PDF HORIZONTAL SCREW CONVEYOR CAPACITY & SPEED CALCULATION: Visit the online engineering guide for assistance with using this calculator. Click on symbol for more information. DESIGN CONDITIONS 1. Flowrate(m): lb/hr 2. Density: 3. Loading (K): % SPCL. FLIGHT […]\n\n•",
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"Calculating Conveyor Power for Bulk Handling | …\n\nOriginal Power Calculation Program (free downloadable Excel program, CEMA 4 version) Online Application Data Sheet (linked to our engineers) Application Data Sheet (downloadable pdf file you can send to us) We use a modified version of the Conveyor Equipment Manufacturers Association guidelines. The primary equation for Effective Tension, Te, is as follows: Te = LKt (Kx + KyWb + …\n\n•",
null,
"Motor Sizing Calculations\n\nCalculation for the Effective Load Torque ( Trms ) for Servo Motors and BX Series Brushless Motors. When the required torque for the motor varies over time, determine if the motor can be used by calculating the effective load torque. The effective load torque becomes particularly important for operating patterns such as fast-cycle operations ...\n\n•",
null,
"Power calculation for belt conveyor | Tecnitude\n\nPower calculation. We provide this calculation form to assist you with assessing the required power for your belt conveyor, depending on the weight carried. You can also use our product configurator to view your tailored conveyor. Feel free to contact us for any of your projects. Tecnitude's team is at your disposal. Name . Company . Email . Telephone . Country +33 (0)3 89 60 34 40. Rent ...\n\n•",
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"roller conveyor calculations? - Commercial …\n\n21.02.2005· roller conveyor calculations? bmw318s70 (Electrical) (OP) 1 Feb 05 23:15. hi. I am trying to buil a spreadsheet for calculation the required HP of roller conveyors. The setup of the conveyor is te following: Motor connected to a gearbox. Output sprocket linked to a roller sprocket. All rollers linked toghether. This conveyor should move a pallet-load of X lbs, at Y feet/minutes. The various ...\n\n•",
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"Understanding Conveyor Belt Calculations | …\n\nUnderstanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system. We use cookies to personalize content and analyze traffic. We also share information about your use of our site with our social media, advertising, and analytics partners who may combine it with other information that you've provided or that we have ...\n\n•",
null,
"torque - Sizing a motor for a conveyor - Electrical ...\n\nCompanies that manufacture conveyor systems probably have software tools that calculate the motor size for them. But for someone who doesn't have this software, how does one go about determining what size motor they need to drive a 6000lbs load over a conveyor of length 20 feet. Let's assume 1800RPM since that's what most conveyor motors at our ...\n\n•",
null,
"How to Calculate 3 Phase Motor Power …\n\nCalculate three-phase motor power consumption by multiplying amps by volts by the square root of three (W = AV (sqrt 3). For example, if the motor is drawing 30 amps at 250 volts, you have 30 x 250 x sqrt 3 (about 1.73) = 12,975 watts). Convert watts to kilowatts by dividing the number of watts by 1,000."
] | [
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https://www.nneellinvest.com/post/average-directional-index | [
"Search\n\n# Average Directional Index",
null,
"The ADX or Average Directional Index is a technical indicator used to measure the overall strength of a trend.\n\nDeveloped by J. Welles Wilder, the Average Directional Index (ADX) helps traders measure how strongly price is trending and whether its momentum is increasing or falling.\n\nIt’s important to emphasize that while ADX measures the strength of a trend, it does NOT identify the trend’s direction.\n\nIt can be used to find out whether the market is ranging or starting a new trend.\n\nThe oscillator ranges between 0 and 100 with high readings indicating a strong trend and low readings indicating a weak trend. How to Calculate ADX:\n\nThe ADX is derived from two directional indicators, known as DI+ and DI-:\n\n• The positive directional indicator (+DI)\n\n• The negative directional indicator (-DI)\n\nThese two indicators are derived from the Directional Movement Index (DMI).\n\nADX is calculated by finding the difference between DI+ and DI-, as well as the sum of DI+ and DI-.\n\nThe difference is divided by the sum, and the resulting number multiplied by 100.\n\nThe result is known as the Directional Index or DX.\n\nA moving average is then taken of DX, typically over a fourteen-day period (although any number of periods can be used.)\n\nThis final moving average is the ADX."
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null
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https://www.pndhs.org/course/topics-in-college-algebra-and-trig/ | [
"contact@pndhs.org (309) 691-8741\n249\n\n# Topics in College Algebra and Trig\n\nCourse ID\n249\nLevel\n12\nSemester\n1, 2\nCredit\n1.000\nMethod\nRegular\n\nThis course will take a more detailed examination into topics discussed in 243 Algebra II and then introduces topics essential to the understanding of trigonometry. In Algebra, topics include linear equations, quadratic equations, polynomial equations, logarithms, exponentials, rational equations and conics. Trigonometry topics will include the trigonometric ratios and functions, right triangle trigonometry, use of the unit circle, graphing trigonometric functions, and solving trigonometric equations. A calculator is required. The math faculty recommends the TI 34 series calculator.\n\n#### Prerequisites\n\nCompletion of 243 Algebra II. An academic course review will be conducted to determine student’s academic placement. Course satisfies a requirement for graduation."
] | [
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https://foxoyo.com/profile/102/dikshant-bansal/answer | [
"",
null,
"# Dikshant Bansal\n\nLet A = the number be abc Let B = the reverse cba Now write he value of each number (1) A = 100*a + 10*b + c and (2) B = 100*c + 10*b + a Now get the differnce, D, (3) D = 100*a + 10*b + c - 100*c - 10*b - a or (4) D = 100*(a - c) + (c - a) or (5) D = 100*(a - c) - (a - c) or (6) D = 99*(a - c) Answer: The greatest number that divides the difference of a 3 digit number and its reverse i\nAug 31, 2018 08:35"
] | [
null,
"https://lh5.googleusercontent.com/-NaXVN-GVB2o/AAAAAAAAAAI/AAAAAAAAC1g/P2kf_vXaK3Q/photo.jpg",
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https://madison.uiu.edu/academics/courses/math-105/ | [
"",
null,
"",
null,
"",
null,
"Skip to content\n\n# MATH 105 College Mathematics/Applications\n\nCredits: 3\nPrerequisites:\n• Pass MATH 095 or ACT math score = 19 or an alternative placement mechanism as approved by the math department or instructor approval.\nDistance Learning Options:\n\nThis course is a survey of mathematical applications of functions. Topics that will be covered include: fundamental concepts of algebra, algebraic equations and inequalities; functions and graphs; zeros of polynomial functions; exponential and logarithmic functions; systems of equations and inequalities. The mathematics of personal finance will also be studied."
] | [
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"https://insight.adsrvr.org/track/pxl/",
null,
"https://insight.adsrvr.org/track/pxl/",
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https://stupefied-davinci-2941e0.netlify.app/divisibility-rules-worksheet-grade-4.html | [
"# Divisibility Rules Worksheet Grade 4",
null,
"### Image result for divisibility rules Divisibility rules",
null,
"### Label the number as 'divisible' or 'not divisible' based on the remainder, in this collection of divisibility test pdf worksheets for grade 4 and grade 5.\n\nDivisibility rules worksheet grade 4. Divisibility rules or divisibility tests have been mentioned to make the division procedure easier and quicker. 3030 is divisible by 5, because it ends in 0. Instruct students to finish the challenge independently, using their rules chart as a guide.\n\nThis free product introduces divisibility rules divisibility tests with a fun to use poem that your studen divisibility rules teaching mathematics math poems. Circulate the room as students work and provide support as needed. 25, 100, 365 9 the sum of the digits of the number is divisible by 9.\n\nB) circle the numbers that are divisible by 4. Direct students' attention to the divisibility challenge on the divisibility rules worksheet. The sum of the digits of 1,485 is 1+4+8+5 = 18 which is divisible by 9, so 1,485 is divisible by 9.\n\nFill in the digits to make this number divisible by 4 & 6. Divisibility rules add to my workbooks (18) embed in my website or blog add to google classroom add to microsoft teams 645 964,726 42,391 8,862 250,860 25,368 53,420 5,232 27,568 186 17 105,743 7,526 3,892 301,570 86,340 4,900 4,584 78 41,346\n\nDivisibility rules worksheet with answers to practice & learn 6th grade math problems on finding divisors is available online for free in printable & downloadable (pdf & image) format. Some of the worksheets for this concept are divisibility rules work, divisibility rules, divisibility rules practice directions use divisibility, divisibility rules workbook, divisibility rule, divisibility rule 1, math mammoth grade 5 a worktext, divisibility rules. Test if the numbers are divisible by 4, by dividing the last 2 digits of the number by 4.\n\nSome of the worksheets for this concept are divisibility work, divisibility and factors, divisibility rules workbook, divisibility rules practice directions use divisibility, divisibility rules, gr 7 divisibility, divisibility rules, divisibility rules a. If you answer incorrectly, your ship is shot by the pirate. Check divisibility (4) a) 3466 b) 1288 c) 39804 d) 64 684 21.",
null,
"### Working 4 the Classroom “Flipping” for a Year End Review",
null,
"### Divisibility Rules Worksheet 6th Grade Divisibility Test",
null,
"### Divisibility Rules Poster Large Printable Bulletin Board",
null,
"### FREE HandsOn Divisibiltiy Rules Worksheet For 2, 3, 6, 5",
null,
"### Divisibility Rules Reference Sheet & Poster Kraus Math",
null,
"### Number Theory, Divisibility Rules, Prime Factorization",
null,
"### Divisibility Rules Reference Sheet & Poster Kraus Math",
null,
"### Multiplication Strategies & Divisibility Rules Cheat",
null,
"### Random Posts",
null,
""
] | [
null,
"https://i.pinimg.com/originals/b3/d0/dc/b3d0dcbc541864dc6b08bc68cc10c76a.png",
null,
"https://i.pinimg.com/originals/d8/d3/dd/d8d3dd0c256d10b9ade9ffcdf6de8b15.jpg",
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"https://i.pinimg.com/originals/41/ba/39/41ba39adf89014e293e20a975085c513.jpg",
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"https://i.pinimg.com/originals/89/f9/1f/89f91f6a8fec674c35baeb678f57747a.jpg",
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"https://i.pinimg.com/originals/9f/5e/9e/9f5e9e821edc9f91dc239d48300e8508.jpg",
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"https://i.pinimg.com/originals/6e/b5/f4/6eb5f4d7a8a6cc3fc98ef67bb5774a3f.jpg",
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"https://i.pinimg.com/736x/4b/9d/b9/4b9db9db20676a88574eeb3d43710299.jpg",
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"https://i.pinimg.com/originals/b8/73/38/b873389190feaf7d3fd05d6a4b00edd9.png",
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"https://i.pinimg.com/originals/98/dd/39/98dd39def8ed643928c8913b781f593e.png",
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"https://i.pinimg.com/736x/6e/73/6a/6e736abcf14f8545e6ff6004c9b7fcf4.jpg",
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"https://i.pinimg.com/originals/1a/7e/c8/1a7ec878a61363e20395306f90a2b4cf.jpg",
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"https://3.bp.blogspot.com/-ZZSacDHLWlM/VhvlKTMjbLI/AAAAAAAAF2M/UDzU4rrvcaI/s1600/btn_close.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8974629,"math_prob":0.82942337,"size":2240,"snap":"2021-43-2021-49","text_gpt3_token_len":564,"char_repetition_ratio":0.269678,"word_repetition_ratio":0.05027933,"special_character_ratio":0.26785713,"punctuation_ratio":0.13302752,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9505015,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,6,null,2,null,2,null,2,null,4,null,4,null,4,null,7,null,4,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T12:14:21Z\",\"WARC-Record-ID\":\"<urn:uuid:210bdf68-7453-4fe4-bd92-439a16e629f0>\",\"Content-Length\":\"33773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8594cfed-355a-4e4d-8153-a2fa14790b08>\",\"WARC-Concurrent-To\":\"<urn:uuid:519247ae-a594-4860-8895-ef154c649a51>\",\"WARC-IP-Address\":\"67.207.81.229\",\"WARC-Target-URI\":\"https://stupefied-davinci-2941e0.netlify.app/divisibility-rules-worksheet-grade-4.html\",\"WARC-Payload-Digest\":\"sha1:H73OFB2CXCFN7453VK3IJSJD3S7VBECA\",\"WARC-Block-Digest\":\"sha1:G33K2W3IJEDZITEO2EJIXDH5Q5DUXXIP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358520.50_warc_CC-MAIN-20211128103924-20211128133924-00595.warc.gz\"}"} |
https://indico.mitp.uni-mainz.de/event/56/contributions/1873/ | [
"# 54. International Winter Meeting on Nuclear Physics\n\nJan 25 – 29, 2016\nBormio, Italy\nEurope/Berlin timezone\n\n## Measurement of neutral mesons in pp and Pb-Pb collisions at midrapidity with the ALICE experiment at the LHC\n\nJan 29, 2016, 5:20 PM\n20m\nBormio, Italy\n\n#### Bormio, Italy\n\nShort Contribution Relativistic Heavy Ion Physics\n\n### Speaker\n\nMs Lucia Leardini (Physikalisches Institut (PI) Heidelberg)\n\n### Description\n\nNeutral mesons, such as $\\pi^{0}$ and $\\eta$, are probes for the study of the energy loss of partons traversing the hot and dense medium, the Quark-Gluon Plasma, that is formed in heavy-ions collisions. Moreover, they represent the largest background for the direct photon measurement and an accurate estimate is therefore necessary to determine the decay photon contribution. The ALICE experiment measures $\\pi^{0}$ and $\\eta$ mesons via the two-gamma decay channel. The photon detection can either be direct, using the electromagnetic calorimeters EMCal and PHOS, or by reconstructing the electron-positron pairs from photon conversions in the detector material (photon conversion method, PCM). With the PCM, photons are reconstructed using the ALICE Inner Tracking System (ITS) and the Time Projection Chamber (TPC). This method has full azimuthal coverage, which compensates for the small conversion probability. It provides a precise measurement at low transverse momentum. The calorimeters have reduced acceptance but trigger capabilities and provide the high $p_{\\mbox{\\tiny{T}}}$ measurement, where they also have good energy resolution. The errors from these measurements are independent thus their comparison is a good cross check and their combination gives a more precise result. In this presentation, we will show the $\\pi^{0}$ and $\\eta$ spectra in Pb--Pb collisions at $\\sqrt{s_{\\mbox{\\tiny NN}}}$ = 2.76~TeV and pp collisions at different center of mass energy obtained over a wide transverse momentum range.\n\n### Primary author\n\nMs Lucia Leardini (Physikalisches Institut (PI) Heidelberg)\n\n Slides"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8298804,"math_prob":0.9279587,"size":1679,"snap":"2023-40-2023-50","text_gpt3_token_len":361,"char_repetition_ratio":0.09313433,"word_repetition_ratio":0.025531914,"special_character_ratio":0.20190589,"punctuation_ratio":0.08303249,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96146464,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T19:56:28Z\",\"WARC-Record-ID\":\"<urn:uuid:d02dbb25-3057-498d-a1f5-8772493c7b8c>\",\"Content-Length\":\"53125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68ffaac6-c763-4aca-944f-af28b5ba7def>\",\"WARC-Concurrent-To\":\"<urn:uuid:e618dc71-59ec-4510-8a1c-70d7b0675573>\",\"WARC-IP-Address\":\"134.93.175.73\",\"WARC-Target-URI\":\"https://indico.mitp.uni-mainz.de/event/56/contributions/1873/\",\"WARC-Payload-Digest\":\"sha1:V24SWSFF537WW77EL5QHKONSXD4NBJFH\",\"WARC-Block-Digest\":\"sha1:CVWAX2H4HAOKG5LC4DQTB7R2WGKABENG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.53_warc_CC-MAIN-20231203193127-20231203223127-00885.warc.gz\"}"} |
https://es.mathworks.com/matlabcentral/answers/485383-if-statement-with-or-condition?s_tid=prof_contriblnk | [
"# If statement with or condition\n\n4 views (last 30 days)\nluca on 15 Oct 2019\nAnswered: Fabio Freschi on 15 Oct 2019\nHi given the following code\nBDR= [175 175 175 175 175 175 175 175 175 175 175 175 175];\nSETTIMANA=[190 130 120 140 100 160 175 165 157 140 130 175 140 ];\nNEXTpro = [25 60 50 40 30 30 30 35 10 34 23 45 12];\nif (SETTIMANA(1) < BDR (1)) | (SETTIMANA(2)< BDR(2)) | (SETTIMANA(3)< BDR(3)) | (SETTIMANA(4)< BDR(4))| (SETTIMANA(5)< BDR(5))| (SETTIMANA(6)< BDR(6))| (SETTIMANA(7)< BDR(7))| (SETTIMANA(8)< BDR(8))| (SETTIMANA(9)< BDR(9)) | (SETTIMANA(10)< BDR (10)) | (SETTIMANA(11)< BDR(11))| (SETTIMANA(12)< BDR(12)) | (SETTIMANA(13)< BDR(13));\nY=NEXTpro\nelse\nY=NEXTpro\nSETT1 = SETTIMANA(1:numel(BDR)); % Equalise Vectors\nY = Y(1:numel(BDR)); % Equalise Vectors\nidxy = SETT1 <= BDR; % Logical Index Vecto\nY=BDR - SETT1;\nY=Y.*idxy\nend\nI cannot understand why the if condition cannot read the or operator in the right way.\nThe conditin is: if I have a value in SETTIMANA that exceed the value in BDR in the same column, then switch to condition else.\n##### 2 CommentsShowHide 1 older comment\nKALYAN ACHARJYA on 15 Oct 2019\nIts perfectly working as you mentioned\nIf any element of SETTIMANA< correcponding NEXTpro\n%...^...Note here\n% do\nelse\n%do\nend\nor\nIf all element of SETTIMANA< correcponding NEXTpro\n%...^...Note here\n% do\nelse\n%do\nend\nWhich one\nAlso you can use logical indexing directly without mentioning one by one of SETTIMANA and NEXTpro\n\nFabio Freschi on 15 Oct 2019\nif any(settimana < bdr)\n...\nelse\n...\nend\n\nR2019b\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5572963,"math_prob":0.9131758,"size":1652,"snap":"2021-43-2021-49","text_gpt3_token_len":596,"char_repetition_ratio":0.20145631,"word_repetition_ratio":0.0858209,"special_character_ratio":0.37288135,"punctuation_ratio":0.096875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9666276,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T19:36:31Z\",\"WARC-Record-ID\":\"<urn:uuid:289bc540-08c7-477e-b5de-2e2a57ff6f3c>\",\"Content-Length\":\"117331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0263933d-c29c-407a-8b7a-6b3a8955f616>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f459e1f-1c59-48bc-a241-14df15e166c1>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://es.mathworks.com/matlabcentral/answers/485383-if-statement-with-or-condition?s_tid=prof_contriblnk\",\"WARC-Payload-Digest\":\"sha1:DFIMZWTVQST5MSABVODJYQRKLHUIDIBY\",\"WARC-Block-Digest\":\"sha1:MYJ6Z2TQAFOLNH66C25QQXXVTEK2VUM7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363216.90_warc_CC-MAIN-20211205191620-20211205221620-00292.warc.gz\"}"} |
http://www.infogalactic.com/info/Linear_interpolation | [
"# Linear interpolation",
null,
"Given the two red points, the blue line is the linear interpolant between the points, and the value y at x may be found by linear interpolation.\n\nIn mathematics, linear interpolation is a method of curve fitting using linear polynomials.\n\n## Linear interpolation between two known points",
null,
"In this geometric visualisation, the value at the green circle multiplied by the distance between the red and blue circles is equal to the sum of the value at the red circle multiplied by the distance between the green and blue circles, and the value at the blue circle multiplied by the distance between the green and red circles.\n\nIf the two known points are given by the coordinates",
null,
"$(x_0,y_0)$ and",
null,
"$(x_1,y_1)$, the linear interpolant is the straight line between these points. For a value x in the interval",
null,
"$(x_0, x_1)$, the value y along the straight line is given from the equation",
null,
"$\\frac{y - y_0}{x - x_0} = \\frac{y_1 - y_0}{x_1 - x_0}$\n\nwhich can be derived geometrically from the figure on the right. It is a special case of polynomial interpolation with n = 1.\n\nSolving this equation for y, which is the unknown value at x, gives",
null,
"$y = y_0 + (y_1-y_0)\\frac{x - x_0}{x_1-x_0}$\n\nwhich is the formula for linear interpolation in the interval",
null,
"$(x_0,x_1)$. Outside this interval, the formula is identical to linear extrapolation.\n\nThis formula can also be understood as a weighted average. The weights are inversely related to the distance from the end points to the unknown point; the closer point has more influence than the farther point. Thus, the weights are",
null,
"${\\textstyle \\frac{x-x_0}{x_1-x_0}}$ and",
null,
"${\\textstyle \\frac{x_1-x}{x_1-x_0}}$, which are normalized distances between the unknown point and each of the end points. Because these sum to 1,",
null,
"$y = y_0 * (1-\\frac{x - x_0}{x_1-x_0}) + y_1 * (1-\\frac{x_1 - x}{x_1-x_0}) = y_0 * (1-\\frac{x - x_0}{x_1-x_0}) + y_1 * (\\frac{x - x_0}{x_1-x_0})$\n\nwhich yields the formula for linear interpolation given above.\n\n## Interpolation of a data set",
null,
"Linear interpolation on a data set (red points) consists of pieces of linear interpolants (blue lines).\n\nLinear interpolation on a set of data points (x0, y0), (x1, y1), ..., (xn, yn) is defined as the concatenation of linear interpolants between each pair of data points. This results in a continuous curve, with a discontinuous derivative (in general), thus of differentiability class",
null,
"$C^0$.\n\n## Linear interpolation as approximation\n\nLinear interpolation is often used to approximate a value of some function f using two known values of that function at other points. The error of this approximation is defined as",
null,
"$R_T = f(x) - p(x) \\,\\!$\n\nwhere p denotes the linear interpolation polynomial defined above",
null,
"$p(x) = f(x_0) + \\frac{f(x_1)-f(x_0)}{x_1-x_0}(x-x_0). \\,\\!$\n\nIt can be proven using Rolle's theorem that if f has a continuous second derivative, the error is bounded by",
null,
"$|R_T| \\leq \\frac{(x_1-x_0)^2}{8} \\max_{x_0 \\leq x \\leq x_1} |f''(x)|. \\,\\!$\n\nAs you see, the approximation between two points on a given function gets worse with the second derivative of the function that is approximated. This is intuitively correct as well: the \"curvier\" the function is, the worse the approximations made with simple linear interpolation.\n\n## Applications\n\nLinear interpolation is often used to fill the gaps in a table. Suppose that one has a table listing the population of some country in 1970, 1980, 1990 and 2000, and that one wanted to estimate the population in 1994. Linear interpolation is an easy way to do this.\n\nThe basic operation of linear interpolation between two values is commonly used in computer graphics. In that field's jargon it is sometimes called a lerp. The term can be used as a verb or noun for the operation. e.g. \"Bresenham's algorithm lerps incrementally between the two endpoints of the line.\"\n\nLerp operations are built into the hardware of all modern computer graphics processors. They are often used as building blocks for more complex operations: for example, a bilinear interpolation can be accomplished in three lerps. Because this operation is cheap, it's also a good way to implement accurate lookup tables with quick lookup for smooth functions without having too many table entries.\n\n## Extensions\n\n### Accuracy\n\nIf a C0 function is insufficient, for example if the process that has produced the data points is known be smoother than C0, it is common to replace linear interpolation with spline interpolation, or even polynomial interpolation in some cases.\n\n### Multivariate\n\nLinear interpolation as described here is for data points in one spatial dimension. For two spatial dimensions, the extension of linear interpolation is called bilinear interpolation, and in three dimensions, trilinear interpolation. Notice, though, that these interpolants are no longer linear functions of the spatial coordinates, rather products of linear functions; this is illustrated by the clearly non-linear example of bilinear interpolation in the figure below. Other extensions of linear interpolation can be applied to other kinds of mesh such as triangular and tetrahedral meshes, including Bézier surfaces. These may be defined as indeed higher-dimensional piecewise linear function (see second figure below).",
null,
"Example of bilinear interpolation on the unit square with the z-values 0, 1, 1 and 0.5 as indicated. Interpolated values in between represented by colour.",
null,
"A piecewise linear function in two dimensions (top) and the convex polytopes on which it is linear (bottom).\n\n## History\n\nLinear interpolation has been used since antiquity for filling the gaps in tables, often with astronomical data. It is believed that it was used by Babylonian astronomers and mathematicians in Seleucid Mesopotamia (last three centuries BC), and by the Greek astronomer and mathematician, Hipparchus (2nd century BC). A description of linear interpolation can be found in the Almagest (2nd century AD) by Ptolemy.\n\n## Programming language support\n\nMany libraries and shading languages have a 'lerp' helper-function, returning an interpolation between two inputs (v0,v1) for a parameter (t) in the closed unit interval [0,1]:\n\n// Imprecise method which does not guarantee v = v1 when t = 1,\n// due to floating-point arithmetic error.\nfloat lerp(float v0, float v1, float t) {\nreturn v0 + t*(v1-v0);\n}\n\n// Precise method which guarantees v = v1 when t = 1.\nfloat lerp(float v0, float v1, float t) {\nreturn (1-t)*v0 + t*v1;\n}\n\n\nThis function is used for alpha blending (the parameter 't' is the 'alpha value'), and the formula may be extended to blend multiple components of a vector (such as spatial x,y,z axes, or r,g,b colour components) in parallel."
] | [
null,
"http://www.infogalactic.com/w/images/thumb/d/dd/LinearInterpolation.svg/300px-LinearInterpolation.svg.png",
null,
"http://www.infogalactic.com/w/images/thumb/a/aa/Linear_interpolation_visualisation.svg/300px-Linear_interpolation_visualisation.svg.png",
null,
"http://www.infogalactic.com/w/images/math/f/c/6/fc6c713e7eb34f6cb756d754fc2f61db.png ",
null,
"http://www.infogalactic.com/w/images/math/6/1/3/613c8cdd5c639e212bb058608712c542.png ",
null,
"http://www.infogalactic.com/w/images/math/f/0/4/f04efc83eb32f8a8960968b395f9f0b2.png ",
null,
"http://www.infogalactic.com/w/images/math/9/f/1/9f13c74f9736bd5305576e6af8b3148d.png ",
null,
"http://www.infogalactic.com/w/images/math/f/4/7/f472ceedb1868e147ebe2c1197bf8748.png ",
null,
"http://www.infogalactic.com/w/images/math/f/0/4/f04efc83eb32f8a8960968b395f9f0b2.png ",
null,
"http://www.infogalactic.com/w/images/math/2/a/7/2a7121408784f784e04ef3a459c57be1.png ",
null,
"http://www.infogalactic.com/w/images/math/5/d/c/5dc84302fc2fdd3438241ffc8e1a1506.png ",
null,
"http://www.infogalactic.com/w/images/math/0/b/a/0ba7f0b630b8623c583d97d782e0dd14.png ",
null,
"http://www.infogalactic.com/w/images/thumb/6/67/Interpolation_example_linear.svg/300px-Interpolation_example_linear.svg.png",
null,
"http://www.infogalactic.com/w/images/math/2/e/4/2e4d68dd3f69136c2a2b743d41da1329.png ",
null,
"http://www.infogalactic.com/w/images/math/c/c/0/cc0d6a9f15f662e6b49893c99fabad82.png ",
null,
"http://www.infogalactic.com/w/images/math/0/a/6/0a69d597543f20cb8440429488cb4451.png ",
null,
"http://www.infogalactic.com/w/images/math/4/5/4/4542c6ab840bf9c584e23f6ad95b6a99.png ",
null,
"http://www.infogalactic.com/w/images/thumb/c/c6/Bilininterp.png/300px-Bilininterp.png",
null,
"http://www.infogalactic.com/w/images/thumb/6/6d/Piecewise_linear_function2D.svg/300px-Piecewise_linear_function2D.svg.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86525524,"math_prob":0.99744064,"size":6217,"snap":"2019-26-2019-30","text_gpt3_token_len":1343,"char_repetition_ratio":0.1781748,"word_repetition_ratio":0.033826638,"special_character_ratio":0.20894322,"punctuation_ratio":0.11527905,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997273,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,2,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T01:32:32Z\",\"WARC-Record-ID\":\"<urn:uuid:f1a28743-2a8e-4d82-95ed-3b6aee62263d>\",\"Content-Length\":\"40442\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74d6f217-41df-45f8-9849-4fc3d01a62ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:c364cd31-fab1-45a7-840f-3def984484d5>\",\"WARC-IP-Address\":\"85.195.95.72\",\"WARC-Target-URI\":\"http://www.infogalactic.com/info/Linear_interpolation\",\"WARC-Payload-Digest\":\"sha1:ZNG7EJFX6VPBFWUFL7IZZ4BOGELOCQ2Z\",\"WARC-Block-Digest\":\"sha1:7CTDZOGDOQRB4D7Z3EXERRJ6JC5MI5G5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525004.24_warc_CC-MAIN-20190717001433-20190717023433-00159.warc.gz\"}"} |
https://www.aaai.org/Library/ICML/2003/icml03-089.php | [
"# td(0) Converges Provably Faster than the Residual Gradient Algorithm\n\nRalf Schoknecht and Artur Merke\n\nIn Reinforcement Learning (RL) there has been some experimental evidence that the residual gradient algorithm converges slower than the td(0) algorithm. In this paper, we use the concept of asymptotic convergence rate to prove that under certain conditions the synchronous off-policy td(0) algorithm converges faster than the synchronous off-policy residual gradient algorithm if the value function is represented in tabular form. This is the first theoretical result comparing the convergence behaviour of two RL algorithms. We also show that as soon as linear function approximation is involved no general statement concerning the superiority of one of the algorithms can be made."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9030069,"math_prob":0.9809322,"size":872,"snap":"2021-43-2021-49","text_gpt3_token_len":165,"char_repetition_ratio":0.122119814,"word_repetition_ratio":0.0,"special_character_ratio":0.17201835,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9550162,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T04:35:30Z\",\"WARC-Record-ID\":\"<urn:uuid:4a1026f8-99c5-4c67-98cf-665f2f81d3c3>\",\"Content-Length\":\"2551\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71d15534-0d27-4e2e-924e-f92408ed7787>\",\"WARC-Concurrent-To\":\"<urn:uuid:bfb2bbdb-ff70-44ca-b8f9-af67c39a8a02>\",\"WARC-IP-Address\":\"144.208.67.177\",\"WARC-Target-URI\":\"https://www.aaai.org/Library/ICML/2003/icml03-089.php\",\"WARC-Payload-Digest\":\"sha1:XHGQR4PCRUBHASHV7BW6SPYKNT6764RQ\",\"WARC-Block-Digest\":\"sha1:N4QEREREMPP5O3YUZMTP6LHWZWSN4ECR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585196.73_warc_CC-MAIN-20211018031901-20211018061901-00144.warc.gz\"}"} |
https://bird.st/learn-python-2/ | [
"Python 是一個高階的編程語言。基本上就是你可以在一個 runtime 下直接執行一些程式碼。 而且很多人也覺得 Python 的程式碼看上去很直覺,很像虛擬碼,因此初學者會很好上手。上一篇文章有提到了,Python 在資料科學、機器學習、深度學習方面有很多很棒的套件。如: Tensorflow、Keras、PyTorch 等... 但是在進入這些套件之前,讓我們先來了解一些 Python 的基礎。正如 C、Java 等課程,大家都會先從一些基本的 Data Type 開時。\n\n### 數字 Numbers\n\n# 這個是註解\nx = 2\ny = 3.2\nprint(x) # \"2\"\nprint(type(y)) # \"<class 'float'>\"\nprint(y ** x) # \"10.24\"\nx += 1\nprint(x) # \"3\"\n\n### 字串 Strings\n\nPython 的字串可以用 '\" 來表示字串。在 Python 裡面一個很特別的功能是,字串可以用 + 來做 concatenation。 一點值得注意的是,字串和數字是無法做 concatenation 的,因此要用 str() 先把數字轉成字串。字串有很多 function 可以在文字處理上應用,大家未來在做文字分析的時候可能會很常用到哦!\n\nh = 'hello'\nw = \"world\"\nhw = h + ' ' + w\nprint(hw) # \"hello world\"\nprint(\"OM\" + str(0)) # \"OM0\"\nhw1 = '%s %s %d' % (hello, world, 1) # sprintf style string formatting\nhw2 = '{} {} {}'.format(hello, world, '2') # format style string formatting\nprint(hw1) # \"hello world 1\"\nprint(hw2) # \"hello world 2\"\nstr = \"bird \"\nprint(len(w)) # 印出字串的長度 \"5\"\nprint(s.capitalize()) # 將第一個字轉為大寫 \"Bird \"\nprint(s.upper()) # 將整個字串轉為大寫 \"BIRD \", s.lower 則為轉小寫\nprint(s.replace('bi', '(ne)')) # 將字串中與第一組字串相似的組合換成第二組字串 \"(ne)rd \"\nprint(s.strip()) # 將字串前後的空格去掉 \"bird\"\nprint(s.split('i')) # 將字串遇到 'i' 的時候切段,並組成 list \"rd\"\n\n### 布林 Boolean\n\nt = True\nf = False\nprint(t and f) # Logical AND \"False\" (兩個都要 True 或 False 才會回傳 True)\nprint(t or f) # Logical OR \"True\" (其中一個 True 就會回傳 True)\nprint(not t) # Logical NOT \"False\" (not True = False, not False = True)\nprint(t != f) # Logical XOR \"True\" (兩者要不一樣才會回傳 True)"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.83644515,"math_prob":0.59940934,"size":1898,"snap":"2021-43-2021-49","text_gpt3_token_len":1219,"char_repetition_ratio":0.12777191,"word_repetition_ratio":0.0,"special_character_ratio":0.32455215,"punctuation_ratio":0.06309148,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96329296,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T17:44:38Z\",\"WARC-Record-ID\":\"<urn:uuid:c20e5edb-d518-4d18-b5c9-f905c430dc3b>\",\"Content-Length\":\"44148\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ab18ccf-6c8a-4199-9b1e-29e41f754825>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f957a3b-9734-4c70-a1d5-e68fb30a160f>\",\"WARC-IP-Address\":\"188.166.24.60\",\"WARC-Target-URI\":\"https://bird.st/learn-python-2/\",\"WARC-Payload-Digest\":\"sha1:CIDX2I3VEK3AFWDJBRWYJPTEFGDHK6WL\",\"WARC-Block-Digest\":\"sha1:7A6LX6NOA25HCFH2V4L6MHCXZWZRVSOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585737.45_warc_CC-MAIN-20211023162040-20211023192040-00204.warc.gz\"}"} |
https://library.automationdirect.com/implementing-a-plc-calibration-routine-to-ensure-accurate-instrument-readings-issue-2-2004/ | [
"# Implementing a PLC Calibration Routine to Ensure Accurate Instrument Readings\n\n## Why Calibrate?\n\nMany applications call for a means to take accurate analog measurements. Force, pressure, electrical current, lengths, positions and other analog values that are measured must be done so with a degree of accuracy. Factory and lab personnel use calibration procedures to ensure that equipment is reading accurately. If the readings are not as accurate as required, a calibration procedure should allow for the implementation of a correction factor. Personnel operating the equipment can perform their jobs more readily if the measurement system can easily be checked for accuracy and calibrated. Ideally, the process would be transparent to operators who do not need to understand the details of the operating system and only need to see and understand the measurement results reported by the equipment.\n\nAn engineer designing a measurement system will generally examine the voltage or current to be provided by the measuring transducer and the “counts” that should result after the analog input card converts the electrical signal to a number. The conversion factors can be set up to convert counts to real world units such as pressure or force. These theoretical or ideal factory published numbers are suitable for specifying equipment, but when the system is set up and running, the actual readings and numbers will differ from the theoretical. Wires will add resistances that cannot be calculated in advance and temperature changes will cause variations that must be accounted for. No analog system will run exactly per “book values” and thus a calibration routine must correct for the differences.\n\n## Calibration Method\n\nThe calibration method described here can be applied to any analog measurement such as pressure, amperage, weight, length or force. Let’s consider an example where the length of an object is being measured. We will assume that a device has been built that converts the length of the object to numerical readings via the analog input. The type of instrumentation used may be a laser measuring device, an LVDT, or any device that provides an analog signal that will vary with the length of the object being measured. We will assume the following:\n\n1. The object being measured ranges from 2 to 9 inches in length and the analog device is capable of measuring over this range.\n\n2. The analog device provides a voltage signal to the analog input card. The voltage is converted to a number or “counts” by the analog card. The “count” resides in real time in a register in the PLC and the ladder logic can perform math functions using this number. For example, let’s assume that an object is being measured 2 inches in length and the count is 100, and when an object 9 inches in length is measured the count is 3900.\n\nThe nice thing about the method described here is there is no need to be concerned with the amount of voltage provided by the transducer or received by the analog input card. By utilizing the counts and then converting the counts directly to length, the voltage becomes irrelevant as long as any change in counts is linear with any corresponding change in length. This provides the advantage of not needing to be concerned with calibration or setup of the transducer and its amplifier. Too often, time is spent setting the “zero” and “span” of amplifiers and signal conditioners where a person might determine: “This object measures 2.3 inches, therefore my voltage should be 1.2 volts”. Then they proceed to set the zero and span to achieve an exact voltage for the corresponding length. By using the calibration method described here, this process is no longer necessary and the exact voltage at any given length is of no concern. Time is saved because the calibration of the amplifier’s voltage is now rolled into the software calibration of the entire system. Additionally, if someone comes along later and adjusts the zero or span, the calibration procedure described below can be run and the PLC program will compensate for the changes.\n\n## Calibration Procedure\n\nCalibration routines are often set up to simply add in a correction factor; but this “one point” method leaves room for error over a range of readings. A car speedometer that is permanently stuck at 60 mph will appear accurate if checked when the vehicle is moving at 60 mph. A better calibration method uses two points; a “low point” and a “high point” and then calculates a line that goes through both points. We will use a 2-point calibration, which requires two calibration units of specific lengths. We will call these units “calibration masters”. In our example, they can be any length within the readable linear range. Let’s assume that one is 4.2 inches long and the other is 6.8 inches long.\n\nFour values will be captured during the calibration procedure:\n\na. The actual length of the “low” calibration master (4.2 inches)\n\nb. The analog reading (counts) when the low master is\nmeasured\n\nc. The length of the “high” cali- bration master (6.8 inches)\n\nd. The analog reading (counts) when the high master is\nmeasured\n\nFrom this point forward the four values, a through d, will be used to set up the math required to covert counts to inches. We will call the resulting value “e”.\n\nThe change in length for any corresponding change in counts is calculated as follows:\n\n(c-a)/(d-b) = e\n\nThe person performing the calibration enters the lengths of the high and low calibration masters (values a and c) via the operator interface and these values are stored in PLC registers. The corresponding analog counts for the high and low calibration masters (values b and d) are captured during the calibration routine and are also stored in PLC registers.\n\nWe will use the following data registers:\n\na = “low” calibration master = Register V3000 / V3001\n\nb = low master analog reading = Register V3002 / V3003\n\nc = “high” calibration master = Register V3004 / V3005\n\nd = high master analog reading = Register V3006 / V3007\n\nThe following ladder logic will calculate “e” and store this value in V3010/V3011. Data register V3014 is used to store temporary math results while the accumulator is in use performing other math. See Figure 1",
null,
"## Length Conversion\n\nAfter the calibration routine has been run and the values a through e are stored in the PLC, the formula for converting counts to length is:\n\n((counts – b) x e) + a\n\nThe following ladder logic will calculate the length from the counts and store the resulting length in V3012/V3013. The analog reading from the device being measured is stored in register V2000. See Figure 2",
null,
"Some of the advantages of this calibration procedure are:\n\nBecause the user chooses the calibration points, calibration masters can be at any length. Calibration masters no longer need to be made to a specified predetermined length.\n\nIf a specific range is the most critical, the calibration can be done at that specific range. In the example above, if the most critical readings are between 3.5 and 4.2 inches of length, the calibration masters can be approximately 3.5 and 4.2 inches. In this way, calculating the straight line over the critical range minimizes any non-linearity in the system and provides the most accurate readings in the critical range.\n\nBy displaying the actual calculated length and the analog counts on the operator interface, anyone who desires can follow along with the PLC and observe the math in action. This makes the system’s internal workings clear to anyone using the equipment.\n\nBy Gary Multer,\nMulteX Automation\n\nOriginally Published: Sept. 1, 2004"
] | [
null,
"https://library.automationdirect.com/wp-content/uploads/2013/09/Calibrating-ladder-logic.jpg",
null,
"https://library.automationdirect.com/wp-content/uploads/2013/09/ladder-logic-length.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91254866,"math_prob":0.95563567,"size":7569,"snap":"2020-34-2020-40","text_gpt3_token_len":1549,"char_repetition_ratio":0.15730338,"word_repetition_ratio":0.017391304,"special_character_ratio":0.2065002,"punctuation_ratio":0.08069164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9726173,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T14:39:15Z\",\"WARC-Record-ID\":\"<urn:uuid:56b882c1-c015-40ed-a391-522acdc7d72f>\",\"Content-Length\":\"100513\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:35514142-6383-42c3-b749-29368c8e93f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc02095a-11ca-46a6-a226-4ba8e34a5716>\",\"WARC-IP-Address\":\"205.151.114.62\",\"WARC-Target-URI\":\"https://library.automationdirect.com/implementing-a-plc-calibration-routine-to-ensure-accurate-instrument-readings-issue-2-2004/\",\"WARC-Payload-Digest\":\"sha1:SIJ7DYH32JFOUOV4LHLWERJJ5VOLSOJL\",\"WARC-Block-Digest\":\"sha1:ZNGHK2VPPGTWLV5HS5HUDBP3JQJSYHSP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735867.94_warc_CC-MAIN-20200804131928-20200804161928-00397.warc.gz\"}"} |
https://earth-planets-space.springeropen.com/articles/10.5047/eps.2009.11.003 | [
"We’d like to understand how you use our websites in order to improve them. Register your interest.\n\n# Earth magnetic field modeling from Oersted and Champ data\n\n## Abstract\n\nWe present a method to model geomagnetic field requiring only a restricted number of measurements on magnetic survey satellite orbits. These points are chosen in an optimal—or close to optimal—manner relying on recent developments in the problem of numerical integration over spheres. The method allows us to compute a series of models at short time intervals, namely 10 days in the present study. At each of these close dates several models are computed from independent sets of data; their redundancy in turn provides a control of results thanks to which the selection of data—for example, as a function of magnetic activity or latitude—may be reduced. We find that the internal low degree Gauss coefficients derived from Oersted and Champ data, respectively, differ from one another by 1 or 2 nT. We then take as a second example of the method application a brief study of the so-called external field. We compare the first-degree axisymmetric field with the Dst index.\n\n## 1. Introduction\n\nThe problem of computing a model of magnetic field that first fits ground observations, then satellite observations (since the years 1960) goes back to Gauss in the 1830s. The most recent models using Oersted and Champ data rely on a least squares technique providing the spherical harmonic coefficients, and some of these solve the Euler angles of the sensor attitude at the same time (Olsen et al., 2006). There is probably no need to abandon spherical harmonic expansion, which is so practical for all applications of the models. Nevertheless, even when keeping to this classical method, different options exist. We will describe and use one of these, the choice of which is guided by a few considerations.\n\nFirst, it is desirable to retain as much data as possible from high (>55°) latitude regions, despite the large disturbances which are present in those regions. Second, the key issue in spherical harmonic expansion modeling, i.e. the computation of Gauss coefficients, both internal and external, is the geographical distribution of data; integral orthogonality properties of the harmonics over the data points must be strongly adhered to, not loosely. Third, we call for a flexible algorithm, using the minimum but sufficient number of data points to compute the internal and external fields, at short time intervals, for night hours or day hours, different universal times, and different conditions of activity, etc.\n\nExcellent models have been computed by different teams according to their own methods: using data from Pogo, Magsat, Oersted and Champ satellites as well as from ground magnetic observatories (Sabaka et al., 2002) and Sabaka et al. (2004) have constructed comprehensive models covering the time interval (1960-2002). Maps of the internal field (core + crustal) and the external field (ionospheric and magnetospheric) are made available with a short time sampling. Other teams have produced models of the different ingredients of the geomagnetic field from satellite data covering the time interval (2000, 2005) (Maus et al., 2006; Olsen et al., 2006).\n\nWe present the general features of our method in the main text reserving more mathematical considerations for Appendix. Its efficiency will be demonstrated in the following sections.\n\n## 2. The Model\n\n### 2.1 Optimal arrays of points\n\nLet us consider a sphere S of unit radius. An optimal array—in the sense to be defined below—of N points on S is made of the points whose colatitude θ k and longitude ϕ k , k = 1,… N, are given by the formulae below",
null,
"A subroutine computes the longitudes ϕ k and colatitudes θ k of the optimal array, N being given (Appendix). In this paper, we will take N = 1000, 2000, 3000, 5000, 10000.\n\nLet the surface harmonic functions",
null,
"be ranked in the usual lexicographic order, and u j be the corresponding jth harmonic, internal or external (see Eq. (2)), in the corresponding series. The gradients u i , U j are orthogonal on the set of N points Q l , l = 1,… N. (see Appendix)",
null,
"((1))\n\nwith C i being the corresponding norm of u i .\n\n### 2.2 Array of points close to an optimal array\n\nLet us compute, as an example, a model based on night values of a given period of time. We first retain all measurement points P k satisfying this local time condition. Let E (Pk) = E (P k (r,θ,ϕ)) be this set of points. We pick up in E (P k ) the point P i which is the closest to the point Q i of a given optimal array, in the sense that the angle P i OQ i (O being the sphere center) is the smallest (Fig. 1). The number N of points P i , equal to the number of points Q i of the optimal array, is much smaller than the number of points P k . After this operation is made for each point Q i , we are left with an N array close to the optimal: E N (P i (r i ,θ i , ϕ i )). Now, let the geomagnetic potential to be computed be written in the usual form:",
null,
"((2))\n\nwith",
null,
", and K[ext] being the degrees of the internal and external expansions.\n\nLet B⃗(P l ) = (X(P l ), Y(P l ), Z(P l )) be the vectorial measurement of B⃗ at P l ε E N (P). We want to compute the Gauss coefficients g and γ such that",
null,
"((3))\n\nis minimal.\n\nV contains K[tot] = K[int](K[int] + 2) + K[ext] (K[ext] + 2) unknown coefficients, and there are 3N equations (3). In this paper we will address the core field and, briefly, a component of the so-called external field, with K[int] = 165, and K[ext] = 2. For such values, N = 1000 already provides a largely overdetermined set of equations. The system (3) will be solved by the usual least squares technique, and the inversion of the resulting normal matrix by the Singular Value Decomposition algorithm. When computing this matrix, the following approximations of scalar products in Eq. (1) appear:",
null,
"with, for example,",
null,
"which allows for computing the following angles between the K[tot] column vectors of the matrix:",
null,
"Looking at the values of β ij is key to the method. Indeed, u j vectors are not expected to be strictly orthogonal on the {P i } array: their projections on the sphere (r = a) are not located exactly at points of the optimal array and, furthermore, radii of P i points may vary within 200 kilometers (see next paragraph).\n\nRemark We could correct B⃗(P i ) for the difference B⃗(P i )-B⃗(Q i ) using an a priori model. However, here we prefer to present a self-contained algorithm.\n\n### 2.3 A synthetic example\n\nFirst, we pick up a model {g j , γ j } from the literature (Langlais et al., 2003; K[int] = 16, K[ext] = 2 ) and compute the values X, Y, Z of the model field at M points which are “real” points in the sense that there are points of the orbits of Champ where real measurements were made (specifically the data points corresponding to 130 days centered at 2003.0). We select N points P l from the whole set, the one closest to an optimal array of 1000 or 3000 points {Q l }, as explained in Fig. 1, and compute back the Gauss coefficients {g j , γ j }. The differences between the initial Gauss coefficients and the recovered ones, g j and γ k , j = 1,… 288 k = 1,… 8, for N = 1000 and N = 3000, are of the order of 10-5 nT.\n\nLet us take the opportunity with this synthetic example— but with real orbit points—to be more specific about the orthogonality of u j vectors, taking the case N = 1000 (1000 points in array {Q l }). For K[int] = 16 and K[ext] = 2, we find that 99% of couples u i , u j vectors make angles",
null,
"with ε < 0.01 radian. The value of ε does not change much with the degree: from 0.005 for the low harmonic couples to 0.03 for the high harmonics. Simulations show that for such values of the departure ε from orthogonality, the coefficients g j and γ j can be recovered independently of one another with the required accuracy. Such will be the case for all the computations in the paper.\n\n### 2.4 The time sampling of the modeling\n\nA huge advantage of satellite data is their high density (108 per year!), which allows for a massive number of computations which could not be dreamed of in the pre-satellite times; this density makes it possible to split the data set into many subsets for various applications. In particular, it is possible to compute models at a succession of close time moments; in this paper, we will compute a model every 10 days, at times t k = t0 + k × 10 days. Nevertheless, those models are not genuine instantaneous models at t k . Indeed, to get a uniform enough distribution of local times when computing a model of the main field Eq. (2), it is necessary to consider data spanning a time interval τ around t k ;the duration τ is different for Champ and Oersted. So, our model at time t k is a model computed from data in a time interval of length τ centered on t k . But we keep g(t), h(t) (with a daily sampling) in the following temporal series, and not the decimated series with a sampling of τ days (τ = 130 in case of Champ; see below); a surprising amount of information on short term (τ) features is preserved in the process of computing the model from data in a window of length τ (e.g., Blanter et al., 2005).\n\n### 2.5 The choice of data amount\n\nWe have addressed the question of the number N of data points required to compute a model (an expansion) to degree d with an empirical point of view; indeed, not only the distribution of data points, but also the noise intervene in a broad sense, on the data. We compute a large number (100) of models {g i , h j } from suitable random simulations of data; we infer a mean value m and error bar σ on each coefficient. Each random simulation at N points is obtained by biasing the initial vector field (calculated from a 16-degree model) by a Gaussian random vector whose distribution parameters were estimated at each point using real data records. And we repeated those computations varying N from 700 to 6000 in 25 steps. For the large majority of coefficients, m is approximately stabilized at N = 2400 (Fig. 2). The value of σ decreases from 0.75 nT for N = 700 to 0.2–0.4 nT (depending on the coefficient) for N = 2400 and keeps decreasing slowly thereafter. From these results we retained N = 2400 in most of the computations of the present paper, for models of degree 16.\n\n### 2.6 Selecting the data as a function of magnetic activity We now switch to real data.\n\nIn most studies, measurements are selected according to the magnetic situation at the time of the measurement, as characterized by the value of the planetary Kp index. We made a series of experiments to evaluate the influence of magnetic activity on our modeling. We take advantage of redundant observations. Let us first retain all of the data, without any selection, versus magnetic activity. For each date t k (multiple of 10 days) we build a number of quasi-optimal arrays {P i } close to the same optimal array {Q i } (with, in general, N = 2400), from different disjoint sets of data; we then compute as many coefficient sets or models. Some of these appear to be perturbed by big magnetic storms; but it is always possible to find some which are not. The trend of the representative curves allows us to easily discard perturbed values due to the tightness of time sampling. Figure 3(a) has been chosen to illustrate the situation; it represents the evolution of the g 11 estimate. Crosses are for estimates computed from 130 days of Champ measurements without any selection in the function of magnetic activity. A segment of the curve, in the second half of 2002, is shifted 12 nT below the general trend. And a blank is observed in the first months of 2004. Both segments (S. Maus, personal communication) are characterized by a relatively low number of usable measurement points, which makes it harder to find points close to optimal data sets in the 130 days of data, especially if high magnetic activity is present during those intervals; in this situation, steps in (g, h) estimate may occur. In fact, due to the abundance of data, it is always possible—except in the case of long gaps—to find close to optimal subsets from which g, h estimates fill the gaps of the graphs of Fig. 3, and are on the general trend. Those conclusions hold for all low degree coefficients of the main field. We kept those gaps here for illustration. Nevertheless, it would be awkward not to take advantage of the large redundancy of data to avoid computations from data corresponding to high activity, e.g., a m > 20 nT (Mayaud, 1980). An extra verification is easily obtained by changing the threshold for a m . Coefficients computed with this condition are represented by dots in Fig. 3 together with those estimated without selection. Along the same lines, we keep the vectorial measurements in high latitudes (as Maus et al , 2006).\n\n## 3. Fitting Oersted and Champ Data\n\n### 3.1 The data\n\nWe use Champ data provided by the German team in the form of a list t, X(t), Y(t), Z(t), r⃗(t), r⃗(t) being the current point and τ being the time of the measurement on the orbit, counted in seconds from June 2001 to December 2004, and Oersted data provided by the Danish team in the same form, τ being then a multiple of 1.3 sec (see Stolle et al., 2006), running from March 1999 to June 2003. Note that these data are transformed data. To obtain geocentric components X, Y, Z, Euler angles of the sensor attitude have been determined by the teams in charge. We do not discuss this determination.\n\n### 3.2 Fitting the data\n\nA first opportunity to check the efficiency and accuracy of the algorithm is to look at how coefficients issued from Oersted and Champ fit together, paying special attention to the overlapping period (June 2001–June 2003). We treat the data from both satellites exactly the same way, except for the time span τ required to build a model from data reasonably uniformly distributed in local time; indeed, 130 days are required in the case of Champ, while 2 years are needed in the case of Oersted to get the same performance; actually we use a 90-day time span for Oersted to the cost of a less strict condition on the uniformity of local time distribution.\n\nResults are illustrated, for a few Gauss coefficients in Fig. 4. Let us look, for example, at h 11 (Fig. 4(b)); at each time t k there are eight estimates for Champ, five for Oersted. We will systematically use this control in all our computations. Clearly, h 11 values derived from Oersted and Champ agree within 2 nT most of the time, without any further averaging. Other examples are for h 35 and h 45 (mind the enlarged scale). the mean values from Oersted and Champ coincide within a nT. In the computations of this section, we retained data subsets built in such a way that local times are reasonably uniformly distributed. The agreement between Oersted- and Champ-derived coefficients is pretty good but not perfect. There is a physical limitation to this agreement. the ionospheric field is not sampled in exactly the same way by the two satellites. Resulting departures are small and quite variable with the coefficient. For example, for coefficient g 01 , a tiny drift of Oersted estimates with respect to Champ ones is observed Fig. 4(a) (the differences remaining smaller than 2 nT).\n\nWe also computed the dipole moment and the angle between the geographical axis and the dipole axis over the 1999–2004 time span (Fig. 5(a)). Except for some trouble at the beginning of Oersted life, the linear decreasing trend appears almost perfect.\n\n## 4. Internal and External Field\n\nThe satellite sees as a field of internal origin the sum of the main field generated by the dynamo, the lithospheric field, and the field generated by electric currents flowing in the ionosphere below the perigee of the satellite, especially according to the classical views, in the E layer at 110 km altitude (Ratcliffe, 1972). The satellites also encounter electric currents since they are flying in the upper part of the F layer. In particular, the fields associated with field-aligned electric currents in the polar regions can be large. But, as discussed in Section 2.6, despite their large magnitude, they do not severely affect the aimed modeling (the computation of internal low ( 13) degree Gauss coefficients from data covering a given time interval). So, we may reasonably assume that the essential part of the field which is external to the solid Earth but internal to the satellites orbits is generated by currents in the E layer, i.e., currents driven by the atmospheric dynamo and currents driven in high latitude regions by forces originating high in the ionosphere (the polar current system) (Ratcliffe, 1972; Encrenaz et al., 2004).\n\nTo obtain a model of the main field, we have to get rid of this ionospheric field. As is well known, this is not an easy task (Olsen, 1996; Thomson, 2000). It is nevertheless possible to estimate the magnitude of the ionospheric contribution to the internal Gauss coefficients. We will not develop this question at length in the present paper. A classical method is to select data on the basis of local time, e.g., all local times, only day times or only night times (e.g., 6:00–18:00 or 18:00–6:00 LT). Our results will be illustrated by a few graphs. Figure 6(a) shows estimates of g 01 , over a time span of 4 years, derived from Champ data, computed respectively from data at all local times, day times, and night times. The difference between night time estimates and the all times estimates is almost everywhere smaller than 3 nT (2 nT in the second half of 2004). Results for g 22 are shown in Fig. 6(b) (mind the scale). Generally, the amplitudes of the differences between the different estimates as well as their evolution in time depend on the considered coefficient in a way which is not straightforward to understand. For all of the coefficients, differences between different estimates do not exceed 2 or 3 nT. A last graph Fig. 7 shows two estimates of g 01 using all local times; for the first one, no selection is made in the function of activity; for the second one, only measurements corresponding to a m 20 nT are retained. A significant difference of 2 nT shows up, but the trend is the same.\n\n### 4.1 The internal field\n\nTo obtain the main (dynamo) field model, we first take the average of models of the internal field (sources within the sphere r = rperigee) computing on all the universal times (practically at 0:00 UT, 1:00 UT to 23:00 UT). That comes down to compute a model I from points uniformly distributed in longitude; to check it we choose points Q i (see Section 2.1) whose measurements correspond to local times drawn randomly in the interval 0:00 LT–24:00 LT and compute a model II derived from this set of points. Figure 8(c) shows that the two models are indeed identical. Of course, adopting such a model for the main field means that the 24-h averaged ionospheric field is supposed to be zero—when averaged over a full day. It is such a model that we compute at days t k = t0 + k × 10 days, from data in a 130-day time span τ centered on t k .\n\nWe compare our two models, relative to 2003.0, with the POMME-3 model (Maus et al., 2006) relative to the same epoch, in two ways. First, we compare the coefficients of our models and those of the POMME-3 model by computing their differences as well as the mean and standard deviation of these differences (Fig. 8(a) and 8(b)). Except for g 01 andg 03 , the models can be said to be very similar. Our computation of the main field supposes we eliminate the ionospheric field by averaging in longitude, i.e. that this latter field has no zonal component, or, for approximate symmetry reasons, no component g 02k+1 . Choosing only night values leads to slightly different estimates (Fig. 6(a)). We then draw the maps of the vertical components of models I and POMME-3 at the core-mantle boundary (CMB) more precisely on a sphere of radius 3480 km. We also map their differences (Fig. 9). We note that, despite the smallness of difference in coefficients illustrated by Fig. 8, a few small scale anomalies may reach a notable amplitude. The geometrical factor",
null,
"is indeed equal to 4500; the downward continuation to the CMB of a core field contaminated by crustal anomalies is known to require some precautions. As expected, the largest discrepancies are observed in high lattitudes, due to field aligned currents.\n\n### 4.2 The external (ring current) field\n\nWe keep here a rather formal point of view, without addressing the physical nature of the field. This is the way the question has been treated for decades—the external field was globally called the ring current field up to recent modelings of satellite data (Olsen et al., 2000). Only external field coefficients of the first degree were considered to be safely determined. So we compute, with the same sampling interval (every 10 days), from the same sets of data, coefficients (we note them γ, ν for simplicity, γ for the cosine and ν for sine term) of the external field, i.e., external to the sphere containing the satellites’ orbits (r > rapogee max). It is useful to recall again that we are not computing an instantaneous field, but, at day t k ,a field based on a data set extracted from 130 days of data centered at t k (for Champ). For each data set we check that the orthogonality conditions required in Section 2 are verified: the VM[ext] are orthogonal, with a high accuracy, to one another, and orthogonal to all the VM[int] (Section 3). There is no contamination of the external coefficients by the internal ones, i.e., no contamination of the external field by the internal field. The results depend on what is being looked for. For example, y° value is much larger when computed only from night (18:00-06:00 LT) values than when computed from day (06:00–18:00) values. This asymmetry was pointed out as early as in 1970 by Olson (1970).\n\nWe retain all the local times—which comes down to averaging in longitude—and focus on the axisymmetric coefficients γ 0 k . The graph of Fig. 10 represents the variation from April 1999 to December 2004 of γ 01 . Data from both satellites, Champ and Oersted, are used. Again, the fitofthe models derived from the two data sets is excellent, within 1 or 2 nT. γ 01 displays variations with time constants of a few months and amplitudes of some 15 nT. At the bottom of Fig. 10 the evolution of Dst index is presented, averaged over a running window of 90 days to make the two graphs comparable (such a comparison is not a new idea (e.g. Cain et al., 1967). Let us recall that Dst (Sugiura, 1964) is “the disturbance field which is axially symmetric with respect to the dipole axis, and which is regarded as a function of storm time”. The correlation between Dst and γ 01 is good up to the end of 2002; amplitudes of γ 01 variations are smaller. An interesting observation concerns the base level; it is zero by construction for Dst index, while γ 01 evolves between 15 and 40 nT.\n\nMaus and Lühr recently performed a study of the mag-netospheric field during magnetically quet times (Maus and Luhr, 2005) using Oersted and Champ data from the years 1999–2004. The field is decomposed into contributions from sources in the solar-magnetic frame, and those in the geocentric-solar-magnetospheric frame. Such a separation is probably necessary for a coherent study of the external field. We pointed out at the beginning of this section the limited scope of our study of the “external field“.\n\n## 5. Conclusion\n\nThe general objective of the analysis we presented here is to model different ingredients of the field altogether. The method reported here is the realization. This paper also has also a methodological character; we computed Gauss coefficients of the main (dynamo) field, and computed coefficients of the so-called ring current field. The characteristics of the analysis are the following. To compute a given model of the main field, for example, it is possible to use for each τ-interval centered on day τ a number of disjoint data sets, each including a rather small (1000, 2400) number of points. In this way, spurious values of coefficient estimates are made conspicuous; this provides a control of the model (g, h) which allows us to release data selection. An advantage of a close time-spacing of the model is to provide time series whose trends can be studied in the usual way.\n\n## References\n\n1. Blanter, E., M. Shnirman, and J. L. Le Mouël, Solar variability, evaluation of correlation properties, J. Atmos. Terr. Phys., 67, 521–534, 2005.\n\n2. Cain, J. C., S. J. Hendricks, R. A. Langel, and W. V. Hudson, A proposed model for the International Geomagnetic Reference Field-1965, J. Geomag. Geoelectr., 19, 335–355, 1967.\n\n3. Delsarte, P., J. Goethais, and J. Seidel, Spherical codes and designs, Geom. Dedicata, 6, 363–388, 1977.\n\n4. Encrenaz, T., J. Bibring, M. Blanc, M. Barucci, P. Zarka, and F. Roques, Le système solaire, CNRS editions, 2004.\n\n5. Habicht, W. and B. L. van der Waerden, Lagerung von Punkten auf der Kugel, Math. Ann., 123, 223–234, 1951.\n\n6. Hardin, D. P. and E. B. Saff, Discretizing manifolds via minimum energy points, Notices of AMS, 51(N10), 1186–1194, 2004.\n\n7. Hardin, R. and N. Sloane, McLaren/rss improved snub cube and other new spherical designs in three dimensions, /avmalgin. livejournal.com/1203330.html, Disc. Comp. Geom., 15, 429–442, 1996.\n\n8. Langlais, B., M. Mandea, and P. Ultre-Guerard, High-resolution magnetic field modeling: application to Magsat and Ørsted data, Phys. Earth Planet. Inter., 135,77–91, 2003.\n\n9. Maus, S. and H. Luhr, Signature of the quiet-time magnetospheric magnetic field and its electromagnetic induction in the rotating Earth, Geophys. J. Int., 162, 755–763, 2005.\n\n10. Maus, S., M. Rother, C. Stolle, W. Mai, S. Choi, H. Lühr, D. Cooke, and C. Roth, Third generation of the Potsdam Magnetic Model of the Earth (POMME), G3, 7, N7, 2006.\n\n11. Mayaud, P. N., Derivation, meaning and use of geomagnetic indices, Geophysical Monograph, 22, AGU, Washington D.C., 1980.\n\n12. Olson, W. P., Variations in the Earth’s surface magnetic field from the magnetopause current system, Planet. Space Sci., 18,1471–1484,1970.\n\n13. Olsen, N., A new tool for determining ionospheric currents from magnetic satellite data, Geophys. Res. Lett., 23, 3635–3638, 1996.\n\n14. Olsen, N., T. J. Sabaka, and L. Toffner-Clausen, Determination of the IGRF 2000 model, Earth Planets Space, 52, 1175–1182, 2000.\n\n15. Olsen, N., H. Lühr, T. J. Sabaka, M. Mandea, M. Rother, L. Toffner-Clausen, and S. Choi, CHAOS—a model of the Earths magnetic field derived from CHAMP, Oersted, and SAC-C magnetic satellite data, Geophys. J. Int., 166,67–75, 2006.\n\n16. Rakhmanov, E., E. Saff, and Y. Zhou, Minimal discrete energy on the sphere, Math. Res. Lett., 1, 647–662, 1994.\n\n17. Ratcliffe, J. A., An Introduction to the Ionosphere and Magnetosphere, Cabridge University Press, 1972.\n\n18. Sabaka, T. J., N. Olsen, and R. A. Langel, A comprehensive model of the quiet-time near-Earth magnetic field: Phase 3, Geophys. J. Int., 151,32–68, 2002.\n\n19. Sabaka, T. J., N. Olsen, and M. Purucker, Extending comprehensive models of the Earth’s magnetic field with Oersted and CHAMP data, Geophys. J. Int., 159, 521–547, 2004.\n\n20. Stolle, C., H. Lühr, M. Rother, and G. Balasis, Magnetic signatures of equatorial spread F as observed by the CHAMP satellite, J. Geophys. Res., 111, A02304, 2006.\n\n21. Sugiura, M., Hourly values of equatorial Dst for IGY, in Annals of the International Geophysical Year, 35, 945–948, Pergamon Press, Oxford, 1964.\n\n22. Thomson, A. W. P., Improving the modelling of the geomagnetic mainfield: Isolating the average ionospheric field in satellite data, Earth Planets Space, 52, 1199–1206, 2000.\n\n## Acknowledgments\n\nWe acknowledge fruitful discussions with S. Maus. We deeply thank the two referees, S. Macmillan and H. Utada, for the attention they paid to the first version of the paper and the resulting numerous and helpful comments and recommendations.\n\nAuthors\n\n## Appendix A. The Numerical Integration Over Spheres\n\n### Appendix A. The Numerical Integration Over Spheres\n\nThe question of distributing uniformly N points on a surface—we consider here the case of a sphere—is by no means a trivial one; it has been of interest to mathematicians since Antiquity, and is still the object of research (Hardin and Saff, 2004).\n\nWe want an algorithm which distributes a set of N points on a sphere in such a way that the distribution of these points converges to an uniform distribution when N gets large. But uniformity makes sense only for infinite sets. For N finite, we look for a configuration which is optimal with respect to some property, but which may be poor with respect to another property.\n\nFor example, we may want to approximate the integral of a function over the sphere S2 by an arithmetic sum (without weights) of values ƒ (⃗k) at some N well-chosen points ⃗k on S2, in other words we want the difference",
null,
"((A.1))\n\nto be small for a large class of functions. Such a configuration of points will be called optimal for the evaluation of the integral. The problem has not yet received a general precise solution, but explicit particular solutions have been found. For example Delsarte et al (1977) considered so-called {N, t} spherical designs: configurations of N points ⃗k such that for all polynomials Pm(⃗) (i.e., polynomials in three variables (x, y, z) = r⃗) of degree m t, the difference in Eq. (A.1) is equal to zero. By means of a computer search, Hardin and Sloane (1996) found spherical designs for all t 13 with a minimal number of points; for instance, they produced a (94, 13)-design. A variety of algorithms have also been proposed for explicitly constructing asymptotically uniform distributions of points on S2. The most recent approach is to look for configurations corresponding to the minimum potential energy of N repelling points (repelling force needs to be redefined). The problem with the minimum energy configurations algorithm is that it is long and cumbersome. That is why we consider in this paper much more straightforward “spiral sets” which are almost as good as the “polynomial adapted” or “energy sets” refered to above. They are also good with respect to the property of orthogonality of spherical harmonics; this essential property will be simply checked.\n\nLet us now describe the spiral set devised by Rakhmanov, Saff, and Zhou (Rakhmanov et al., 1994), used in the present study. In spherical coordinates (θ,ϕ), 0 θ π, 0 ϕ 2π 0 ≤ϕ ≤2π, we take the following coordinates of the N points as:",
null,
"((A.2))\n\nThe estimate of the maximum diameter of N nonoverlapping disks on the spherical surface (the so-called “best packing argument” (Habicht and van der Waerden, 1951)) suggests the constant in Eq. (A.2) to be chosen such that",
null,
"The following version of the latter construction with a good choice of the constant is easy to remember: we first generate N points (x, y) belonging to the unit square 0 x, y 1 and then use the cylindrical equal-area projection (i.e., θ = arcsin(2y - 1), ϕ = 2πx) onto the spherical surface. The generation of the initial sequence in the square is also simple: xk = {(k - 1)sfϕ} (here braces denote the non-integer part of the real value) and γk = (k - N - 1). The good value of the constant in Eq. (A.2) corresponds to",
null,
"(the reciprocal of the so-called “golden number”). This spiral set is illustrated in the main text for N = 1000 and N = 3000 (Fig. 1).\n\nLet the surface harmonic functions Pmn (cos θ) cos , Pmn (cos θ) sin be ranked in the usual lexicographic order, let uj be the jth surface harmonic in the corresponding series. Their gradients ui, uj are orthogonal to one another on the set of N points r⃗l, l = 1,… N:",
null,
"((A.3))\n\nThis ensures that computing Gauss coefficients g mn , h mn in the main text through non-weighted averages taken on the points r⃗l of the spiral set (Ql in the main text) is valid. To keep a practical point of view we checked that this orthogonality property is satisfied with a sufficient accuracy for our needs in the present study (see main text).\n\nAs we do not know of any other algorithm as simple to operate as this spiral set one, while producing better results for the problem at hand, we did not extend the analysis to other constructions.\n\n## Rights and permissions\n\nReprints and Permissions\n\nLe Mouël, J.L., Shebalin, P. & Khokhlov, A. Earth magnetic field modeling from Oersted and Champ data. Earth Planet Sp 62, 277–286 (2010). https://doi.org/10.5047/eps.2009.11.003\n\n• Revised:\n\n• Accepted:\n\n• Published:\n\n• Issue Date:\n\n### Key words\n\n• Geomagnetic field model\n• satellite\n• numerical integration over spheres",
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http://algebralab.org/lessons/lesson.aspx?file=Algebra_LinearEqGraphing.xml | [
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"Graphing Linear Functions\nThis lesson will make use of slopes and y-intercepts on a graph. If you need to review these topics, click here for slope (linear equations slope.doc) or click here for y-intercepts. (linear equations intercepts.doc)\n\nLet's start with a line that has a positive slope.\nSuppose we are given the function y = 2x + 5. Because this is in slope-intercept form of a line, we can see that the slope is 2 and the y-intercept is 5.\n\nThis is enough information to graph the function. Because we know the y-intercept is 5, we can start by placing a point at (0, 5) on the coordinate system.",
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"The slope of 2 gives us the additional information we need to complete the graph. Remember that slope is “rise over run” or “the change in y over the change in x”. In any case, we always need to think about slope as a fraction. Since we know the slope is 2, we should think of it in the fraction form of",
null,
". This says that every time we change the y-value by 2, we must also change the x-value by 1. Most of the time this is done by moving up 2 (changing in the y-direction) and over 1 (changing in the x-direction) and then placing the new point. You can see on the graph below that the new point is at (1, 7).",
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"Once you know two points on a graph, you can connect those two points with a line and you then have the graph of the equation.",
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"To find other points on the line, simply choose a starting point, and then move right one space and up two spaces to satisfy the slope of",
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". You should know that you can also move to the left one space and down two spaces and still find another point on the line. Why would this work? Think about this. When you move to the left one you are changing the x-value by -1. When you move down two you are changing the y-value by -2. This will give a slope of",
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"which is what our line indicates. Look at the graph below to verify moving left 1 and down 2 will give you the point (-1, 3) which is on the graph.",
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"Let’s now consider what should happen if we have a negative slope.\nLook at the function",
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". The y-intercept for this graph will be at the point (0, -3). This will be our starting point for finding other points on the graph.\n\nSince the slope is",
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"there are several ways to approach this problem. We know that we will be moving in the y-direction two spaces and in the x-direction three spaces. We just need to figure out whether those moves are left, right, up or down.\n\nWe need to remember that when there is a negative fraction, as there is in this case, it can be written in one of three ways.",
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"• This indicates that if we move down 2, we should move right three because of",
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".\n• It also says that if we move up 2, we should move left 3 because of",
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".\nLet’s find two more points.\n• Using",
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", we start from (0, -3) move right (x-direction) 3 and down (y-direction) 2. This will put a new point at (0+3, -3-2) = (3, -5)\n• Using",
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", we again start from (0, -3) move left (x-direction) 3 and up (y-direction) 2. This will put a new point at (0-3, -3+2) = (-3, -1)\nThe graph of",
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"is shown below. You should now verify these three points on the graph.",
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"Examples\nGraph each line by plotting three points.",
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"What is your answer?",
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"What is your answer?",
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https://calc17.com/what-is-95-percent-in-money | [
"# What is 95 percent in money? How much?\n\n95 percent is just 95*Money/100 of the total amount of money.\n\nMoney\nPercent: %\n\n 20 dollars = 100%. 95% = 19 dollars 100 dollars = 100%. 95% = 95 dollars 3500 dollars = 100%. 95% = 3325 dollars 3500 dollars = 95%. 100% = 3684.21 dollars\nHTML code:\nBB code:"
] | [
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https://www.geteasysolution.com/1.50x+8.45=26.95 | [
"# 1.50x+8.45=26.95\n\n## Simple and best practice solution for 1.50x+8.45=26.95 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.\n\nIf it's not what You are looking for type in the equation solver your own equation and let us solve it.\n\n## Solution for 1.50x+8.45=26.95 equation:\n\n1.50x+8.45=26.95\nWe move all terms to the left:\n1.50x+8.45-(26.95)=0\nWe add all the numbers together, and all the variables\n1.50x-18.5=0\nWe move all terms containing x to the left, all other terms to the right\n1.50x=18.5\nx=18.5/1.50\nx=12+0.5/1.50\n\n`"
] | [
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https://answers.everydaycalculation.com/compare-fractions/4-7-and-1-50 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Compare 4/7 and 1/50\n\n4/7 is greater than 1/50\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 7 and 50 is 350\n2. For the 1st fraction, since 7 × 50 = 350,\n4/7 = 4 × 50/7 × 50 = 200/350\n3. Likewise, for the 2nd fraction, since 50 × 7 = 350,\n1/50 = 1 × 7/50 × 7 = 7/350\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 200/350 > 7/350 or 4/7 > 1/50\n\n#### Compare Fractions Calculator\n\nand\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS\n\n© everydaycalculation.com"
] | [
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https://brainmass.com/chemistry/stoichiometry/density-metal-calculated-23695 | [
"Explore BrainMass\n\n# Density of a Metal\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nA graduated cylinder weighs 46.3 g when empty and holds 35.1 ml water (d = 1.00 g/cm3). An unknown metal is submerged into the water and brings the total volume to 58.1 ml and the mass of the cylinder and contents to 332.3 g. What is the density of the metal?\n\nhttps://brainmass.com/chemistry/stoichiometry/density-metal-calculated-23695\n\n## SOLUTION This solution is FREE courtesy of BrainMass!\n\nPlease see the attached Word document for all calculations and detailed explanation of this exercise.\n\nA graduated cylinder weighs 46.3 g when empty and holds 35.1 ml water (d = 1.00 g/cm3). An unknown metal is submerged into the water and brings the total volume to 58.1 ml and the mass of the cylinder and contents to 332.3 g. What is the density of the metal?\n\nThe density of the metal is the mass of the metal divided by the volume of the metal (from the definition of density).\n\nFirst recall that one milliliter equals one cubic centimeter.\n\n1 ml = 1\n\nThe mass of the water in the cylinder is:\n\nThen it's simple to calculate the mass of the metal since:\n\ng\n\nSimilarly you calculate the volume of the metal:\n\nThen use the equation for density:\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!"
] | [
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http://rsujskf.s602.xrea.com/?atcoder_regular_contest_092_b | [
"# AtCoder Regular Contest 092/AtCoder Beginner Contest 091 D問題 - Two Sequences\n\n## cLay(version 20191111-1)のコード\n\nC++に変換後のコードはこちら\n\nint N, A[2d5], B[2d5];\n\nint a0s, a0[2d5], a1s, a1[2d5];\nint b0s, b0[2d5], b1s, b1[2d5];\n\n{\nint res = 0, all;\nrd(N,A(N),B(N));\nrrep(bt,30){\nall = (1<<(bt+1)) - 1;\na0s = a1s = b0s = b1s = 0;\nrep(i,N){\nif(BIT_ith(A[i],bt)) a1[a1s++] = (A[i]&all) ^ BIT_ith(bt);\nelse a0[a0s++] = (A[i]&all);\nif(BIT_ith(B[i],bt)) b1[b1s++] = (B[i]&all) ^ BIT_ith(bt);\nelse b0[b0s++] = (B[i]&all);\n}\nsortA(a0s, a0); sortA(a1s, a1);\nsortA(b0s, b0); sortA(b1s, b1);\nif(counterSumIsLT(a0s, a0, b1s, b1, BIT_ith(bt)) % 2) res ^= BIT_ith(bt);\nif(counterSumIsLT(a1s, a1, b0s, b0, BIT_ith(bt)) % 2) res ^= BIT_ith(bt);\nif(((ll) a0s * b0s - counterSumIsLT(a0s, a0, b0s, b0, BIT_ith(bt))) % 2) res ^= BIT_ith(bt);\nif(((ll) a1s * b1s - counterSumIsLT(a1s, a1, b1s, b1, BIT_ith(bt))) % 2) res ^= BIT_ith(bt);\n}\nwt(res);\n}\n\n\nCurrent time: 2021年09月28日08時14分39秒"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.545922,"math_prob":0.9998246,"size":1140,"snap":"2021-31-2021-39","text_gpt3_token_len":556,"char_repetition_ratio":0.15140845,"word_repetition_ratio":0.06451613,"special_character_ratio":0.4622807,"punctuation_ratio":0.236,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T23:14:39Z\",\"WARC-Record-ID\":\"<urn:uuid:72677ca7-dfd0-4345-b94a-9f0b852ff0ab>\",\"Content-Length\":\"10832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c35fcf4f-c6a6-4457-a3b1-9b76454908ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:346d9b37-19a4-4018-a3cf-6f89cd8c46f8>\",\"WARC-IP-Address\":\"150.95.9.216\",\"WARC-Target-URI\":\"http://rsujskf.s602.xrea.com/?atcoder_regular_contest_092_b\",\"WARC-Payload-Digest\":\"sha1:V4QVZFDWFYFNHOCDZXUTKKDEXVHXV33X\",\"WARC-Block-Digest\":\"sha1:4TJF5VPNW622SGZ23OBFJTGVJZR575CG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058552.54_warc_CC-MAIN-20210927211955-20210928001955-00566.warc.gz\"}"} |
https://modulocalculator.com/12-over-13-mod-49 | [
"Home » Modulo Division » mod 49 » 12/13 mod 49\n\n# 12/13 mod 49\n\nWelcome to 12/13 mod 49, our post which explains the mathematical operation 12/13 modulus 49.\n\nThis is also known as remainder of 12/13 divided by 49, and if you have been looking for 12/13 modulo 49, then you are right here, too.\n\nRead on to find the 12/13 mod 49 value as well as the math in a nutshell.\n\nReset\n\n## 12/13 Modulo 49\n\n12/13 modulus 49 stands for the Euclidean division discussed, defined and explained in full detail on our home page.\n\nThe result of this modulo operation is:\n\n12/13 mod 49 = 0.(923076)\n\n12/13 is the dividend, 49 is the divisor (modulo), 0 is the quotient explained below, and 0.(923076) is called the remainder.\n\nThe division rest of 12/13 by 49 equals 0.(923076), and the value of the quotient is 0.\n\nProof: 12/13 = (49×0) + 0.(923076).\n\nNote that there is no other quotient q than 0, and that there is no other remainder r than 0.(923076) which solves the equation 12/13 = (49×q) + r and 0 ≤ r < 49; r ∈ set of real numbers R; q ∈ set of whole numbers Z.\n\nNow that you understand what 12/13 mod 49 means, it’s time to zoom in on how this modulo operation is actually calculated.\n\nStep by step, easy and straight to the point.\n\nYou can find the math the next part of this post.\n\n## How is 12/13 mod 49 Calculated?\n\nTo obtain 12/13mod49 conduct these three steps:\n\n1. Integer division (result without fractional part) of dividend by modulus: 12/13 / 49 = 0\n2. Multiplication of the result right above (0) by the divisor (49): 0 × 49 = 0\n3. Subtraction of the result right above (0) from the dividend (12/13): 12/13 – 0 = 0.(923076).\n\nCalculation examples similar to the modulo division 12/13%49, but more detailed, can be found in our article in the header menu.\n\nOther operations belonging to the modulo 49 division category include, for example:\n\nNote that you can locate all of our calculations, including 12/13 modulus 49, quickly, by filling in the search box placed in the header and sidebar; the result page contains all relevant posts.\n\nAhead is the summary of our information.\n\n## Remainder of 12/13 Divided by 49\n\nYou have reached the final section of this post, and you should be able to answer questions like what is 12/13 mod 49?, compute is value, and name its part.\n\nHowever, if you are in doubt about something related to the quotient and remainder of 12/13 by 49, or if you like to leave feedback, then simply use the comment form at the bottom of this article.\n\nAlternatively, send us an email with a meaningful title such as 12/13 modulus 49 division.\n\nEither way you let us know your question, we will get back to you as soon as possible.\n\nIn conclusion:\n\nIf our 12/13 remainder 49 math has been of help to you, hit the sharing buttons and place a bookmark in your browser.\n\nWe recommend to you installing our absolutely free PWA app (see menu or sidebar).\n\nThanks for visiting our post about 12/13 modulo 49.\n\nSubmitting..."
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https://www.colorhexa.com/9fada4 | [
"In a RGB color space, hex #9fada4 is composed of 62.4% red, 67.8% green and 64.3% blue. Whereas in a CMYK color space, it is composed of 8.1% cyan, 0% magenta, 5.2% yellow and 32.2% black. It has a hue angle of 141.4 degrees, a saturation of 7.9% and a lightness of 65.1%. #9fada4 color hex could be obtained by blending #ffffff with #3f5b49. Closest websafe color is: #999999.\n\n• R 62\n• G 68\n• B 64\nRGB color chart\n• C 8\n• M 0\n• Y 5\n• K 32\nCMYK color chart\n\n#9fada4 color description : Dark grayish cyan - lime green.\n\nThe hexadecimal color #9fada4 has RGB values of R:159, G:173, B:164 and CMYK values of C:0.08, M:0, Y:0.05, K:0.32. Its decimal value is 10464676.\n\nHex triplet RGB Decimal 9fada4 `#9fada4` 159, 173, 164 `rgb(159,173,164)` 62.4, 67.8, 64.3 `rgb(62.4%,67.8%,64.3%)` 8, 0, 5, 32 141.4°, 7.9, 65.1 `hsl(141.4,7.9%,65.1%)` 141.4°, 8.1, 67.8 999999 `#999999`\nCIE-LAB 69.425, -6.647, 2.938 35.941, 39.938, 40.935 0.308, 0.342, 39.938 69.425, 7.267, 156.156 69.425, -7.339, 5.394 63.197, -9.078, 5.833 10011111, 10101101, 10100100\n\n``#9fada4` `rgb(159,173,164)``\n``#ad9fa8` `rgb(173,159,168)``\nComplementary Color\n``#a1ad9f` `rgb(161,173,159)``\n``#9fada4` `rgb(159,173,164)``\n``#9fadab` `rgb(159,173,171)``\nAnalogous Color\n``#ad9fa1` `rgb(173,159,161)``\n``#9fada4` `rgb(159,173,164)``\n``#ab9fad` `rgb(171,159,173)``\nSplit Complementary Color\n``#ada49f` `rgb(173,164,159)``\n``#9fada4` `rgb(159,173,164)``\n``#a49fad` `rgb(164,159,173)``\n``#a8ad9f` `rgb(168,173,159)``\n``#9fada4` `rgb(159,173,164)``\n``#a49fad` `rgb(164,159,173)``\n``#ad9fa8` `rgb(173,159,168)``\n• #768a7d\n``#768a7d` `rgb(118,138,125)``\n• #83968a\n``#83968a` `rgb(131,150,138)``\n• #91a197\n``#91a197` `rgb(145,161,151)``\n``#9fada4` `rgb(159,173,164)``\n``#adb9b1` `rgb(173,185,177)``\n• #bbc4be\n``#bbc4be` `rgb(187,196,190)``\n• #c8d0cb\n``#c8d0cb` `rgb(200,208,203)``\nMonochromatic Color\n\nBelow, you can see some colors close to #9fada4. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n``#9fada1` `rgb(159,173,161)``\n``#9fada2` `rgb(159,173,162)``\n``#9fada3` `rgb(159,173,163)``\n``#9fada4` `rgb(159,173,164)``\n``#9fada5` `rgb(159,173,165)``\n``#9fada6` `rgb(159,173,166)``\n``#9fada8` `rgb(159,173,168)``\nSimilar Colors\n\nThis text has a font color of #9fada4.\n\n``<span style=\"color:#9fada4;\">Text here</span>``\n\nThis paragraph has a background color of #9fada4.\n\n``<p style=\"background-color:#9fada4;\">Content here</p>``\n\nThis element has a border color of #9fada4.\n\n``<div style=\"border:1px solid #9fada4;\">Content here</div>``\nCSS codes\n``.text {color:#9fada4;}``\n``.background {background-color:#9fada4;}``\n``.border {border:1px solid #9fada4;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #080a09 is the darkest color, while #fefefe is the lightest one.\n\n• #080a09\n``#080a09` `rgb(8,10,9)``\n• #111412\n``#111412` `rgb(17,20,18)``\n• #1a1f1c\n``#1a1f1c` `rgb(26,31,28)``\n• #232a26\n``#232a26` `rgb(35,42,38)``\n• #2d342f\n``#2d342f` `rgb(45,52,47)``\n• #363f39\n``#363f39` `rgb(54,63,57)``\n• #3f4942\n``#3f4942` `rgb(63,73,66)``\n• #48544c\n``#48544c` `rgb(72,84,76)``\n• #515e56\n``#515e56` `rgb(81,94,86)``\n• #5a695f\n``#5a695f` `rgb(90,105,95)``\n• #637469\n``#637469` `rgb(99,116,105)``\n• #6c7e72\n``#6c7e72` `rgb(108,126,114)``\n• #75897c\n``#75897c` `rgb(117,137,124)``\n• #7f9286\n``#7f9286` `rgb(127,146,134)``\n• #8a9b90\n``#8a9b90` `rgb(138,155,144)``\n• #94a49a\n``#94a49a` `rgb(148,164,154)``\n``#9fada4` `rgb(159,173,164)``\n• #aab6ae\n``#aab6ae` `rgb(170,182,174)``\n• #b4bfb8\n``#b4bfb8` `rgb(180,191,184)``\n• #bfc8c2\n``#bfc8c2` `rgb(191,200,194)``\n• #c9d1cc\n``#c9d1cc` `rgb(201,209,204)``\n``#d4dad6` `rgb(212,218,214)``\n• #dee3e0\n``#dee3e0` `rgb(222,227,224)``\n• #e9ecea\n``#e9ecea` `rgb(233,236,234)``\n• #f4f5f4\n``#f4f5f4` `rgb(244,245,244)``\n• #fefefe\n``#fefefe` `rgb(254,254,254)``\nTint Color Variation\n\nA tone is produced by adding gray to any pure hue. In this case, #a6a6a6 is the less saturated color, while #54f88e is the most saturated one.\n\n• #a6a6a6\n``#a6a6a6` `rgb(166,166,166)``\n``#9fada4` `rgb(159,173,164)``\n• #98b4a2\n``#98b4a2` `rgb(152,180,162)``\n• #91bba0\n``#91bba0` `rgb(145,187,160)``\n• #8ac29e\n``#8ac29e` `rgb(138,194,158)``\n• #84c89c\n``#84c89c` `rgb(132,200,156)``\n• #7dcf9a\n``#7dcf9a` `rgb(125,207,154)``\n• #76d698\n``#76d698` `rgb(118,214,152)``\n• #6fdd96\n``#6fdd96` `rgb(111,221,150)``\n• #68e494\n``#68e494` `rgb(104,228,148)``\n• #61eb92\n``#61eb92` `rgb(97,235,146)``\n• #5bf190\n``#5bf190` `rgb(91,241,144)``\n• #54f88e\n``#54f88e` `rgb(84,248,142)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #9fada4 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://face2ai.com/math-statistics-2-2-de-moivre-naive-result/ | [
"# 狄莫弗的初步结果\n\n$$b(m)\\sim 2.168\\frac{(1-\\frac{1}{N})^N}{\\sqrt{N-1}}\\tag{5}$$\n$$\\text{log}(\\frac{b(m)}{b(m+d)})\\sim (m+d-\\frac{1}{2})\\text{log}(m+d-1)\\\\ +(m-d+\\frac{1}{2})\\text{log}(m-d+1)-2m\\text{log}m+\\text{log}(\\frac{m+d}{m})\\tag{6}$$\n\n(5)的结论现在我们已经在书上见不到了,因为我们有更好的近似结论了,而且(5)对狄莫弗后续研究也没啥作用,\n\n## 总结\n\n### 说点什么",
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"Subscribe"
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"https://secure.gravatar.com/avatar/",
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http://talkstats.com/threads/constant-variable.13580/ | [
"# Constant Variable\n\n#### blastStu\n\n##### New Member\nHi folks,\nI was wondering if you could help me understand a problem I have with a regression. Basically, SPSS excludes one of my variables because it claims that the variable is constant, but I just don't believe it!\n\nI've got four variables that I'm concerned with, all are binary:\n\nisSpeaking: 0 = No, 1 = Yes\nisNodding: 0 = No, 1 = Yes\nTaskRole: 0 = Instructor, 1 = Learner\nRecipientRole: 0 = Secondary, 1 = Primary.\n\nEach data point is a time point per person. isSpeaking and RecipientRole are exclusive, by that I mean if isSpeaking is 1, then there will be no data for RecipientRole (as a person can't be a speaker and a recipient at the same time here). If isSpeaking is 0, then there should be a value for RecipientRole. So a person may be speaking, nodding, a learner and no entry for recipient role.\n\nWhen I perform a correlation analysis there is an interaction between isNodding and isSpeaking (and lots of other interactions):",
null,
"isSpeaking is constant w.r.t. RecipientRole as expected.\n\nHowever, when I try to predict isNodding (in a binary logistic regression), the isSpeaking variable is excluded all together (see point b):",
null,
"However, it's just not constant for the selected cases! It's constant w.r.t to RecipientRole in that whenever there is a value in the RecipientRole variable there is a 0 in isSpeaking.. but surely this just means there's no interaction between them?\n\nAm I missing something with how a regression works here?\n\nThanks,\n\nRegards\nStuart\n\n#### blastStu\n\n##### New Member\nOK. I have part of the answer now. I didn't realise that if there is a missing value for a variable then SPSS excludes that entire case. In my situation the isSpeaking and RecipientRoles variables are such that, if isSpeaking is 0, then RecipientRole will be 1 or 0. If isSpeaking is 1, then there will be no value for Recipient Role.\n\nAs this is the case, SPSS is excluding all cases where isSpeaking = 1, and hence isSpeaking is always a constant 0.\n\nIs there a way to force SPSS to analyse all the data, even when there's a missing value in a datapoint?\n\nThanks,\nS\n\n#### Mean Joe\n\n##### TS Contributor\nI'm not much help since I don't know SPSS.\n\nBut is there a command to tell SPSS to analyze subjects \"where isSpeaking=0\"? Or can you make a subset, where isSpeaking=0, and then analyze that subset? Or just drop the isSpeaking variable from your analysis?"
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"http://www.eecs.qmul.ac.uk/~stuart/moz-screenshot-15.png",
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https://number.academy/112 | [
"# Number 112\n\nNumber 112 spell 🔊, write in words: one hundred and twelve . Ordinal number 112th is said 🔊 and write: one hundred and twelfth. The meaning of number 112 in Maths: Is Prime? Factorization and dividers. The square root and cube root of 112. The meaning in computer science, numerology, codes and images, writing and naming in other languages, other interesting facts.\n\n## Useful information about number 112\n\nThe decimal (Arabic) number 112 converted to a Roman number is CXII. Roman and decimal number conversions.\n The number 112 converted to a Mayan number is",
null,
"Decimal and Mayan number conversions.\n\n#### Weight conversion\n\n112 kilograms (kg) = 246.9 pounds (lb)\n112 pounds (lb) = 50.8 kilograms (kg)\n\n#### Length conversion\n\n112 kilometers (km) equals to 69.594 miles (mi).\n112 miles (mi) equals to 180.247 kilometers (km).\n112 meters (m) equals to 180.247 feet (ft).\n112 feet (ft) equals 34.138 meters (m).\n112 centimeters (cm) equals to 44.1 inches (in).\n112 inches (in) equals to 284.5 centimeters (cm).\n\n#### Temperature conversion\n\n112° Fahrenheit (°F) equals to 44.4° Celsius (°C)\n112° Celsius (°C) equals to 233.6° Fahrenheit (°F)\n\n#### Power conversion\n\n112 Horsepower (hp) equals to 82.36 kilowatts (kW)\n112 kilowatts (kW) equals to 152.30 horsepower (hp)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n112 seconds equals to 1 minute, 52 seconds\n112 minutes equals to 1 hour, 52 minutes\n\n### Codes and images of the number 112\n\nNumber 112 morse code: .---- .---- ..---\nSign language for number 112:",
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"",
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"",
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"Images of the number Image (1) of the number Image (2) of the number",
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"There is no copyright for these images. Number 112 infographic. More images, other sizes, codes and colors ...\n\n### Gregorian, Hebrew, Islamic, Persian and Buddhist year (calendar)\n\nGregorian year 112 is Buddhist year 655.\nBuddhist year 112 is Gregorian year 431 a. C.\nGregorian year 112 is Islamic year -526 or -525.\nIslamic year 112 is Gregorian year 730 or 731.\nGregorian year 112 is Persian year -511 or -510.\nPersian year 112 is Gregorian 733 or 734.\nGregorian year 112 is Hebrew year 3872 or 3873.\nHebrew year 112 is Gregorian year 3648 a. C.\nThe Buddhist calendar is used in Sri Lanka, Cambodia, Laos, Thailand, and Burma. The Persian calendar is official in Iran and Afghanistan.\n\n## Share in social networks",
null,
"## Mathematics of no. 112\n\n### Multiplications\n\n#### Multiplication table of 112\n\n112 multiplied by two equals 224 (112 x 2 = 224).\n112 multiplied by three equals 336 (112 x 3 = 336).\n112 multiplied by four equals 448 (112 x 4 = 448).\n112 multiplied by five equals 560 (112 x 5 = 560).\n112 multiplied by six equals 672 (112 x 6 = 672).\n112 multiplied by seven equals 784 (112 x 7 = 784).\n112 multiplied by eight equals 896 (112 x 8 = 896).\n112 multiplied by nine equals 1008 (112 x 9 = 1008).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 112\n\nHalf of 112 is 56 (112 / 2 = 56).\nOne third of 112 is 37,3333 (112 / 3 = 37,3333 = 37 1/3).\nOne quarter of 112 is 28 (112 / 4 = 28).\nOne fifth of 112 is 22,4 (112 / 5 = 22,4 = 22 2/5).\nOne sixth of 112 is 18,6667 (112 / 6 = 18,6667 = 18 2/3).\nOne seventh of 112 is 16 (112 / 7 = 16).\nOne eighth of 112 is 14 (112 / 8 = 14).\nOne ninth of 112 is 12,4444 (112 / 9 = 12,4444 = 12 4/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 112\n\n#### Is Prime?\n\nThe number 112 is not a prime number. The closest prime numbers are 109, 113.\n112th prime number in order is 613.\n\n#### Factorization and dividers\n\nFactorization 2 * 2 * 2 * 2 * 7 = 112\nDivisors of number 112 are 1 , 2 , 4 , 7 , 8 , 14 , 16 , 28 , 56 , 112\nTotal Divisors 10.\nSum of Divisors 248.\n\n#### Powers\n\nThe second power of 1122 is 12.544.\nThe third power of 1123 is 1.404.928.\n\n#### Roots\n\nThe square root √112 is 10,583005.\nThe cube root of 3112 is 4,820285.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 112 = loge 112 = 4,718499.\nThe logarithm to base 10 of No. log10 112 = 2,049218.\nThe Napierian logarithm of No. log1/e 112 = -4,718499.\n\n### Trigonometric functions\n\nThe cosine of 112 is 0,455969.\nThe sine of 112 is -0,889996.\nThe tangent of 112 is -1,951877.\n\n### Properties of the number 112\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 112 in Computer Science\n\nCode typeCode value\nAscii112 ascii is character p\nUnix timeUnix time 112 is equal to Thursday Jan. 1, 1970, 12:01:52 a.m. GMT\nIPv4, IPv6Number 112 internet address in dotted format v4 0.0.0.112, v6 ::70\n112 Decimal = 1110000 Binary\n112 Decimal = 11011 Ternary\n112 Decimal = 160 Octal\n112 Decimal = 70 Hexadecimal (0x70 hex)\n112 BASE64MTEy\n112 MD57f6ffaa6bb0b408017b62254211691b5\n112 SHA1601ca99d55f00a2e8e736676b606a4d31d374fdd\n112 SHA224a4067cf5e783771c4a958ecc76f0a6976bc666e899e4c8cb40c6f7b0\n112 SHA384a701b50984e65e1d718b8fb937889d9e7586656155120c3c4bb4f0b119220bf6c421957c9e89afba502227cafb78440d\nMore SHA codes related to the number 112 ...\n\nIf you know something interesting about the 112 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 112\n\n### The meaning of the number 1 (one), numerology 1\n\nCharacter frequency 1: 2\n\nNumber one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.\nMore about the meaning of the number 1 (one), numerology 1 ...\n\n### The meaning of the number 2 (two), numerology 2\n\nCharacter frequency 2: 1\n\nThe number two (2) needs above all to feel and to be. It represents the couple, duality, family, private and social life. He/she really enjoys home life and family gatherings. The number 2 denotes a sociable, hospitable, friendly, caring and affectionate person. It is the sign of empathy, cooperation, adaptability, consideration for others, supersensitivity towards the needs of others. The number 2 (two) is also the symbol of balance, togetherness and receptivity. He/she is a good partner, colleague or companion; he/she also plays a wonderful role as a referee or mediator. Number 2 person is modest, sincere, spiritually influenced and a good diplomat. It represents intuition and vulnerability.\nMore about the meaning of the number 2 (two), numerology 2 ...\n\n## Interesting facts about the number 112\n\n### Asteroids\n\n• (112) Iphigenia is asteroid number 112. It was discovered by C. H. F. Peters from Litchfield Observatory, Clinton on 9/19/1870.\n\n### Aircrafts and flights\n\n• 05/05/1972 the Alitalia plane, flight number 112, McDonnell Douglas DC-8-43 crashed in Cinisi, Sicily, Italy 108 passengers and 7 crew members died.\n\n### Areas, mountains and surfaces\n\n• The total area of Bolshoy Lyakhovsky Island is 1,991 square miles (5,157 square km). Country Russia (Sakha Republic). 112th largest island in the world.\n\n### Distances between cities\n\n• There is a 112 miles (180 km) direct distance between Āgra (India) and Delhi (India).\n• There is a 112 miles (180 km) direct distance between Al Jīzah (Egypt) and Alexandria (Egypt).\n• There is a 112 miles (179 km) direct distance between Allahābād (India) and Lucknow (India).\n• There is a 70 miles (112 km) direct distance between Anshan (China) and Fushun (China).\n• There is a 112 miles (179 km) direct distance between Bareilly (India) and Meerut (India).\n• There is a 70 miles (112 km) direct distance between Dombivli (India) and Pune (India).\n• There is a 70 miles (112 km) direct distance between Foshan (China) and Yunfu (China).\n• There is a 112 miles (179 km) direct distance between Gorakhpur (India) and Chandigarh (India).\n• There is a 112 miles (179 km) direct distance between Goyang-si (South Korea) and Pyongyang (North Korea).\n• There is a 112 miles (179 km) direct distance between Los Angeles (USA) and San Diego (USA).\n• There is a 112 miles (180 km) direct distance between Mendoza (Argentina) and Santiago (Chile).\n• There is a 112 miles (180 km) direct distance between Mysore (India) and Salem (India).\n• There is a 112 miles (179 km) direct distance between Teni (India) and Thiruvananthapuram (India).\n\n### Games\n\n• Pokémon Rhydon (Saidon, Sidon) is a Pokémon number # 112 of the National Pokédex. Rhydon is ground and rock type Pokémon in the first generation. It is a Monster and Field egg group Pokémon. The other Rhydon indexes are Johto index 207 , Hoenn index 170 , Sinnoh index 187 , Kalos Centro index 051 - Coastal Zone Pokédex.\n\n### In other fields\n\n• 112 is the emergency telephone number in the European Union. Calls to this number are free.\n\n### Mathematics\n\n• 112 is the side of the smallest square that can be tiled with distinct integer-sided squares.\n\n### Science\n\n• Copernicium is the chemical element in the periodic table that has the symbol Cn and atomic number 112.\n\n## Number 112 in other languages\n\nHow to say or write the number one hundred and twelve in Spanish, German, French and other languages.\n Spanish: 🔊 (número 112) ciento doce German: 🔊 (Anzahl 112) einhundertzwölf French: 🔊 (nombre 112) cent douze Portuguese: 🔊 (número 112) cento e doze Chinese: 🔊 (数 112) 一百一十二 Arabian: 🔊 (عدد 112) مائة و اثنا عشر Czech: 🔊 (číslo 112) sto dvanáct Korean: 🔊 (번호 112) 백십이 Danish: 🔊 (nummer 112) ethundrede og tolv Hebrew: (מספר 112) מאה ושנים עשרה Dutch: 🔊 (nummer 112) honderdtwaalf Japanese: 🔊 (数 112) 百十二 Indonesian: 🔊 (jumlah 112) seratus dua belas Italian: 🔊 (numero 112) centododici Norwegian: 🔊 (nummer 112) en hundre og tolv Polish: 🔊 (liczba 112) sto dwanaście Russian: 🔊 (номер 112) сто двенадцать Turkish: 🔊 (numara 112) yüzoniki Thai: 🔊 (จำนวน 112) หนึ่งร้อยสิบสอง Ukrainian: 🔊 (номер 112) сто дванадцять Vietnamese: 🔊 (con số 112) một trăm mười hai Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 112 or any natural number (positive integer) please write us here or on facebook."
] | [
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"https://numero.wiki/s/numeros-mayas/numero-maya-112.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png",
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"https://numero.wiki/s/senas/lenguaje-de-senas-numero-2.png",
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"https://numero.wiki/img/a-112.jpg",
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https://agrat.cat/multiplication-and-division-word-problems-4th-grade-pdf | [
"",
null,
"",
null,
"Multiplication And Division Word Problems 4Th Grade Pdf. Fastt math is proven effective for struggling students. If the nurses are equally assigned to the doctors, how many nurses are assigned to most.\n\nThese math word problems may require multiplication or division to solve. The following collection of free 4th grade maths word problems worksheets cover topics including addition, subtraction, multiplication, division, mixed operations, fractions, and decimals. Each nurse is assigned to assist one doctor.\n\n### We Have 100 Pictures About Multiplication Word Problem Worksheets 3Rd Grade Like Multiplication Word Problem Worksheets 3Rd Grade, Multiplication Word Problems 2 Interactive Worksheet And Also 428 Addition Worksheets For You To Print Right Now.\n\nUse multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem. Multiplying numbers in columns is a math skill which requires lots of practice. Kids use details from the word problems on this third grade math worksheet to construct and solve multiplication problems.\n\n### The Following Collection Of Free 4Th Grade Maths Word Problems Worksheets Cover Topics Including Addition, Subtraction, Multiplication, Division, Mixed Operations, Fractions, And Decimals.\n\nThese worksheets, sorted by grade level, cover a mix of skills from the curriculum. Math worksheet based on mixed operations division and Each pizza has 4 slices."
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https://wpcalc.com/en/cost-of-goods-sold/ | [
"# Cost of Goods Sold\n\nCost of goods sold (COGS) denotes the carrying value of goods and raw materials sold to customers during specific period.\n\n### Calculator of Cost of Goods Sold\n\n Select Currency = Beginning inventory = Purchases = Ending inventory = COGS =\n\n### Formula of Cost of Goods Sold\n\nCost of goods sold = Beginning inventory + Purchases – Ending inventory\n\n### Example of Cost of Goods Sold\n\nBeginning inventory of a company was \\$16000 and the company purchased new inventory for the cost of \\$5000. Ending inventory of the company was \\$10000.Calculate the cost of goods sold in a year.\n\n#### Given:\n\nBeginning inventory = \\$16000\nPurchases = \\$5000\nEnding inventory = \\$10000\n\n### To Find Cost of goods sold\n\n#### Solution:\n\nCOGS = Beginning inventory + Purchases – Ending inventory\n= 16000 + 5000 – 10000\n= 21000 – 10000\n= \\$11000",
null,
""
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"https://secure.gravatar.com/avatar/",
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https://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.interpolate.KroghInterpolator.html | [
"# scipy.interpolate.KroghInterpolator¶\n\nclass scipy.interpolate.KroghInterpolator(xi, yi, axis=0)[source]\n\nInterpolating polynomial for a set of points.\n\nThe polynomial passes through all the pairs (xi,yi). One may additionally specify a number of derivatives at each point xi; this is done by repeating the value xi and specifying the derivatives as successive yi values.\n\nAllows evaluation of the polynomial and all its derivatives. For reasons of numerical stability, this function does not compute the coefficients of the polynomial, although they can be obtained by evaluating all the derivatives.\n\nParameters: xi : array_like, length N Known x-coordinates. Must be sorted in increasing order. yi : array_like Known y-coordinates. When an xi occurs two or more times in a row, the corresponding yi’s represent derivative values. axis : int, optional Axis in the yi array corresponding to the x-coordinate values.\n\nNotes\n\nBe aware that the algorithms implemented here are not necessarily the most numerically stable known. Moreover, even in a world of exact computation, unless the x coordinates are chosen very carefully - Chebyshev zeros (e.g. cos(i*pi/n)) are a good choice - polynomial interpolation itself is a very ill-conditioned process due to the Runge phenomenon. In general, even with well-chosen x values, degrees higher than about thirty cause problems with numerical instability in this code.\n\nBased on [R44].\n\nReferences\n\n [R44] (1, 2) Krogh, “Efficient Algorithms for Polynomial Interpolation and Numerical Differentiation”, 1970.\n\nExamples\n\nTo produce a polynomial that is zero at 0 and 1 and has derivative 2 at 0, call\n\n>>> KroghInterpolator([0,0,1],[0,2,0])\n\n\nThis constructs the quadratic 2*X**2-2*X. The derivative condition is indicated by the repeated zero in the xi array; the corresponding yi values are 0, the function value, and 2, the derivative value.\n\nFor another example, given xi, yi, and a derivative ypi for each point, appropriate arrays can be constructed as:\n\n>>> xi_k, yi_k = np.repeat(xi, 2), np.ravel(np.dstack((yi,ypi)))\n>>> KroghInterpolator(xi_k, yi_k)\n\n\nTo produce a vector-valued polynomial, supply a higher-dimensional array for yi:\n\n>>> KroghInterpolator([0,1],[[2,3],[4,5]])\n\n\nThis constructs a linear polynomial giving (2,3) at 0 and (4,5) at 1.\n\nMethods\n\n __call__(x) Evaluate the interpolant :Parameters: x : array_like Points to evaluate the interpolant at. derivative(x[, der]) Evaluate one derivative of the polynomial at the point x :Parameters: x : array_like Point or points at which to evaluate the derivatives der : integer, optional Which derivative to extract. derivatives(x[, der]) Evaluate many derivatives of the polynomial at the point x Produce an array of all derivative values at the point x.\n\n#### Previous topic\n\nscipy.interpolate.BarycentricInterpolator.set_yi\n\n#### Next topic\n\nscipy.interpolate.KroghInterpolator.__call__"
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https://www.shuanghei.com/article/2022/09/6311f5a3e8a1e00a1c25525d.html | [
" mongodb中使用lookup联合查询及分组group排序sort_双黑个人网站\n\n# mongodb中使用lookup联合查询及分组group排序sort\n\n0 0条评论\n\npaper表\n\n``````const schema = baseSchema().add({\n/**\n* 会员ID\n*/\nuser: {\ntype: mongoose.Schema.Types.ObjectId,\nref: 'user',\nrequired: true\n},\n/**\n* 得分\n*/\nscore: { type: Number, required: true, index: true },\n/**\n* 所用时间,单位秒\n*/\ntime: { type: Number, required: true, index: true },\n/**\n* 答题时间\n*/\ncreatetime: { type: Number }\n})``````\n\nuser表\n\n(两张表实际项目中列肯定不止那么少,我只是找几个重点罗列)\n\n`````` Model.aggregate().lookup({\nfrom: 'users', localField: 'user',\nforeignField: '_id', as: 'users'\n})\n.sort('-score time createtime')\n.group({\n_id: '\\$user', score: { \\$first: '\\$score' }, time: { \\$first: '\\$time' },\ncreatetime: { \\$first: '\\$createtime' }, users: { \\$first: '\\$users' }\n})\n.project({\n_id: 0, score: 1, time: 1, createtime: 1,\nnickname: { \\$min: '\\$users.nickname' }\n})\n.sort('-score time createtime').limit(10)``````"
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https://stats.stackexchange.com/questions/409398/show-statistical-significance-or-not-between-means | [
"# Show statistical significance (or not) between means\n\nSuppose I have two experiments A and B. Each experiment is composed of N elements that are clearly separated into two groups. In experiment B one of the groups has a higher mean than the same group in experiment A. This difference (shown in red in the image) is what I am trying to understand. I need to show that it is significant, or not, determined by any value. p-value? a test?",
null,
"To show statistical significance (that is, to show that the results of your experiments are not just due to chance), you first need to define your hypotheses (the null and the alternative hypothesis). In this case, the null hypothesis, $$H_0$$, could be \"There is no difference between the mean of the groups in experiments $$A$$ and $$B$$\". The alternative hypothesis, $$H_a$$, could be \"There is a difference\".\nYou now perform a statistical test. If the resulting p-value (returned by the statistical test) is less than a threshold value (often called the \"significance level\" and it is e.g. $$0.05$$), then you reject the null hypothesis and you say that this result (that is, the p-value less than the significance level) is statistical significant."
] | [
null,
"https://i.stack.imgur.com/V8b1j.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93197656,"math_prob":0.9693672,"size":1697,"snap":"2019-51-2020-05","text_gpt3_token_len":386,"char_repetition_ratio":0.12758417,"word_repetition_ratio":0.0,"special_character_ratio":0.23217443,"punctuation_ratio":0.109144546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985309,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T22:40:53Z\",\"WARC-Record-ID\":\"<urn:uuid:19240516-7f1f-42ae-8322-46f3468f7765>\",\"Content-Length\":\"140682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d33cd74-ff1f-4e3d-bca0-701907dce547>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2c1277a-832f-49d4-bc03-bd3bc372c397>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/409398/show-statistical-significance-or-not-between-means\",\"WARC-Payload-Digest\":\"sha1:TNM2EG6VHJ4EFCRUVQ4Y4UXFESVSTYMO\",\"WARC-Block-Digest\":\"sha1:D2GF2OAOYGQYVGRJL4775EAKPHQOUTCZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250591234.15_warc_CC-MAIN-20200117205732-20200117233732-00549.warc.gz\"}"} |
https://machinelearninginterview.com/topics/machine-learning/maximum-likelihood-estimate/ | [
"# What is the Maximum Likelihood Estimate (MLE)?\n\nProbabilistic Models help us capture the inherant uncertainity in real life situations. Examples of probabilistic models are Logistic Regression, Naive Bayes Classifier and so on.. Typically we fit (find parameters) of such probabilistic models from the training data, and estimate the parameters. The learnt model can then be used on unseen data to make predictions.\n\nOne way to find the parameters of a probabilistic model (learn the model) is to use the MLE estimate or the maximum likelihood estimate.\n\nMaximum likelihood estimate is that value for the parameters that maximizes the likelihood of the data.\n\nWhat are some examples of the parameters of models we want to find? What exactly is the likelihood? How do we find parameters that maximize the likelihood? These are some questions answered by the video.",
null,
"##### What are some examples of parameters we might want to estimate?\n\nConsider the Gaussian distribution. The mathematical form of the pdf is shown below. The parameters of the Gaussian distribution are the mean and the variance (or the standard deviation). Given a set of points, the MLE estimate can be used to estimate the parameters of the Gaussian distribution.",
null,
"",
null,
"Consider the Bernoulli distribution. A good example to relate to the Bernoulli distribution is modeling the probability of heads (p) when we toss a coin. The probability of heads is p, the probability of tails is (1-p). By observing a bunch of coin tosses, one can use the maximum likelihood estimate to find the value of p.\n\n##### What is the likelihood of a probabilistic model?\n\nThe likelihood is the joined probability distribution of the observed data given the parameters.\n\nFor instance, if we consider the Bernoulli distribution for a coin toss with probability of heads as p. Suppose we toss the coin four times, and get H, T, T, H.\n\nThe likelihood of the observed data is the joint probability distribution of the observed data. Hence:",
null,
"##### How do we find the parameter that maximizes likelihood?\n\nThe MLE estimator is that value of the parameter which maximizes likelihood of the data. This is an optimization problem.",
null,
"For instance for the coin toss example, the MLE estimate would be to find that p such that p (1-p) (1-p) p is maximized.",
null,
""
] | [
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/MLE_Estimate_Thumbnail.png",
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/normal_dist_formula-e1616395155211-300x172.jpg",
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/normal-distribution-probability-300x108.jpg",
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/bernoulli_distribution.png",
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/picture_mle.jpeg",
null,
"https://machinelearninginterview.com/wp-content/uploads/2021/03/finding_MLE_coin_toss.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8478182,"math_prob":0.9874127,"size":2777,"snap":"2022-40-2023-06","text_gpt3_token_len":562,"char_repetition_ratio":0.18391633,"word_repetition_ratio":0.012958963,"special_character_ratio":0.19409434,"punctuation_ratio":0.09356725,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99941194,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T05:47:54Z\",\"WARC-Record-ID\":\"<urn:uuid:c7f2511b-d449-4314-8f93-2ba712c83624>\",\"Content-Length\":\"53887\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0893c553-eb17-486e-a685-6a6deaa4b935>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e09e9fd-b395-4389-be17-940413bf9641>\",\"WARC-IP-Address\":\"63.250.43.9\",\"WARC-Target-URI\":\"https://machinelearninginterview.com/topics/machine-learning/maximum-likelihood-estimate/\",\"WARC-Payload-Digest\":\"sha1:ARFBQK4EVULKNM5AD7KNDJCOURYTMH4N\",\"WARC-Block-Digest\":\"sha1:77RYOOEPZATNKN5F4DTIIJKGT4BYNW74\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500094.26_warc_CC-MAIN-20230204044030-20230204074030-00359.warc.gz\"}"} |
https://www.instructables.com/Ben-A-Light-Following-Breadboard-Arduino-Robot/ | [
"# Ben - a Light Following Breadboard Arduino Robot\n\n11,269\n\n29\n\n15\n\n## Introduction: Ben - a Light Following Breadboard Arduino Robot\n\nBen the Light Following Breadboard Arduino Robot is the second robot I have made to help teach robotics to high school students in a class I teach voluntarily. The first robot also has its own Instructable which can be found here: https://www.instructables.com/id/James-Your-first-Arduino-Robot/\nThe video shows the construction process however it is in fast motion and I will cover some of the more complicated things here more in depth.\n\nSo to make ben you will need the following components:\n· A small sheet of acrylic\n· An Arduino Nano\n· Two Continuous Rotation Servos\n· Two hobby wheels (I used model aeroplane wheels)\n· A castor wheel\n· 9V battery\n· 4.8V rechargeable battery pack (or just 4 AA’s in a battery case)\n· Two Light Dependant Resistors\n· Two 10,000 ohm Resistors\n· Double Sided Foam Tape\n· A Power Switch (not necessary but would be handy)\n\nYou won’t need any tools for the construction however you could replace the Double Sided Foam Tape with Hot Glue in which case you would need a Hot Glue Gun.\n\nNow the first thing that may need further explanation is the use of the Light Dependant Resistors. Light Dependant Resistors (or LDR’s) are resistors whose value changes depending on the amount of ambient light, but how can we detect resistance with Arduino? Well you can’t really, however you can detect voltage levels using the analog pins, which can measure (in basic use) between 0-5V. Now you may be asking “Well how do we convert resistance values into voltage changes?”, it’s simple, we make a voltage divider. A voltage divider takes in a voltage and then outputs a fraction of that voltage proportional to the input voltage and the ratio of the two values of resistors used. The equation for which is:\nOutput Voltage = Input Voltage * ( R2 / (R1 + R2) )\nWhere R1 is the value of the first resistor and R2 is the value of the second.\n\nThe circuit schematic for which looks like this\nA diagram of this in our situation looks a little something like this\n\nNow this still begs the question “But what resistance values does the LDR have?”, good question. The less amount of ambient light the higher the resistance, more ambient light means a lower resistance. Now for the particular LDR’s I used their resistance range was from 200 – 10 kilo ohms, but this changes for different ones so make sure to look up where you bought them from and try to find a datasheet or something of the sort.\n\nNow in this case R1 is actually our LDR, so let’s bring back that equation and do some math-e-magic (mathematical electrical magic).\n\nNow first we need to convert those kilo ohm values into ohms:\n200 kilo-ohms = 200,000 ohms\n10 kilo-ohms = 10,000 ohms\nSo to find what the output voltage is when we are in pitch black we plug in the following numbers:\n5 * ( 10000 / (200000 + 10000) )\nThe input is 5V as that is what we are getting from the Arduino.\nThe above gives 0.24V (rounded off).\nNow we find what the output voltage is in peak brightness by using the following numbers:\n5 * ( 10000 / (10000 + 10000) )\nAnd this gives us 2.5V exactly.\n\nSo these are the voltage values that we are going to get into the Arduino’s analog pins, but these are not the values that will be seen in the program, “But why?” you may ask. The Arduino uses an Analog to Digital Chip which converts the analog voltage into usable digital data. Unlike the digital pins on the Arduino that can only read a HIGH or LOW state being 0 and 5V the analog pins can read from 0-5V and convert this into a number range of 0-1023.\n\nNow with some more math-e-magic we can actually calculate what values the Arduino will actually read. Because this will be a linear function we can use the following formula:\nY = mX + C\nWhere; Y = Digital Value\nWhere; m = slope, (rise / run), (digital value / analog value)\nWhere; C = Y intercept\nThe Y intercept is 0 so that gives us:\nY = mX\nm = 1023 / 5 = 204.6\nTherefore:\nDigital value = 204.6 * Analog value\n\nSo in pitch black the digital value will be:\n204.6 * 0.24\nWhich gives approximately 49.\n\nAnd in peak brightness it will be:\n204.6 * 2.5\nWhich gives approximately 511.\n\nNow with two of these set up on two analog pins we can create two integer variables to store their values two and do comparison operators to see which one has the lowest value, turning the robot in that direction.\n____________________________________________________________________________________________________\n\nNow that was probably the most complex thing about the whole robot build however there is just one more thing that I would like to mention and it’s to do with using servos with Arduino.\n\nThere are several tutorials and diagrams on the internet showing that you must connect the voltage in of the servo up to the 5V rail of the Arduino and the ground of the servo to the ground of the Arduino, this is dangerous! Servos can draw a lot of current, and in most cases that current draw will be more than the voltage regulator on the Arduino can supply, this will lead to bad things happening. The proper way to hook servos to your Arduino is to use an external power supply. In Bens cause I am running the continuous rotation servos of a 4.8V rechargeable Ni-Cd battery pack, this is ideal as the servos operate well from 4.8-6V, 6V being the peak charge voltage of the battery pack.\n\nNow you may be tempted to just hook up V+ of the battery to the V+ of the servos and the GND of the battery pack to the GND of servos and the signal pins to the Arduino, this won’t work either! You need to remember that electricity needs to flow from one ‘point’ back to its original point, not connecting the ground of the servos and battery pack to the Arduino’s ground won’t allow the electricity to flow from the signal pins.\n\nHere is a diagram showing the proper circuitry\n_____________________________________________________________________________________________________\n\nCombining the two diagrams shown earlier gives the complete circuitry required to make Ben.\nNow I won’t explain the code as it is heavily commented and should pretty much explain itself.\nCode",
null,
"Participated in the\nToy Contest",
null,
"Participated in the\nKit Contest",
null,
"Participated in the\nArduino Contest\n\n## Recommendations\n\nArduino: 1.6.7 (Windows 10), Board: \"Arduino Nano, ATmega168\"\n\nBuild options changed, rebuilding all\nMultiple libraries were found for \"Servo.h\"\nUsed: C:\\Users\\Shane\\Documents\\Arduino\\libraries\\Servo\nNot used: D:\\Shane\\Arduino\\libraries\\Servo\n\nSketch uses 2,552 bytes (17%) of program storage space. Maximum is 14,336 bytes.\nGlobal variables use 65 bytes (6%) of dynamic memory, leaving 959 bytes for local variables. Maximum is 1,024 bytes.\navrdude: ser_open(): can't open device \"\\\\.\\COM1\": The system cannot find the file specified."
] | [
null,
"https://www.instructables.com/assets/img/pixel.png",
null,
"https://www.instructables.com/assets/img/pixel.png",
null,
"https://www.instructables.com/assets/img/pixel.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8911713,"math_prob":0.9168142,"size":8842,"snap":"2022-05-2022-21","text_gpt3_token_len":2102,"char_repetition_ratio":0.14381082,"word_repetition_ratio":0.023316063,"special_character_ratio":0.25209227,"punctuation_ratio":0.08454811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98989534,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T23:32:15Z\",\"WARC-Record-ID\":\"<urn:uuid:aed8cbf6-8171-4ba2-8bc3-e7e3df1c5127>\",\"Content-Length\":\"94014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea049479-e1b5-4cd7-ba38-da3927bd8918>\",\"WARC-Concurrent-To\":\"<urn:uuid:acbb57b9-358a-4c3d-a9b3-f836cb2ceec5>\",\"WARC-IP-Address\":\"146.75.33.105\",\"WARC-Target-URI\":\"https://www.instructables.com/Ben-A-Light-Following-Breadboard-Arduino-Robot/\",\"WARC-Payload-Digest\":\"sha1:VXQ4LELMWWRK6GWAYXS3SGTW7M3KOFX7\",\"WARC-Block-Digest\":\"sha1:5XB27YQBCYOA27AG2UFMC4LCEYO44R65\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662541747.38_warc_CC-MAIN-20220521205757-20220521235757-00228.warc.gz\"}"} |
http://uribo.github.io/rpkg_showcase/dataset/wakefield.html | [
"# wakefield: Generate Random Data Sets",
null,
"``````> library(wakefield)\n``````\n``````\nAttaching package: 'wakefield'\n\nThe following object is masked from 'package:raster':\n\narea\n``````\n\nバージョン: 0.2.0\n\n`age` Generate Random Vector of Ages\n`animal` Generate Random Vector of animals\n`animal_list` Animal List\n`answer` Generate Random Vector of Answers (Yes/No)\n`area` Generate Random Vector of Areas\n`as_integer` Convert a Factor Data Frame to Integer\n`car` Generate Random Vector of Cars\n`children` Generate Random Vector of Number of Children\n`coin` Generate Random Vector of Coin Flips\n`color` Generate Random Vector of Colors\n`date_stamp` Generate Random Vector of Dates\n`death` Generate Random Vector of Deaths Outcomes\n`dice` Generate Random Vector of Dice Throws\n`dna` Generate Random Vector of DNA Nucleobases\n`dob` Generate Random Vector of Birth Dates\n`dummy` Generate Random Dummy Coded Vector\n`education` Generate Random Vector of Educational Attainment Level\n`employment` Generate Random Vector of Employment Statuses\n`eye` Generate Random Vector of Eye Colors\n`grade` Generate Random Vector of Grades\n`grade_level` Generate Random Vector of Grade Levels\n`grady_augmented` Augmented List of Grady Ward's English Words and Mark Kantrowitz's Names List\n`group` Generate Random Vector of Control/Treatment Groups\n`hair` Generate Random Vector of Hair Colors\n`height` Generate Random Vector of Heights\n`hour` Generate a Random Sequence of H:M:S Times\n`id` Identification Numbers\n`income` Generate Random Gamma Vector of Incomes\n`internet_browser` Generate Random Vector of Internet Browsers\n`interval` Cut Numeric Into Factor\n`iq` Generate Random Vector of Intelligence Quotients (IQs)\n`language` Generate Random Vector of Languages\n`languages` Languages of the World\n`level` Generate Random Vector of Levels\n`likert` Generate Random Vector of Likert-Type Responses\n`lorem_ipsum` Generate Random Lorem Ipsum Strings\n`marital` Generate Random Vector of Marital Statuses\n`military` Generate Random Vector of Military Branches\n`minute` Generate a Random Sequence of Minutes in H:M:S Format\n`month` Generate Random Vector of Months\n`name` Generate Random Vector of Names\n`name_neutral` Gender Neutral Names\n`normal` Generate Random Normal Vector\n`peek` Data Frame Viewing\n`plot.tbl_df` Plots a tbl_df Object\n`political` Generate Random Vector of Political Parties\n`presidential_debates_2012` 2012 U.S. Presidential Debate Dialogue\n`print.available` Prints an available Object.\n`print.variable` Prints a variable Object\n`probs` Generate a Random Vector of Probabilities.\n`r_data` Pre-Selected Column Data Set\n`r_data_frame` Data Frame Production (From Variable Functions)\n`r_dummy` Title\n`r_insert` Insert Data Frames Into 'r_data_frame'\n`r_list` List Production (From Variable Functions)\n`r_na` Title\n`r_sample` Generate Random Vector\n`r_sample_binary` Generate Random Binary Vector\n`r_sample_factor` Generate Random Factor Vector\n`r_sample_integer` Generate Random Integer Vector\n`r_sample_logical` Generate Random Logical Vector\n`r_sample_ordered` Generate Random Ordered Factor Vector\n`r_sample_replace` Generate Random Vector (Without Replacement)\n`r_series` Data Frame Series (Repeated Measures)\n`race` Generate Random Vector of Races\n`relate` Create Related Numeric Columns\n`religion` Generate Random Vector of Religions\n`sat` Generate Random Vector of Scholastic Aptitude Test (SATs)\n`second` Generate a Random Sequence of Seconds in H:M:S Format\n`sentence` Generate Random Vector of Sentences\n`seriesname` Add Internal Name to Data Frame\n`sex` Generate Random Vector of Genders\n`sex_inclusive` Generate Random Vector of Non-Binary Genders\n`smokes` Generate Random Logical Smokes Vector\n`speed` Generate Random Vector of Speeds\n`state` Generate Random Vector of states\n`state_populations` State Populations (2010)\n`string` Generate Random Vector of Strings\n`table_heat` View Data Table Column Types as Heat Map\n`time_stamp` Generate a Random Sequence of Times in H:M:S Format\n`upper` Generate Random Letter Vector\n`valid` Generate Random Logical Vector\n`variables` Available Variable Functions\n`varname` Add Internal Name to Vector\n`wakefield` Generate Random Data Sets\n`year` Generate Random Vector of Years\n`zip_code` Generate Random Vector of Zip Codes\n\n## age\n\n``````> age(10)\n``````\n`````` 28 28 34 21 34 29 24 33 32 23\n``````"
] | [
null,
"http://www.r-pkg.org/badges/version/wakefield",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5324604,"math_prob":0.60248643,"size":4485,"snap":"2020-34-2020-40","text_gpt3_token_len":1063,"char_repetition_ratio":0.35393885,"word_repetition_ratio":0.081203006,"special_character_ratio":0.2238573,"punctuation_ratio":0.0466563,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.97003263,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T15:21:12Z\",\"WARC-Record-ID\":\"<urn:uuid:654cb6b3-43dc-434d-a8f6-5889f716d2cf>\",\"Content-Length\":\"238823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4250770b-9493-4e20-a7a1-a1e752ef1133>\",\"WARC-Concurrent-To\":\"<urn:uuid:35fe7ca1-2a1e-49c0-98a7-1a08e10411c2>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"http://uribo.github.io/rpkg_showcase/dataset/wakefield.html\",\"WARC-Payload-Digest\":\"sha1:5CANFUJTZPAIFW2VBSPDS5BREUJLKKC7\",\"WARC-Block-Digest\":\"sha1:NE45IFDGMXKK6RBQYEOPV5SJ6CEVNBOO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737883.59_warc_CC-MAIN-20200808135620-20200808165620-00283.warc.gz\"}"} |
InfiMM-WebMath-40B Dataset
InfiMM-WebMath-40B is a large-scale, open-source multimodal dataset specifically designed for mathematical reasoning tasks. It incorporates both text and images, extracted from web documents, to advance the pre-training of Multimodal Large Language Models (MLLMs). The dataset is tailored to support sophisticated reasoning tasks that involve understanding both text and visual elements like diagrams, figures, and geometric plots.
Dataset Overview
The InfiMM-WebMath-40B dataset includes:
- 24 million web documents.
- 85 million image URLs.
- 40 billion text tokens.
These documents were sourced from Common Crawl data snapshots (2019–2023), filtered to focus on high-quality mathematical and scientific content in both English and Chinese.
Data Structure
The dataset is organized in a format that captures both text and images in their original order, ensuring accurate interleaving between the two modalities. The structure is as follows:
{
"URL": "...", # The URL of the source document.
"text_list": [...], # List of extracted text segments, None if the element is an image.
"image_list": [...], # List of image URLs, None if the element is a text segment.
"metadata": {...} # Metadata containing information about the extraction process (e.g., processing details, timestamps).
"metadata": { # Metadata containing information about the extraction process (e.g., processing details, timestamps).
"ft_lang_label", # Type of languages detected by fastText
"ft_lang_prob", # Probability of type of language detected by fastText
"math_prob", # First round math content detection with high recal fastText model
"size",
"snap", # Timestamp of Common Crawl snapshot
"text_gpt3_token_len",
"char_repetition_ratio",
"word_repetition_ratio",
"special_character_ratio",
"punctuation_ratio",
"nsfw_num_words", # Number of words which are NSFW
"has_unicode_error", # If there's any unicode error exists
"math_prob_llama3", # Probability of second round math detection with high precision fastText model
}
}
Interleaved Text and Images
The text_list and image_list are designed as parallel arrays, maintaining the sequence of the document. This interleaving structure allows models to reconstruct the flow of the original document:
- If
text_list[i]
contains text, thenimage_list[i]
isNone
, indicating that the content at this position is text. - If
text_list[i]
isNone
, thenimage_list[i]
contains a URL to an image at that position in the document.
This interleaving of text and images ensures that models trained on this dataset can process the content in the same way a human would, following the logical flow between text explanations and accompanying visual aids.
Data Collection and Filtering Pipeline
The InfiMM-WebMath-40B dataset was created through a comprehensive multi-stage filtering and extraction process, starting with over 120 billion web pages from the Common Crawl repository. The key steps in this pipeline are outlined below::
- Language Filtering: The first step involved filtering for English and Chinese content. We utilized Trafilatura to extract text from web pages, and LangDetect to efficiently identify the language, ensuring only relevant multilingual content was retained..
- High Recall Math Filtering: To capture as much math-related content as possible, we employed a modified version of Resiliparse for HTML parsing. In conjunction with a FastText model optimized for high recall, this phase ensured any potential mathematical data are preserved.
- Deduplication: MinHash were used for fuzzy text deduplication and web page URL exact matching for neighboring Common Crawl snapshots.
- Rule-Based Filtering: This step applied specific filtering rules to remove irrelevant or low-quality content, such as documents containing NSFW material or boilerplate “lorem ipsum,” enhancing the dataset’s overall quality.
- High Precision Math Filtering: A second pass was performed using a FastText model, this time tuned for high precision, to ensure only highly relevant mathematical content remained in the dataset. This refinement step further improved the dataset’s focus and relevance for mathematical reasoning tasks.
- Image Filtering: Finally, rule-based filtering was applied to images, removing irrelevant or extraneous visuals (e.g., logos, banners) to ensure that the remaining images were aligned with the mathematical content.
How to Use the Dataset
- Base Text Download: The dataset is available for download as a set of web documents with interleaved text and image URLs.
- Image Download: Users need to download images according to the image URLs provided.
Note
If you want more data with more precision, you can always use higher thresholds with math_prob
and math_prob_llama3
fields in metadata
.
License
InfiMM-WebMath-40B is made available under an ODC-By 1.0 license; users should also abide by the CommonCrawl ToU: https://commoncrawl.org/terms-of-use/. We do not alter the license of any of the underlying data.
Citation
@misc{han2024infimmwebmath40badvancingmultimodalpretraining,
title={InfiMM-WebMath-40B: Advancing Multimodal Pre-Training for Enhanced Mathematical Reasoning},
author={Xiaotian Han and Yiren Jian and Xuefeng Hu and Haogeng Liu and Yiqi Wang and Qihang Fan and Yuang Ai and Huaibo Huang and Ran He and Zhenheng Yang and Quanzeng You},
year={2024},
eprint={2409.12568},
archivePrefix={arXiv},
primaryClass={cs.CV},
url={https://arxiv.org/abs/2409.12568},
}
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