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https://fr.slideserve.com/dava/chapter-2-linear-equations-and-functions	1,624,601,974,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00301.warc.gz	247,542,457	21,711	Chapter 2: Linear equations and functions # Chapter 2: Linear equations and functions ## Chapter 2: Linear equations and functions - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Chapter 2:Linear equations and functions BIG IDEAS: Representing relations and functions Graphing linear equations and inequalities in two variables Writing linear equations and inequalities in two variables 2. On a blank notebook page, please complete the Prerequisite Skills on PG 70 #2-12 Even 3. Lesson 1: represent relations and functions 4. Essential question How do you graph relations and functions? 5. Relation: A mapping, or pairing, of input values with output values • Domain: The set of input values of a relation • Range: The set of output values of a relation • Function: A relation for which each input has exactly one output VOCABULARY 6. The domain consists of all the x-coordinates:–2, –1, 1, 2,and3. The rangeconsists of all the y-coordinates:–3, –2, 1, and 3. EXAMPLE 1 Represent relations Consider the relation given by the ordered pair (–2, –3), (–1, 1), (1, 3), (2, –2), and (3, 1). a.Identify the domain and range. SOLUTION 7. b. Represent the relation using a graph and a mapping diagram. EXAMPLE 1 Represent relations SOLUTION b. Graph Mapping Diagram 8. The relation isa function because each input is mapped onto exactly one output. a. Identify functions EXAMPLE 2 Tell whether the relation is a function. Explain. SOLUTION 9. b. The relation isnota function because the input 1 is mapped onto both – 1 and 2. Identify functions EXAMPLE 2 Tell whether the relation is a function. Explain. SOLUTION 10. for Examples 1 and 2 GUIDED PRACTICE 1. Consider the relation given by the ordered pairs (–4, 3), (–2, 1), (0, 3), (1, –2), and (–2, –4) a. Identify the domain and range. SOLUTION The domain consists of all the x-coordinates:–4, –2, 0 and 1, The rangeconsists of all the y-coordinates: 3, 1,–2 and –4 11. for Examples 1 and 2 GUIDED PRACTICE b. Represent the relation using a table and a mapping diagram. SOLUTION 12. 2. Tell whether the relation is a function. Explain. ANSWER Yes; each input has exactly one output. for Examples 1 and 2 GUIDED PRACTICE 13. Use the vertical line test EXAMPLE 3 SOLUTION The team graph does not represent a function because vertical lines at x=28 and x=29 each intersect the graph at more than one point. The graph for Kevin Garnett does represent a function because no vertical line intersects the graph at more than one point. 14. Graph an equation in two variables EXAMPLE 4 Graph the equationy= – 2x–1. SOLUTION STEP1 Construct a table of values. 15. Graph an equation in two variables EXAMPLE 4 STEP 2 Plot the points. Notice that they all lie on a line. STEP3 Connect the points with a line. 16. a. f (x) = –x2 – 2x + 7 The functionfis not linear because it has an x2-term. EXAMPLE 5 Classify and evaluate functions Tell whether the function is linear.Thenevaluate the function when x= – 4. SOLUTION f (x) =–x2– 2x+ 7 Write function. f (–4) =–(– 4)2– 2(–4) + 7 Substitute–4forx. Simplify. =–1 17. The function gis linear because it has the form g(x) = mx + b. b. g(x) = 5x + 8 EXAMPLE 5 Classify and evaluate functions SOLUTION g(x) = 5x+ 8 Write function. Substitute –4 forx. g(–4) = 5(–4) + 8 Simplify. =–12 18. for Examples 4 and 5 GUIDED PRACTICE 4. Graph the equationy = 3x – 2. ANSWER 19. for Examples 4 and 5 GUIDED PRACTICE Tell whether the function is linear. Then evaluate the function when x = –2. 5. f (x) = x – 1 – x3 6. g (x) = –4 – 2x ANSWER ANSWER Not Linear; Thef(x) = 5,whenx = –2 Linear; Thef(x) = 0,whenx = –2 20. Essential question How do you graph relations and functions? Make a table of domain and range values. Then plot the points from the table. If it is a function, connect the dots. 21. An internet company had a profit of \$2.6 million in retail sales over the last five years. What was its average annual profit? 22. Lesson 2: Find Slope and Rate of change 23. Essential question How do you determine whether two nonvertical lines are parallel or perpendicular? 24. Slope: The ratio of vertical change (the rise) to the horizontal change (the run) for a nonvertical line • Parallel: Two lines in the same plane that do not intersect • Perpendicular: Two lines in the same plane that intersect to form a right angle • Rate of change: A comparison of how much one quantity changes, on average, relative to the change in another quantity. VOCABULARY 25. A skateboard ramp has a rise of 15 inches and a run of 54 inches. What is its slope? rise slope = run 5 5 15 18 18 = = 54 ANSWER . The slope of the ramp is EXAMPLE 1 Find slope in real life Skateboarding SOLUTION 26. m = y2–y1 –1 –3 = x2–x1 2–(–1) 4 ANSWER 3 = The correct answer is A. EXAMPLE 2 Standardized Test Practice SOLUTION Let (x1, y1) =(–1, 3) and (x2,y2)=(2, –1). 27. 2. What is the slope of the line passing through the points (–4, 9) and (–8, 3) ? ANSWER The correct answer is D. for Examples 1 and 2 GUIDED PRACTICE 28. 1 5 1 1 – 2 2 3 4 ANSWER ANSWER ANSWER ANSWER for Examples 1 and 2 GUIDED PRACTICE Find the slope of the line passing through the given points. 3. (0, 3), (4, 8) 5. (–3, –2), (6, 1) 4. (– 5, 1), (5, – 4) 6. (7, 3), (–1, 7) 29. a. Line 1: through (–2, 2) and (0, –1) Line 2: through (–4, –1) and (2, 3) b. Line 1: through (1, 2) and (4, –3) Line 2: through (–4, 3) and (–1, –2) –1 – 2 = a. Find the slopes of the two lines. 0– (–2) 3 – 3 2 = – m1 = 2 EXAMPLE 4 Classify parallel and perpendicular lines Tell whether the lines are parallel, perpendicular, or neither. SOLUTION 30. 11.Line1: through(–2, 8)and(2, –4) 12.Line1: through(–4, –2)and(1, 7) Line2:through(–5, 1)and(–2, 2) Line2:through(–1, –4)and(3, 5) for Example 4 GUIDED PRACTICE GUIDED PRACTICE Tell whether the lines are parallel, perpendicular, or neither. ANSWER perpendicular ANSWER neither 31. Essential question How do you determine whether two nonvertical lines are parallel or perpendicular? Calculate the slope: Parallel lines have equal slope Perpendicular lines have slopes that are opposite reciprocals 32. In 2005, Carey’s Pet Shop had a profit of \$55,500. In 2006, profits were \$38,700. In a graph of the data, is the slope of the segment between 2005 and 2006 positive or negative? 33. Lesson 3: Graph equations of lines 34. Essential question How do you graph a linear equation using intercepts? 35. Y-intercept: The y-coordinate of a point where a graph intersects the y-axis • X-intercept: The x-coordinate of a point where a graph intersects the x-axis • Slope-intercept form: y=mx+b • Standard form: Ax + By = C where a ≠ 0 VOCABULARY 36. b. a. y = x + 3 y = 2x a. EXAMPLE 1 Graph linear functions Graph the equation. Compare the graph with the graph of y = x. SOLUTION The graphs ofy = 2x andy = x both have a y-intercept of 0, but the graph ofy = 2x has a slope of2 instead of 1. 37. b. EXAMPLE 1 Graph linear functions The graphs ofy = x + 3 and y = x both have a slope of 1, but the graph ofy = x + 3has a y-intercept of 3 instead of 0. 38. EXAMPLE 2 Graph an equation in slope-intercept form STEP 4 Draw a line through the two points. 39. 2 5 5. y = x + 4 for Examples 1 and 2 GUIDED PRACTICE Graph the equation 4. y = –x + 2 40. 1 2 6. y = x – 3 for Examples 1 and 2 GUIDED PRACTICE Graph the equation 7. y = 5 + x 41. for Examples 1 and 2 GUIDED PRACTICE Graph the equation 8. f (x) = 1 – 3x 9. f (x) = 10 – x 42. EXAMPLE 3 Solve a multi-step problem Biology The body length y (in inches) of a walrus calf can be modeled by y = 5x + 42 where xis the calf’s age (in months). • Graph the equation. • Describe what the slope and y-intercept represent in this situation. • Use the graph to estimate the body length of a calf that is 10 months old. 43. EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Graph the equation. STEP 2 Interpret the slope and y-intercept. The slope, 5,represents the calf’s rate of growth in inches per month. The y-intercept, 42, represents a newborn calf’s body length in inches. 44. EXAMPLE 4 Graph an equation in standard form STEP3 Identify the y-intercept. Lety= 0. 5(0) + 2y = 10 y = 5 Solve for y. The y-intercept is 5. So, plot the point (0, 5). STEP4 Draw a line through the two points. 45. EXAMPLE 5 Graph horizontal and vertical lines Graph (a) y = 2 and (b) x = –3. SOLUTION a. The graph of y = 2 is the horizontal line that passes through the point (0, 2). Notice that every point on the line has a y-coordinate of 2. b. The graph of x = –3 is the vertical line that passes through the point (–3, 0). Notice that every point on the line has an x-coordinate of –3. 46. for Examples 4 and 5 GUIDED PRACTICE Graph the equation. 12. 3x – 2y = 12 11. 2x + 5y = 10 47. for Examples 4 and 5 GUIDED PRACTICE Graph the equation. 13. x = 1 14. y = –4 48. Essential question How do you graph a linear equation using intercepts? To find x: Set y = 0, solve for x, (x,0) To find y: Set x = 0, solve for y, (0,y) Plot each intercept and connect the dots. 49. On a blank piece of paper please complete the Quiz for Lessons 2.1-2.3 on Page 96 #1-9. When finished please turn into the homework bin. 50. Lesson 4: Write equations of lines	2,821	9,269	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.65625	5	CC-MAIN-2021-25	latest	en	0.763168
http://www.traditionaloven.com/tutorials/distance/convert-japan-bu-unit-to-thousandth-of-inch-mil.html	1,529,280,611,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00546.warc.gz	522,045,753	11,690	Convert 分 to mil \| Japanese bu to thou length conversion Amount: 1 Japanese bu (分) of length Equals: 119.30 thou (mil) in length Converting Japanese bu to thou value in the length units scale. TOGGLE : from thou into Japanese bu in the other way around. length from Japanese bu to thou Conversion Results: Enter a New Japanese bu Amount of length to Convert From * Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8) * Precision is how many numbers after decimal point (1 - 9) Enter Amount : Decimal Precision : CONVERT : between other length measuring units - complete list. Conversion calculator for webmasters. Length, Distance, Height & Depth units Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units. Convert length measuring units between Japanese bu (分) and thou (mil) but in the other reverse direction from thou into Japanese bu. conversion result for length: From Symbol Equals Result To Symbol 1 Japanese bu 分 = 119.30 thou mil Converter type: length units This online length from 分 into mil converter is a handy tool not just for certified or experienced professionals. First unit: Japanese bu (分) is used for measuring length. Second: thou (mil) is unit of length. 119.30 mil is converted to 1 of what? The thou unit number 119.30 mil converts to 1 分, one Japanese bu. It is the EQUAL length value of 1 Japanese bu but in the thou length unit alternative. How to convert 2 Japanese bu (分) into thou (mil)? Is there a calculation formula? First divide the two units variables. Then multiply the result by 2 - for example: 119.30326891 * 2 (or divide it by / 0.5) QUESTION: 1 分 = ? mil 1 分 = 119.30 mil Other applications for this length calculator ... With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool: 1. in practicing Japanese bu and thou ( 分 vs. mil ) values exchange. 2. for conversion factors training exercises between unit pairs. 3. work with length's values and properties. International unit symbols for these two length measurements are: Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Japanese bu is: Abbreviation or prefix ( abbr. ) brevis - short unit symbol for thou is: mil One Japanese bu of length converted to thou equals to 119.30 mil How many thou of length are in 1 Japanese bu? The answer is: The change of 1 分 ( Japanese bu ) unit of length measure equals = to 119.30 mil ( thou ) as the equivalent measure for the same length type. In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 分 - Japanese bu for length amount, the rule is that the Japanese bu number gets converted into mil - thou or any other length unit absolutely exactly. Conversion for how many thou ( mil ) of length are contained in a Japanese bu ( 1 分 ). Or, how much in thou of length is in 1 Japanese bu? To link to this length Japanese bu to thou online converter simply cut and paste the following. The link to this tool will appear as: length from Japanese bu (分) to thou (mil) conversion. I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.	826	3,633	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2018-26	latest	en	0.838394
https://cw.fel.cvut.cz/wiki/courses/be5b33kui/semtasks/05_ml1/image	1,726,179,597,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00177.warc.gz	165,020,085	6,283	# Reading .png image into a numerical vector To be able to work with images reasonably (learn from them and classify them), we shall transform each image into a numerical vector. Each number in the vector will correspond to the intensity of certain pixel in the image. ## TLDR With the use of Pillow and Numpy libraries, the whole thing is easy: from PIL import Image import numpy as np impath = 'train_data/img_1112.png' image_vector = np.array(Image.open(impath)).flatten() ## Explanation If you are interested in what the above code actually does, let's do it step by step: from PIL import Image import numpy as np impath = 'train_data/img_1112.png' im = Image.open(impath) print(type(im)) im2d = np.array(im) print(type(im2d), im2d.shape) im1d = im2d.flatten() print(type(im1d), im1d.shape) print(im1d) After running the above code, you should get output similar to this: <class 'PIL.PngImagePlugin.PngImageFile'> <class 'numpy.ndarray'> (10, 10) <class 'numpy.ndarray'> (100,) [230 202 168 139 124 129 147 180 206 221 227 181 126 84 51 49 80 145 206 241 227 169 102 50 18 7 27 96 183 233 212 156 92 40 21 15 25 76 164 224 196 136 78 47 33 32 39 73 141 203 175 118 64 41 42 45 48 66 122 184 151 87 39 24 35 34 28 35 84 156 139 76 36 26 35 37 30 38 69 130 152 106 87 99 116 114 95 77 82 122 198 180 186 217 237 226 199 168 152 172] Explanation: • Function Image.open() reads in the image and returns an instance of class PngImageFile. • Function np.array() converts the image into an instance of class numpy.ndarray, i.e., it creates a 2D array of size 10×10. • Method ndarray.flatten() turns 10×10 matrix into a vector of length 100; its contents are at the end of the output.	584	1,736	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.078125	3	CC-MAIN-2024-38	latest	en	0.559491
https://stackoverflow.com/questions/62679102/how-can-i-calculate-the-acutal-elapsed-time-in-my-timezone-between-two-dates-in/62679509	1,723,221,273,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00230.warc.gz	434,669,497	44,915	# How can I calculate the acutal elapsed time in my timezone between two dates in C#? How do I calculate the actual elapsed time between two `DateTime`/`DateTimeOffset`, both in my timezone, taking daylight saving time, leap years, etc, into account? For example, the following code comparing the difference between 00:00 and 05:00 of March 29th 2020 will output a timespan of 6 hours (05:00 becomes 06:00 after the conversion to local time, since daylight saving time starts 02:00 in my timezone, changing clocks forward from 02:00 to 03:00). ``````var timezone = TimeZoneInfo.Local; var start = new DateTimeOffset(2020, 03, 29, 00, 00, 00, timezone.BaseUtcOffset).LocalDateTime; var end = new DateTimeOffset(2020, 03, 29, 05, 00, 00, timezone.BaseUtcOffset).LocalDateTime; var diff1 = end.Subtract(start); var diff2 = end - start; Console.WriteLine(diff1); // 06:00:00 Console.WriteLine(diff2); // 06:00:00 `````` I'm trying to get an output of 4 hours in the example above since that is the elapsed time between 00:00 and 05:00. ``````00:00 -> 01:00 (+ 1 hr) 01:00 -> 02:00 (+ 1 hr) 02:00 -> 03:00 (0) 03:00 -> 04:00 (+ 1 hr) 04:00 -> 05:00 (+ 1 hr) = 4 hrs `````` I'm clearly missing something here. I've read Comparisons and arithmetic operations with DateTimeOffset values but can't wrap my head around how to do this calculation. I am aware that NodaTime can probably achieve this, but surely there's a way to do this without external libraries? • The elapsed time is 5 full hours, not 4. Assuming DST rules didn't come into force on that date, in which case `end`'s offset would be wrong. Using an offset isn't enough to handle DST, you need to know the actual timezone. The de-facto standard for timezone calculations is the IANA timezone database, used by almost everyone except Windows. You can use it in .NET through the NodaTime library Commented Jul 1, 2020 at 14:32 • @PanagiotisKanavos It says in the question that DST changed on that date, or what do you mean? If I had a stopwatch on this date, started it 00:00 and stopped it 05:00, it would show me that 4 hours elapsed. Commented Jul 1, 2020 at 14:34 • So the offset is wrong. It's no longer the `BaseUtcOffset`, it's a different one. `LocalDateTime` doesn't perform any DST conversions, it returns a `DateTime` based on the offset stored in DateTimeOffset. You could have calculated the differenc simply by subtracting the two `DateTimeOffset` values, you don't gain anything by going through `LocalDateTime` Commented Jul 1, 2020 at 14:35 • To get the correct DateTimeOffset you can use TimeZoneInfo.ConvertTime(DateTime,TimeZoneInfo) to convert a local DateTime to a DateTimeInfo using the offset specified by that timezone's DST rules. I always prefer to use NodaTime with explicit timezone names though. All commercial web services report times using the IANA timezone names, eg `Europe/Berlin` or `Europe/Athens` Commented Jul 1, 2020 at 14:42 • @PanagiotisKanavos If you remove LocalDateTime both diffs become 5 hours, so it clearly does some DST conversions. However, I appreciate the fact that you pointed out the UTC offset, I believe I've figured out how to get the elapsed time. Will post answer. Commented Jul 1, 2020 at 14:43 As Panagiotis Kanavos pointed out, So the offset is wrong. It's no longer the BaseUtcOffset, it's a different one I found TimeZoneInfo.GetUtcOffset and used that. I think it does what I want? Timezones are confusing, but at least I get my output of 4 hours now. ``````var timezone = TimeZoneInfo.Local; var startTime = new DateTime(2020, 03, 29, 00, 00, 00); var endTime = new DateTime(2020, 03, 29, 05, 00, 00); var startTimeOffset = timezone.GetUtcOffset(startTime); var endTimeOffset = timezone.GetUtcOffset(endTime); var start = new DateTimeOffset(2020, 03, 29, 00, 00, 00, startTimeOffset); var end = new DateTimeOffset(2020, 03, 29, 05, 00, 00, endTimeOffset); var diff1 = end.Subtract(start); var diff2 = end - start; Console.WriteLine(diff1); // 04:00:00 Console.WriteLine(diff2); // 04:00:00 `````` • Have you tried `TimeZoneInfo.ConvertTime()` ? Eg `var start=TimeZoneInfo.ConvertTime(startTime,timezone)` ? Commented Jul 1, 2020 at 14:49 • @PanagiotisKanavos That seems to still give me a diff of 5 hours. `startTime = TimeZoneInfo.ConvertTime(startTime, timezone);` Commented Jul 1, 2020 at 14:53 • This answer is fine, but a couple of things to point out. 1) There is a `DateTimeOffset` constructor that accepts a `DateTime` and a `TimeSpan`, so you could just pass the `startTime` and `endTime` into the constructor instead of specifying the integers twice. 2) For `GetUtcOffset` (and really any case where an offset is determined, such as in my answer below also), be aware that in the case of ambiguous or invalid local time (such as occur with DST transitions), .NET will assume you want the standard time offset, even if you may have in fact wanted the daylight time offset. Commented Jul 1, 2020 at 17:29 Your answer is fine, and is the correct approach when the offset is not provided with if the time zone, assuming it can be any arbitrary time zone. However, if you are only working with the local system time zone, you can simplify the code as follows: ``````var start = new DateTimeOffset(new DateTime(2020, 3, 29, 0, 0, 0, DateTimeKind.Local)); var end = new DateTimeOffset(new DateTime(2020, 3, 29, 5, 0, 0, DateTimeKind.Local)); TimeSpan diff = end - start; `````` This works because the `.Kind` property of a `DateTime` is evaluated when passed into the `DateTimeOffset` constructor that takes a single `DateTime` parameter. If it's `DateTimeKind.Local` or `DateTimeKind.Utc`, then the offset is set correctly. It's only in the case of `DateTimeKind.Unspecified` that the offset could be incorrect (because it will assume local, even if that's not the case). See the remarks in the docs here.	1,581	5,866	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2024-33	latest	en	0.879287
https://www.codewars.com/users/Iliort/comments	1,723,677,774,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00420.warc.gz	542,049,908	12,492	• ###### user6356573commented on "Mysterious function" kata Can someone give me at least a semblance of a clue in how to solve this? • ###### LarryAtGUcommented on "Mysterious function" kata I saw the solution, it has no much to do with programming. I think this Kata is very bad designed. :( • ###### RileyHuntercommented on "Mysterious function" kata Please don't post solution code in the discourse without a spoiler tag. If you are posting code of any sort, please format it correctly. Use backticks for inline code snippets `like this` => `like this` Use triple backticks for code blocks ``` Like this ``` => ``````Like this `````` Your solution is close but your logic is not quite correct; add the following test case and see if it helps: `test.assert_equals(get_num(400), 2, "Wrong output!")` • ###### monkruscommented on "Mysterious function" kata This comment is hidden because it contains spoiler information about the solution • ###### servideicommented on "Mysterious function" kata This comment is hidden because it contains spoiler information about the solution • ###### aotearoacommented on "Mysterious function" kata I found more complexed pattern. Unfortunately, it passed only these initial 5 test cases, no further. Such a shame and loss of time. What could be the chances for that. So... If the number has even number of digits: 3479283469 => 34792 \| 83469 => sum A \| sum B => sum B - sum A => 30 - 25 => 5 123321 => 123 \| 321 => sum A \| sum B => sum B - sum A => 6 - 6 => 0 If the number has odd number of digits: 90783 => 9 \| 0783 => count of digits smaller than first digit => 4 300 => 3 \| 00 => count of digits smaller than first digit => 2 89282350306 => 8 \| 9282350306 => count of digits smaller than first digit => 8 . solved. • ###### LS2008commented on "Mysterious function" kata This comment is hidden because it contains spoiler information about the solution • ###### Chrono79commented on "Mysterious function" kata There are some problems with the Java translation. • ###### akar-0commented on "Mysterious function" kata I don't do Java. • ###### Swag_Legendcommented on "Mysterious function" kata Could you review it? • ###### akar-0resolved an issue on "Mysterious function" kata Because nobody reviewed it, and there is no way this is a kata issue (= a bug in the kata). • ###### Swag_Legendcreated an issue for "Mysterious function" kata why is the java translation not approved?	633	2,458	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.15625	3	CC-MAIN-2024-33	latest	en	0.829702
https://www.engineersgarage.com/insight/how-mcb-works?page=10	1,506,235,113,000,000,000	text/html	crawl-data/CC-MAIN-2017-39/segments/1505818689897.78/warc/CC-MAIN-20170924062956-20170924082956-00291.warc.gz	784,742,457	31,797	Working of MCB (Miniature Circuit Breaker) \| How MCB works Close or Esc Key # Insight - How MCB works ### Written By: Arpit Jain Arc quenching in MCB (Miniature Circuit Breakers) Another important point to be considered in the design of circuit breakers is Arc quenching. To understand how MCB carries out the arc quenching, it is important to know about the electric arc and how is it produced. When an overcurrent is interrupted by the circuit breaker by opening its contacts, current tries to bridge the gap. In an attempt to maintain the circuit, the air heats up and becomes a conductor. As a result an arc forms. In general, when air and gases are heated, they become electric conductors. The hotter they get, the better they conduct. The following image shows how an electric arc looks like. The heat from an uncontrolled arc in a circuit breaker can cause a rapid and violent expansion of the nearby air and could severely damage the circuit breaker. Therefore, besides separating the electrical contacts, a circuit breaker also has to quickly extinguish the arc. A number of factors can be employed for quenching the arc, such as, speed, distance, dielectric strength, cooling etc. a) Speed : When the contacts separate rapidly, there is less time for the arc to form and maintain itself. b) Distance : When the distance between opened contacts is more, the arc has to stretch more to maintain the current flow which requires more voltage. c) Cooling : When the arc is forced against a cold material, it absorbs and dissipates the heat. d) Dielectric Strength : When the arc is submerged in a medium with higher dielectric strength than air (sulfur hexafluoride, SF6), the insulating nature of the medium helps in quenching the arc. In an MCB, arc chutes or arc dividers are used for arc quenching. When the contacts of an MCB separate, generating an electrical arc between them through air, the arc is moved into the arch chute where it is divided into small segments. The overall energy level of the arc gets split up which is not sufficient to sustain the arc and therefore it gets dissipated. ### I likeD it..I had seen it I likeD it..I had seen it many times at my home but nver tried to know how it works..thanks to engineersgarage ..evrone shud watch its intresting ### explanation is explanation is strong. thanks. regards, raja giri Amazing... ### gr8 work.. arc quenching gr8 work.. arc quenching explained nicely. now I know what arc chutes are for. ### Arc is like a circle it has Arc is like a circle it has on ends ra erri puka wow, very nice. Awesome.. ### mcb is a best of electrical mcb is a best of electrical component ### nice & informative nice & informative ### Great !!!!!!!!!! nicely Great !!!!!!!!!! nicely Explained!!!! ### do it can get failure in its do it can get failure in its work.... ### Thank you.....it helped me a Thank you.....it helped me a lot.... ### Great work..... Really Thank you so muchhhh.... ### Very Nice presentation.Thank Very Nice presentation. Thank you . thanks Thanks ### VERY GOOD EXPLANATION VERY GOOD EXPLANATION ### very nice & good presentation very nice & good presentation. Thank you . ### nice one thanks a lot nice one thanks a lot ### thank you VERY GOOD thank you VERY GOOD EXPLANATION ### in witch case: solenoid will in witch case: solenoid will trip or , Bimetal will trip ### WITCH???? I'm sorry but do WITCH???? I'm sorry but do you have spelling difficulties,it's spelt as WHICH!!!! ### Really the diagram of MCB Really the diagram of MCB helped me a lot in teaching. ### Thanks so much.Nice and Thanks so much.Nice and simple explanation.Do add some video clips if possible. ### it is really understandable it is really understandable to every one easily YES I really like it very much ### easy explanation which easy explanation which everyone can understand. awesome ,hats off !!... ### Havells 32Amp MCB Can you Havells 32Amp MCB Can you tell me Input and output Line ? ### Many thanks for this piece. Many thanks for this piece. Really helped me a lot!! ### very good information..Really very good information.. Really the Device is great..functionality!! Good explain its very good ### Very good explanation. Thanks Very good explanation. Thanks! ### "protect against over current "protect against over current and over temperature faults" what does this really mean... please explain more on this.. if current flow is more than 220V will it trip, or i a use 6Amp tripper and give more load to it then it will trip.. pls help me to understand. ### I want to know if there is I want to know if there is loose connection in the joint of two wires at that condition mcb will trip or not Excellent ### <p>In my system was falut by <p>In my system was falut by short circuit but the MCB not it triped. i want to know the reason.</p> ### How to calculate what MCB I How to calculate what MCB I need for my main switch of home that takes single phase supply line ? 10AMP, 20 AMP, 15 AMP - how to know ? ### there is a formula to there is a formula to calculate amp rating of mcb that is. amp rating= total load in watt/ supply voltage * power factor generally we assume power factor = 0.8 ### Wonderful Wonderful Really a good article Thnx ### Thanks for explanation but I Thanks for explanation but I didn't get whether it will work for low currents... ### Shall I need stabilizer if I Shall I need stabilizer if I had MCB in my home... Plz anyone give me reply... Thanks ### Very super in The Very super in The information thanks ### ITS AMAZING ZING ZING ITS AMAZING ZING ZING INADEQUATE EXPLANATION. FUSE DOES THE SAME BREAK THE CURRENT. ### Thanks for the breif Thanks for the breif discussion of MCB the information about the MCB is brilliant.. ### why we don't take short why we don't take short circuit protection by bi-metallic strip? very useful ### how the rated current is how the rated current is fixed for every MCB and how it detects the current rating ## Pages ### Filtered HTML • Search Engines will index and follow ONLY links to allowed domains. ### Plain text • No HTML tags allowed.	1,470	6,212	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2017-39	longest	en	0.919334
https://mail.haskell.org/pipermail/haskell/2001-December/008588.html	1,513,417,702,000,000,000	text/html	crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00773.warc.gz	628,477,013	2,289	# gcd definition S.D.Mechveliani mechvel@math.botik.ru Sun, 16 Dec 2001 15:35:57 +0300 ```Simon Peyton-Jones <simonpj@microsoft.com> writes > [..] > If someone could write a sentence or two to explain why gcd 0 0 = 0, > (ideally, brief ones I can put in the report by way of explanation), > I think that might help those of us who have not followed the details > of the discussion. Here it is. gcd n m :: Integer = if n == 0 && m == 0 then 0 else greatest integer that divides both n and m It is set so according to classic definition (by Pythago, Euclid ...) GCD x y = such g that divides both x and y and is a multiple of any g' that divides both x and y. In particular, GCD 0 0 :: Integer = 0 and can be nothing else. Because 0 divides 0 and divides nothing else, everything divides 0 (z0 = 0). People say, D.Knuth writes gcd 0 0 = 0 in his book. And he is a known mathematically reliable person. * Example. gcd 12 18 :: Integer = 6 or -6, because 6 divides 12 and 18, and any other such number divides 6. * The Haskell report says "greatest integer" for domain = Integer, and thus, chooses the sign `+' for the result. This choice is not a mistake and helps to write shorter programs. * All the above agrees with the modern generic theory of ideals (started in \|XX by Kummer, Gauss, Lagrange) for any commutative domain R having the properties of (a /= 0, b /= 0 ==> ab /= 0), existence of unique factorization to primes. According to it gcd x y = least by inclusion ideal of kind (g) that includes the ideal (x, y), where Ideal (a,b..c) =def= {xa + yb +..+ zc \| x,y..z <- R} - a subset in R. `includes` (as set) is a partial order on ideals, and it is true that (1) a divides b <==> (a) includes (b), (2) (a) includes (a). Specializing this definition to Integer, we obtain the source formula. ----------------- Serge Mechveliani mechvel@botik.ru ```	574	1,906	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2017-51	latest	en	0.878398
http://slideplayer.com/slide/3403789/	1,527,432,504,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00404.warc.gz	261,673,321	27,062	# 2 The Equation Learning Objectives ## Presentation on theme: "2 The Equation Learning Objectives"— Presentation transcript: 2 The Equation Learning Objectives Describe and use the fundamental accounting equation. Identify five basic classes of accounts Explain five essential accounting concepts Analysis: Percentage of sales (common size income statements) Unit 2 Objective 2.1: The accounting equation The following slides use Jack & Jill Enterprises to demonstrate the fundamental accounting equation O2.1 Jack and Jill Enterprises Hey Jill! Lets fetch water and make some money You fetch I’ll supervise O2.1 Jack and Jill Enterprises We receive \$1 for each pail of water we fetch And we pay \$1 to replace each shoe we wear out O2.1 First, we can list all of J & J’s assets in a blue box. Cash 3 Pails 2 Total 5 O2.1 Assets Liabilities Equity Cash 3 Mom payable 2 The red box has two parts, creditors and owners. Pails 2 Next, list all the claims against the assets in a red box. Equity Jack & Jill, Capital 3 Total 5 Total 5 O2.1 Assets Liabilities Equity Cash 3 Mom payable 2 Pails 2 The most important thing to remember is the total in the blue box must always equal the total in the red box. A = L + E Equity Jack & Jill, Capital 3 Total 5 Total 5 O2.1 This is the BALANCE SHEET Assets Liabilities Equity Now lets see how this system works for J & J Enterprises Now lets see how this system works for J & J Enterprises. (Remember, Assets must always equal Liabilities Plus Owner’s Equity. O2.1 Jack and Jill Enterprises Assets Liabilities Mom payable 2 3 4 Cash 6 7 5 8 7 Pails 2 Fetch 1 Pail Fetch 2 Pails Fetch 1 Pail Buy 2 Shoes Fetch 3 Pails Buy 1 Shoe Equity 4 3 8 7 6 5 7 J&J, Capital 9 5 10 6 8 5 Total assets 7 9 Total Liab. + Equity 8 9 7 6 10 9 O2.1 The system stayed in balance We increased Equity with each sale We decreased Equity for each expense. The system stayed in balance Assets Liabilities Equity O2.1 Answer: With this simplified system, you would have have to memorize So what are the total sales earned by Jack and Jill so far? Answer: With this simplified system, you would have have to memorize the total sales. Sales ? O2.1 Jack and Jill Enterprises I wish I could remember what our sales were. Assets Liabilities Mom payable 2 3 I wish I could remember what our sales were. 4 Cash 6 7 5 8 7 Pails 2 Equity 4 3 8 6 7 7 5 J&J, Capital 5 10 9 8 6 5 Total assets 7 9 Total Liab. + Equity 7 8 9 10 9 6 O2.1 We need a better method to keep track of sales and expenses for each time period BALANCE SHEET INCOME STATEMENT Solution: The Income Statement O2.1 To keep track of sales and expenses for any given period. Income Statement A separate “temporary” list of Sales and Expenses BALANCE SHEET INCOME STATEMENT O2.1 The Sales and Expenses are collected on the Income Statement BALANCE SHEET INCOME STATEMENT Sales Assets Liabilities Equity Expenses O2.1 The resulting Profit or Loss is calculated BALANCE SHEET INCOME STATEMENT Sales Assets Liabilities Equity Expenses Profit Loss O2.1 At the end of the time period The Profit or Loss is moved to Equity BALANCE SHEET INCOME STATEMENT Sales Assets Liabilities Equity Expenses Let’s try it Profit Loss O2.1 Jack and Jill Enterprises Balance Sheet Income Statement Assets Liabilities + Equity Cash Pails Total assets Mom payable J&J, Capital Total Liab & Equity Revenue 3 6 8 4 2 1 7 7 5 3 7 4 2 Shoe Expense 3 2 3 5 9 10 7 9 5 8 6 Fetch 1 Pail Fetch 2 Pails Fetch 1 Pail Buy 2 Shoes 1 3 2 4 4 5 Fetch 3 Pails Buy 1 Shoe OR O2.1 This is a big improvement This is a big improvement. The Income Statement allows us to tell at a glance what the sales and expense totals are. The Income Statement is used to collect sales and expenses for a given time period -for example a month or a year. O2.1 Jack and Jill Enterprises Balance Sheet Income Statement Assets Liabilities + Equity Cash Pails Total assets Mom payable J&J, Capital Total Liab & Equity Revenue 4 8 6 3 2 1 7 5 7 3 4 7 2 Shoe Expense 3 2 3 5 9 10 7 9 6 5 8 The systems stays balanced –you must consider the change in J&J Capital that is being recorded in the Income Statement 3 1 4 5 4 2 OR O2.1 Assets = Liabilities + Equity + Revenues - Expenses Following are exercises using the fundamental accounting equation Assets = Liabilities + Equity + Revenues - Expenses O2.1 Assets Liabilities Equity What is Total Equity? 78 Total Liabilities Equity ? Total Equity 72 Total Assets 150 Total Liab & Equity 150 O2.1 What are Total Liabilities? Assets Liabilities What are Total Liabilities? ? 1092 Total Liabilities Equity Total Equity 658 Total Assets 1750 Total Liab & Equity 1750 O2.1 Assets Liabilities Equity What are Total Assets? Total Liabilities 6,420 Equity Total Equity 3,430 ? Total Assets 9,850 Total Liab & Equity 9,850 O2.1 What is the total Owner, Capital? Balance Sheet Income Statement Assets Liabilities + Equity Total assets Total Liabilities Owner, Capital (Beginning) Total Liab & Equity Revenue 1 3 4 870 Expenses 2 405 ? 1,275 10 7 9 8 5 6 1,275 3 1 4 2 5 What is the total Owner, Capital? OR O2.1 What is the total Owner, Capital? (consider the Income Statement also) Balance Sheet Income Statement Assets Liabilities + Equity Total assets Total Liabilities Owner, Capital (Beginning) Total Liab & Equity Revenue 1 3 100 4 870 Expenses 25 2 405 ? 1,350 10 7 9 5 8 6 Beg. Owner Capital 405 plus Revenue 100 minus Expenses 25 = 480 Ending Owner, Capital What is the total Owner, Capital? (consider the Income Statement also) 75 3 1 4 2 5 OR O2.1 Objective 2.2: Account Classification The following slides offer an expanded Balance Sheet and Income Statement to help understand the 5 classes of accounts O2.2 There are 5 basic classes of accounts Balance Sheet Income Statement Assets Liabilities + Equity 4. Revenue Cash Accounts Receivable Supplies Inventory Land Buildings Equipment Copyrights Accounts Payable Wages Payable Taxes Payable Unearned Revenue Notes Payable Owner’s, Capital 2. Liabilities Revenue 1. Assets Cost of Goods Sold Wages Expense Taxes Expense Rent Expense Utilities Expense Interest expense 5. Expenses 3. Equity There are 5 basic classes of accounts Profit OR Loss O2.2 There are a number of words used in account descriptions that help us identify the classification. Note the words on the following slide and the classification they indicate O2.2 Remember where these words belong! Earned Income Payable Cash Revenue Sales Receivable Unearned Prepaids Expense Capital Equipment Withdrawals O2.2 Where does this account belong? Accounts Receivable Accounts Receivable O2.2 Where does this account belong? Prepaid Insurance Prepaid Insurance O2.2 Where does this account belong? Accounts Payable Accounts Payable O2.2 Where does this account belong? Sales Sales O2.2 Where does this account belong? Rent expense Rent expense O2.2 Where does this account belong? Taxes payable Taxes payable O2.2 Where does this account belong? J. Black, Capital J. Black, Capital O2.2 Where does this account belong? Interest Income Interest Income O2.2 Where does this account belong? Interest expense Interest expense O2.2 Where does this account belong? Equipment Equipment O2.2 Where does this account belong? Salary expense Salary expense O2.2 Where does this account belong? Fee Income Fee Income O2.2 Where does this account belong? J. Black, Drawing J. Black, Drawing O2.2 Where does this account belong? Unearned Revenue Unearned Revenue O2.2 Where does this account belong? Cost of Goods Sold Cost of Goods Sold O2.2 Where does this account belong? Cash Cash O2.2 Where does this account belong? Dividend Revenue Dividend Revenue O2.2 Objective 2.3: Concepts The following slides present 6 important concepts that guide many accounting practices and decisions. O2.3 (Monetary Unity) Concept Revenue Recognition Concept Concepts Objectivity Concept Unit of Measure (Monetary Unity) Concept Matching Concept Cost Concept Periodicity Concept Revenue Recognition Concept O2.3 (Monetary Unity) Concept Simply put, we must express accounts in monetary units such as dollars. Unit of Measure (Monetary Unity) Concept The monetary unit used is specific to the country in which business activity is conducted. \$ £ ¥ DM O2.3 Information collected in Accounting information should be separated into regular time periods such as months, quarters and years. Periodicity Concept Information collected in time periods allows comparisons that help us track financial progress. O2.3 For any accounting time period, all expenses that were incurred to earn revenues for that period must be matched (recorded as an expense) in that same time period. Matching Concept This concept is one that guides a great deal of accounting activity in the production of accrual financial statements O2.3 Historical cost is used to record values in the accounts (in most cases) rather than current market values or replacement costs Cost Concept This concept is part of the conservative approach to financial reporting that is required by FASB O2.3 When recording values where historical cost is not relevant, objective information is required rather than (subjective) values based on opinion or educated guesses. Objectivity Concept Objective information is provided by reliable, unbiased third parties, generally from market data O2.3 Revenue Recognition Concept Revenue should be recorded when it is earned and receivable. Record revenue when the seller has completed everything necessary to have the right to receive payment O2.3 What concept is involved? Tom buys 12 new kayaks to add to his inventory paying \$425 each from a dealer going out of business. Tom buys the same kayak from the manufacturer for \$675. Cost Concept What value should be recorded in the accounts for this purchase? \$425 each O2.3 What concept is involved? Thao wants to invest a 2 year old delivery van owned personally into her florist business. She saw two identical vans for sale at \$36,000. She paid \$47,000 for the van two years ago. Current published values for the van indicate a value of \$31,000. Objectivity Concept What value should be recorded in the accounts for this investment? \$31,000 O2.3 What concept is involved? It is important to express accounting information in a common denominator of units that is simple, reliable, universally available and understandable. Unit of Measure (Monetary Unity) Concept \$ £ ¥ DM O2.3 What concept is involved? There would be limited value in accounting information that continuously accumulated sales and expenses month after month, year after year without interruption. Periodicity Concept O2.3 What concept is involved? Tom sells a kayak for \$950 and records the revenue. What else should Tom record as a result of this sale? Matching Concept Tom should record the expense (Cost of Goods Sold) for the cost incurred to acquire this inventory item. O2.3 What concept is involved? Mark’s Hydraulic Repair receives a repair job from a credit customer on May 15, completes the job on May 18 and bills the customer. The customer picks up the repair on May 22. When can Mark’s record the revenue? Revenue Recognition Concept Mark’s can record the revenue on the day the work was completed, May 18 O2.3 Objective 2.4: Common size income statement Useful information can be obtained by expressing income statement items in terms of their percentage of sales. O2.4 A standard common size percentage calculated is net profit as a percentage of sales. This allows comparison from one period to the next 46,400/268,000 = 17% O2.4 End of Unit 2	2,846	11,585	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.21875	3	CC-MAIN-2018-22	latest	en	0.828605
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-5-section-5-1-polynomial-functions-and-models-5-1-assess-your-understanding-page-338/24	1,545,164,679,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376829568.86/warc/CC-MAIN-20181218184418-20181218210418-00232.warc.gz	916,964,633	12,414	## College Algebra (10th Edition) By definition, a polynomial is a function containing only terms where x is raised to a positive, integer power or constants. x is raised to the $\frac{1}{2}$ power, which is not an integer, so it is not a polynomial.	60	251	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2018-51	latest	en	0.913107
https://amp.doubtnut.com/question-answer/let-x1-2-3-4-5-6-7-8-9-let-r1-be-a-relation-on-x-given-by-r1x-y-x-y-is-divisible-by-3-and-r2-be-anot-1455604	1,591,383,565,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00428.warc.gz	239,395,201	21,093	IIT-JEE Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now. Click Question to Get Free Answers Watch 1 minute video This browser does not support the video element. Question From class 12 Chapter RELATIONS Let , Let be a relation on given by is divisible by 3} and be another relation on given by or or Show that . Let . Let be a relation in X given by is divisible by 3} and be another relation on X given by 6:23 Let Let R be a relation in given by is divisible by and another on given by Show that 6:23 Let be a continuous function such that Let and be the area of the region bounded by and the x-axis . Then (a) (b) (c) (d) 2:17 Consider the two circles let A be a fixed point on the circle ,say be a variable point on the circle . Theline BA meets the circle again at C. Then The maximum value of is 2:39 <br> . 1:16 Let <br> .Then is equal to 3:36 <br> 3:37 Find the value of x and y, when <br> (i) <br> (ii) <br> (iii) 5:21 In which of the following tables vary inversely: (i) x, 4, 3, 12, 1 y, 6, 8, 2, 24 (ii) x, 5, 20, 10, 4 y, 9, 12, 8, 36 (iii) x, 4, 3, 6, 1 y, 9, 12, 8, 36 (iv) x 9 24 15 3 y 8 3 4 25 3:11 Let and and 4:40 Let and 3:15 Let be real numbers satisfying and then the value of is 5:18 Latest Blog Post NCERT Alternative Academic Calendar for Classes 11 and 12 Released NCERT alternative academic calendar for classes 11 and 12 released. New alternative calendar help teachers on various technological & social media tools to teach students remotely. CBSE 2020: Know How to Change Exam Centre & Eligibility Criteria CBSE has released criteria for applying the change in the board exam centre. Know how to request for change, eligibility criteria, mandatory conditions & more. RBSE 2020 Date Sheet Released for Pending Exams of Class 10 & 12 RBSE 2020 date sheet released for pending exams of class 10 & 12. Exams will be conducted from 18 to 30 July for class 12, and 29 to 30 July 2020 for class 10 students. CISCE Board 2020: Class 10 & 12 Students are Allowed to Change Exam Centres CISCE board 2020 has allowed class 10 & 12 students to change exam centres. know how to apply for change in exam centres, admit card & result. Punjab Board Result 2020 for Class 10, 8 and 5 Announced Punjab board result 2020 for class 10, 8 and 5 announced. Know steps to download the PSEB result and other important details. BITSAT 2020 Exam to Be Held From August 6 -10 BITSAT 2020 exam to be held from August 6 -10. know the complete details regarding the BITSAT 2020 important dates, admit card, cutoff & result. MicroConcepts	776	2,573	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.828125	3	CC-MAIN-2020-24	latest	en	0.880253
https://www.scribd.com/document/237155361/Entropy-Analyses-for-Hyperbolic-Heat-Conduction-Based-on-the-Thermomass-Model	1,540,119,615,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00136.warc.gz	1,077,007,955	49,210	# Entropy analyses for hyperbolic heat conduction based on the thermomass model Y. Dong, Z.Y. Guo Key Laboratory for Thermal Science and Power Engineering of Ministry of Education, Department of Engineering Mechanics, Tsinghua University, Beijing 100084, China a r t i c l e i n f o Article history: Received in revised form 9 January 2011 Accepted 10 January 2011 Available online 19 February 2011 Keywords: Entropy Extended irreversible thermodynamics Hyperbolic heat conduction Thermomass model a b s t r a c t This paper is divided into three major sections with the ﬁrst one introducing the concept of generalized entropy in extended irreversible thermodynamics brieﬂy, that is, the entropy of a non-equilibrium sys- tem depend not only on the classical variables but also on the dissipative ﬂuxes, which makes the hyper- bolic equation of heat conduction based on the Cattaneo–Vernotte model compatible with the second law of thermodynamics. The second section deals with the hyperbolic heat conduction based on the thermo- mass model. According to the Einstein’s mass-energy relation, the phonon gas in dielectrics can be viewed as a kind of weighty compressible ﬂuid, and the momentum equation of the phonon (thermo- mass) gas in the dielectrics, which consists of the driving force, inertia and resistance of phonon (thermo- mass) gas, is just the damped thermal wave equation. In the third section our analyses show that the contribution of the kinetic energy of the phonon gas in the expression of extended entropy based on the thermomass model is identical with that of the heat ﬂux in the expression of generalized entropy in extended irreversible thermodynamics. It implies that the hyperbolic heat conduction based on the thermomass model is compatible with the second law of thermodynamics. 1. Introduction Heat conduction is generally described by the empirical law proposed by Fourier [1] in 1822, which relates linearly the heat ﬂux to the temperature gradient as q ¼ ÀkrT; ð1Þ where T is the temperature, q is the heat ﬂux and k is the thermal conductivity. The Fourier’s law shows well agreement with experi- ments for most practical problems. For fast-transient heat conduc- tion, however, the resulting differential equation for temperature predicts an inﬁnite speed of thermal signal propagation since the diffusion equation is parabolic by nature. On the other hand, with the rapid development of material processing by pulsed sources, Fourier’s law breaks down in modeling laser processing of materials [2,3] or high frequency response in IC chips [4] and heat propagates as wave. The problems caused by the inﬁnite speed of heat propagation have attracted many researchers to remove such a paradox induced by Fourier’s law. Cattaneo [5], Vernotte [6], and Morse and Feshbach [7] proposed a new heat ﬂux model, often termed as C-V model, to replace the Fourier’s law: q þs CV @q @t ¼ ÀkrT; ð2Þ where s CV is the relaxation time and t is time. Taking the divergence of Eq. (2) and combining the result with the energy equation yield s CV @ 2 T @t 2 þ @T @t ¼ ar 2 T; ð3Þ where a is the thermal diffusivity and all thermophysical properties are assumed constant. Eq. (3) is hyperbolic due to the additional term of the second or- der derivative of temperature with respect to time. The nonzero value of s CV makes the heat propagation speed ﬁnite [8]. Heat propagation thus evolves from a diffusion phenomenon to a wave one, with a ﬁnite speed of heat propagation (a/s CV ) 1/2 . Further extensions of the C-V model have also been proposed, such as the single-phase-lagging (SPL) model [9,10] and the dual-phase- lagging (DPL) model [9–11]. In addition, there are many other modern approaches dealing with the problem of heat conduction with ﬁnite speed, which have been reviewed in Ref. [12]. However, a new deep problem arises when the thermodynamic interpretation of the consequence of the C-V model, Eq. (2), is con- sidered, that is, the analyses of Eq. (2) in terms of classical irrevers- ible thermodynamics (CIT) leads to a non-deﬁnite positive value of entropy production. Hence, a new framework of irreversible ther- modynamics, called extended irreversible thermodynamics (EIT), is built [13–21,11], where the generalized entropy also depends on the heat ﬂux, and the deﬁnite positive value of entropy produc- tion can then be guaranteed during the temperature evolution of an isolated system approaching to equilibrium. doi:10.1016/j.ijheatmasstransfer.2011.01.011 Corresponding author. Tel.: +86 10 62782660; fax: +86 10 62783771. International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 Contents lists available at ScienceDirect International Journal of Heat and Mass Transfer j our nal homepage: www. el sevi er . com/ l ocat e/ i j hmt Recently, the general relationship between the temperature gradient and heat ﬂux has been derived based on the thermomass model [22–25], where the phonon gas in the dielectrics can be re- garded as a weighty and compressible ﬂuid according to the Ein- stein’s mass-energy relation. Hence, Newton’s mechanics has been applied to establish the equation of state and the equation of motion for the phonon gas as in ﬂuid mechanics, which is the transport equation of heat with a description of nonlinear and non- local effects. Cimmelli and his colleagues [26] made comparisons between the thermomass model with other phonon gas hydrody- namic models, such as the nonlinear Guyer–Krumhansl equation, which show an interesting and heuristic coincidence. The lineari- zation of momentum equations for the phonon gas leads to a damped thermal wave equation, which is similar to the wave equa- tion based on the C-V model but with a different characteristic time in both physical meaning and magnitude. This paper focuses on the compatibility of the hyperbolic heat conduction based on the thermomass model with the second law of thermodynamics through the comparison with the expression of the generalized en- tropy in EIT. 2. Generalized entropy in EIT [18] The basic concept of EIT is that the entropy of a non-equilibrium system depends not only on the classical variables, but also on the dissipative ﬂuxes, for example, the heat ﬂux in heat conduction problems. Namely the heat ﬂux is considered as an independent variable for describing the thermodynamic state of a system, across which heat is transferred. Then, the differential form of the gener- alized entropy is written as follows: ds EIT ¼ @s EIT @u du þ @s EIT @q Á dq; ð4Þ where s EIT is the generalized entropy in EIT, u is the speciﬁc internal energy, q is the heat ﬂux. In analogy with the classical theory, the non-equilibrium temperature is deﬁned and expanded around the inverse of the local-equilibrium temperature T: h À1 ðu; qÞ ¼ @s EIT =@u ¼ T À1 ðuÞ þa 10 ðuÞq 2 : ð4aÞ The remaining term in Eq. (4) is denoted as @s EIT =@q ¼ ÀT À1 va 10 ðu; qÞ; ð4bÞ where in the minus and the factor T À1 v are introduced for conve- nience. The coefﬁcient a 10 is determined through the comparison with the C-V model. In terms of some derivations with the assumption of the negli- gible second term on the right hand side of Eq. (4a), they obtain ds EIT ¼ T À1 du À s qkT 2 q Á dq; ð5Þ where s is the relaxation time and k the thermal conductivity. After the integration of Eq. (5) with s/qkT 2 assumed to be a constant, they ﬁnd an explicit expression for the generalized entropy outside equi- librium up to second-order terms in q: qs EIT ðu; qÞ ¼ qs eq ðuÞ À 1 2 s kT 2 q Á q: ð6Þ The corresponding entropy production is r s EIT ¼ 1 kT 2 q Á q: ð7Þ The time evolution of entropy in an isolated rigid body with an initial sinusoidal temperature proﬁle predicted by Eq. (6) is plotted in Fig. 1. It can be seen in Fig. 1 that the generalized entropy during the equilibrium has a monotonic increase and concluded that the hyperbolic heat conduction based on the C-V model is compatible with the second law of thermodynamics. For comparison the en- tropy production obtained from the classical thermodynamics, when C-V model is used instead of Fourier’s law, is r s CIT ¼ 1 kT 2 ðq Á q Àsq Á _ qÞ: ð8Þ It can be found clearly in Fig. 1 that the time evolution of the en- tropy exhibits a non-monotonic behavior, namely, the positiveness Nomenclature a thermal diffusivity (m 2 s À1 ) b dimensionless parameter in the thermomass model c speed of light (m s À1 ) C speciﬁc heat (J K À1 kg À1 ) f resistant force (N) k thermal conductivity (W m À1 K À1 ) l length parameter (m) in the thermomass model p pressure (Pa) P v stress tensor or viscous pressure tensor (Pa) q heat ﬂux (W m À2 ) Q heat (J) s speciﬁc entropy (J K À1 kg À1 ) S entropy (J K À1 ) T temperature (K) t time variable (s) u speciﬁc internal energy (J kg À1 ) u h drift velocity of phonon gas (m s À1 ) u hs thermal wave speed in the phonon gas (m s À1 ) U internal energy (J) v velocity (m s À1 ) w mechanical power per unit volume (J m À3 ) x space variable (m) Greek symbols b friction coefﬁcient for motion of phonon gas (kg s À1 ) c Grüneisen constant h non-equilibrium temperature (K) m speciﬁc volume (m 3 kg À1 ) q density (kg m À3 ) r s rate of entropy production per unit volume (J K À1 s À1 m À3 ) s characteristic time (s) Subscripts CIT classical irreversible thermodynamics CV Cattaneo–Vernotte model TIN extended irreversible thermodynamics eq Equilibrium h thermomass (phonon gas) NV thermomass model Superscripts . time derivative À average Y. Dong, Z.Y. Guo / International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 1925 of the entropy production is no longer guaranteed due to the pres- ence of the second term on the right hand side of Eq. (8). An illustrative example was also presented in Ref. [17] to show the different behaviors of the classical entropy and the extended entropy. Let’s consider the heat conduction between two rigid bodies at different temperatures, T 1 and T 2 (T 1 > T 2 ) respectively, which are isolated from the environment and separated by an adi- abatic wall. Heat conduction occurs without work being per- formed, as the adiabatic constraint is removed. According to CIT, the time rate variation of the total entropy gives dS dt ¼ dS 1 dt þ dS 2 dt ¼ T À1 1 dU 1 dt þ T À1 2 dU 2 dt ¼ T À1 1 À T À1 2 _ Q; ð9Þ where dU 1 = qVC 1 dT 1 , dU 2 = qVC 2 dT 2 , with C 1 , C 2 being the speciﬁc heat capacities of the respective subsystems. The combination of Fourier’s law with the conservation equation of thermal energy leads to the time variation of the temperature difference, DT = T 1 À T 2 , as dDT dt ¼ ÀK 0 C eff DT; ð10Þ with K 0 = KT À2 assumed to be a constant and T an intermediate tem- perature of T 1 and T 2 . In view of Eqs. (9) and (10) and if DT << T 2 , the rate of evolution of entropy may be written in terms of DT as dS dt ¼ ÀT À2 C eff dðDTÞ 2 2dt : ð11Þ It is seen that the entropy is a monotonically increasing func- tion of time. If the C-V model is used, the evolution of the temper- ature difference is s d 2 DT dt 2 þ dDT dt þ K 0 C eff DT ¼ 0: ð12Þ The equation is similar to the equation of motion of a damped pendulum. The decay of DT may exhibit an oscillatory behavior when 4sK 0 C eff > 1. In the case of oscillatory decay of DT, the classi- cal entropy behaves a non-monotonic function of time according to Eq. (8), as shown in Fig. 1 (dashed curve). However, since the pro- posed form of the generalized entropy in EIT depends also on the heat ﬂux, the rate of variation of the generalized entropy becomes dS EIT dt ¼ T À1 1 À T À1 2 _ Q À s K _ Q d _ Q dt ; ð13Þ or, in an integrated form, S EIT ¼ S 1 ðU 1 Þ þ S 2 ðU 2 Þ À s 2K _ Q 2 : ð14Þ Eq. (13) can then be rewritten as dS EIT dt ¼ K À1 _ Q 2 ¼ s K C 2 eff dDT dt 2 : ð15Þ This expression is never negative, whose evolution increases mono- tonically as shown in Fig. 1 (solid curve), and the generalized entro- py in EIT is thus suitable as an expression for the second law in hyperbolic heat conduction. Moreover, the expression for extended entropy has been reviewed and commented by F. X. Alvarez, et al [19], who indicated that through several other macroscopic and microscopic approaches, the expression (5) and (6) could also be obtained when the CV model (2) is valid. 3. Thermomass (TM) Model The concept of thermomass comes from Einstein’s theory of special relativity, indicating that thermal energy (heat) can be re- ferred to as its equivalent mass [22–25]. A few words should be gi- ven to make this concept clearer: It is widely accepted that the rest mass (or invariant mass) of a system composed of a number of freely moving particles is greater than the sum of the rest masses of the individual particles[27–31]. The thermal energy includes both kinetic and potential energy of the particles, however, as long as the system could be taken as a whole, its rest mass is greater than the sum of the masses of the atoms contained in this system, by the amount of thermal energy divided by the square of speed of light in vacuum. The term ‘equivalent’ is worth discussion. More precisely, it can be referred as the additional rest mass—or just ‘mass’ according to Okun’s point of view [28]—to the system con- tributed by thermal energy, and is invariant in every inertial frame. Hence, the density of a phonon gas in the dielectrics is q h ¼ qCT c 2 : ð16Þ where q h is the mass density of the phonon gas, q is the density of solid, C the heat capacity, c the speed of light, and T the tempera- ture. The state equation of a phonon gas can be deduced from the Debye’s state equation as p h ¼ cq h CT ¼ cqðCTÞ 2 c 2 ; ð17Þ where c is Grüneisen constant, and p h the thermal pressure of the equivalent mass of phonon gas. Thus, the phonon gas in the dielec- trics could be treated as a kind of weighty ﬂuid with ultra-small mass (at room temperature) proportional to its thermal energy. In view of the fact that the rest mass of lattice acts as a porous frame- work, heat conduction (phonon gas ﬂow) in dielectrics resembles the gas ﬂow in a porous medium. Since the drift velocity of phonon gas is very small relative to the lattice frame, the mass from thermal energy satisﬁes the Newton’s law of motion as long as the lattice frame travels slowly in the inertial reference frame. Therefore, the continuity and momentum equations for a phonon gas can be writ- ten as in ﬂuid mechanics, @q h @t þ r Á ðq h u h Þ ¼ 0; ð18Þ @ðq h u h Þ @t þq h ðu h rÞ Á u h þ rp h þf h ¼ 0; ð19Þ where q h , u h , p h are density, drift velocity and pressure of phonon gas respectively. The velocity of heat motion or the drift velocity of phonon gas can be extracted from the quantity of heat ﬂux S t Classical Entropy Extended Entropy Fig. 1. The evolution of the classical equilibrium entropy S CIT and the extended entropy S EIT . 1926 Y. Dong, Z.Y. Guo / International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 u h ¼ q qCT ; ð20Þ where q is the density of dielectrics and qCT holds for the energy of the phonon gas per unit volume. The resistant force is linearly re- lated to the phonon gas velocity, if u h is not very large, f h ¼ bu h ; ð21Þ with b = 2cC(qCT) 2 /(kc 2 ). The insertion of Eqs. (18), (20), (21) into Eq. (19) gives the con- stitutive equation for the phonon gas in dielectrics s TM @q @t À lqC @T @t þ l @q @x À bk @T @x þ k @T @x þ q ¼ 0; ð22Þ where s TM ¼ k 2cqC 2 T ; l ¼ qk 2cCðqCTÞ 2 ¼ u h s TM ; b ¼ q 2 2cq 2 C 3 T 3 ¼ Ma 2 h : ð23Þ with s TM being the lagging time, l a length parameter, Ma h the Mach number of the drift velocity, u h relative to the thermal wave speed in the phonon gas, u hs . The ﬁrst three terms on the left-hand side of Eq. (22) result from the inertia effect, the fourth term represents the effect from the pressure gradient (driving force), and the last term denotes the resistance as the phonon gas ﬂows through the lattices. So we may rewrite Eq. (22) as s TM @q @t þ2u h s TM @q @x À Ma 2 h k @T @x þ k @T @x þ q ¼ 0: ð24Þ Eliminating heat ﬂux from Eqs. (18) and (24) results in an equation containing temperature, T, alone s TM @ 2 T @t 2 þ2u h s TM @ 2 T @x@t þ aMa 2 h @ 2 T @x 2 À a @ 2 T @x 2 þ @T @t ¼ 0: ð25Þ where a is the thermal diffusivity. For Ma h << 1, the second and third terms on the left hand side of Eqs. (24) and (25), which denote the inertia effect induced by heat ﬂux variation in space, are ne- glected and the constitutive Eq. (20) then reduces to the C-V like model, called the simpliﬁed thermomass model [23]: s TM @q @t þ k @T @x þ q ¼ 0 or @q @t þ k s TM @T @x þ q s TM ¼ 0: ð26Þ Correspondingly, Eq. (25) reduces to a damped wave equation s TM @ 2 T @t 2 À a @ 2 T @x 2 þ @T @t ¼ 0 or @ 2 T @t 2 À a s TM @ 2 T @x 2 þ 1 s TM @T @t ¼ 0: ð27Þ Furthermore, for s TM << t, the ﬁrst term on the left hand side of Eq. (26), which represents the inertia effect induced by the time variation of heat ﬂux, can be neglected, the constitutive Eq. (26) and the damped wave Eq. (27) then further reduces to the Fourier’s law and heat diffusion equation respectively as follows k @T @x þ q ¼ 0; ð28Þ and a @ 2 T @x 2 À @T @t ¼ 0: ð29Þ On the contrary, for s TM >> t, for example, for cases of very high con- ductivity or very low temperature, the second term in Eq. (26), which measures the effect of resistance, is negligible, so that the constitutive Eq. (26) and the damped wave Eq. (27) reduces respec- tively to @q @t þ k s TM @T @x ¼ 0 ð30Þ and @ 2 T @t 2 À a s TM @ 2 T @x 2 ¼ 0 or @ 2 T @t 2 À u hs @ 2 T @x 2 ¼ 0: ð31Þ Eq. (30) is the balance equation between the driving force and the inertial force, and Eq. (31) is the thermal wave equation without damping effect. It is worth noting that the thermomass model, Eq. (22), is similar to the C-V model but with a different characteristic time. The characteristic time s CV in the C-V model is the relaxation time as the energy carriers approaching the thermodynamic equi- librium, while the characteristic time s TM in the thermomass model, on the other hand, describes a lagging time between the heat ﬂux larger than the relaxation time of phonons in the CV-wave model as reported in [23]. 4. Extended entropy based on the TM model 4.1. Extended entropy In view of the fact that a phonon gas is a weighty, compressible ﬂuid, the ﬂowing phonon gas, like a dense gas in motion, is out of mechanical non-equilibrium, rather than thermal equilibrium. Hence, the expressions for the temperature and internal energy in CIT still hold and we need to identify the contribution of kinetic energy to the entropy of phonon gas. If the resistance in the momentumEq. (19) can be neglected, we have Bernoulli’s equation for the inviscid phonon gas u h du h þ dp h q h ¼ 0: ð32Þ The substitution of Eq. (17) into Eq. (32) gives CdT þ 1 2c u h du h ¼ 0: ð33Þ Note that the ﬁrst term represents the energy of thermomass rather than the internal energy, with the parameter 2cCT/c 2 eliminated for convenience. The integrative form of Eq. (33) is CT þ 1 4c u 2 h ¼ CT 0 or CT ¼ CT 0 À 1 4c u 2 h ; ð34Þ which provides the relation of reversible conversion between the potential and kinetic energy of phonon gas. The kinetic energy of a system and the work done by the external force on the system are equal in value. Hence, the kinetic energy divided by tempera- ture represents the negative entropy of phonon gas in motion. We then have the differential form of the entropy for the phonon gas in motion ds EIT ¼ du T À 1 2c u h du h T : ð35Þ Its integration is s EIT ¼ C lnT À 1 4c u 2 h : ð36Þ Substituting Eqs. (20) and (23) into Eqs. (35) and (36) leads to ds EIT ¼ du T À s qkT 2 q Á dq; ð37Þ and qs EIT ¼ qs eq À 1 2 s qkT 2 q Á q: ð38Þ It can be found clearly that the expressions for the extended entro- py based on the thermomass model, Eqs. (35) and (36), are identical Y. Dong, Z.Y. Guo / International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 1927 with those for the generalized entropy, Eqs. (5) and (6), respectively. Consider nowonce again the illustrative example [17] discussed in the preceding section from the viewpoint of the thermomass model. We have two rigid bodies at different temperatures, T 1 and T 2 (T 1 > T 2 ), which are isolated from the environment and sep- arated by an adiabatic wall. Heat conduction occurs between two rigid bodies without work interaction with the environment, as the adiabatic constraint is removed. The conservation equation for heat (thermomass) is _ Q ¼ À dU 1 dt ¼ dU 2 dt ; ð39Þ dDT dt ¼ dðT 1 À T 2 Þ dt ¼ À 2 _ Q qVC for C 1 ¼ C 2 ¼ C ð40Þ The heat ﬂow can be expressed in terms of the drift velocity of pho- non gas [19]: _ Q ¼ S _ q ¼ SqCTu h : ð41Þ The simpliﬁed thermomass model, Eq. (26), for heat conduction be- tween two rigid bodies gives s TM d _ Q dt þ _ Q ¼ gðT 1 À T 2 Þ; ð42Þ where s TM ¼ k=2cqC 2 T, T ¼ ðT 10 þ T 20 Þ=2, g ¼ kA=L, A is the surface area, and L the characteristic length. The time rate of variation of the extended entropy is dS EIT dt ¼ T À1 1 À T À1 2 _ Q À qV 2c u h du h T ; ð43Þ or, in an integrated form, S EIT ¼ S 1 ðU 1 Þ þ S 2 ðU 2 Þ À qV 2 u 2 h : ð44Þ Substituting Eqs. (20) and (23) into Eqs. (43) and (44) leads to dS EIT dt ¼ T À1 1 À T À1 2 _ Q À s K _ Q d _ Q dt ; ð45Þ and S EIT ¼ S 1 ðU 1 Þ þ S 2 ðU 2 Þ À s 2K _ Q 2 : ð46Þ Eqs. (45) and (46) are identical with Eqs. (10) and (11). This implies that the hyperbolic heat conduction on the basis of the thermomass model obeys the second law of thermodynamics. 4.2. Entropy production The entropy production per unit volume and time for viscous gas ﬂows is the production of the generalized thermodynamic force and ﬂux, which equals to the dissipation rate of mechanical power per unit volume divided by temperature [32,33] r s ¼ _ w=T; ð47Þ where _ w is the dissipation rate of mechanical power per unit vol- ume. For instance, in the classical nonequilibrium thermodynamics the entropy production rate caused by the viscous ﬂow is expressed as [32] r s ¼ 1 T ðÀP v : r Á vÞ; ð48Þ where P v is the viscous pressure tensor and v is the velocity. Since heat conduction resembles the gas ﬂow in a porous med- ium based on the thermomass model, the entropy production per unit volume and time in porous dielectrics can be expressed by the dissipated energy of thermomass per unit volume and time di- vided by temperature r s ¼ f h u h =T: ð49Þ The insertion of Eqs. (20), (21) into Eq. (49) yields r s ¼ bu 2 h =T ¼ 2cCðqCTÞ 2 kc 2 u 2 h ¼ 2cC kc 2 q 2 T ¼ 2cCT c 2 q 2 kT 2 : ð50Þ It can be seen that the positiveness of entropy production rate is deﬁnite, which divided by the parameter, 2cCT/c 2 , eliminated in Eq. (33) is exactly the same as Eq. (7) in EIT. 5. Concluding remarks (1) In view of the fact that heat conduction is essentially the ﬂow of a weighty compressible ﬂuid in a porous medium, the ﬂowing phonon gas in a dielectric system, like a dense gas in motion, is still at thermally local equilibrium, but deviates its mechanical equilibrium. (2) The drift velocity of phonon gas can be viewed as an inde- pendent variable in the thermomass model, so that the extended entropy of a dielectric system depends not only on the potential energy (internal energy), but also on the kinetic energy of phonon gas. The ratio of the kinetic energy to the temperature is equivalent to the negative entropy of phonon gas in motion. (3) The expression for the entropy based on the thermomass model for the system out of mechanical equilibrium is iden- tical with that for the generalized entropy in extended irre- versible thermodynamics. This implies that the hyperbolic heat conduction based on the thermomass model is compat- ible with the second law of thermodynamics as well. (4) The entropy production for hyperbolic heat conduction can be directly expressed by the production of the thermody- namic force and thermodynamic ﬂux for the ﬂowing phonon gas based on the thermomass model, which is proportional to the heat ﬂux, and its positiveness can then be guaranteed. References [1] J. Fourier, Analytical Theory of Heat, Dover Publications, New York, 1955. [2] B. Stritzker, A. Pospieszczyk, J.A. Tagle, Measurement of lattice temperature of silicon during pulsed laser annealing, Phys. Rev. Lett. 47 (1981) 356–358. [3] S. Torii, W.J. Yang, Heat transfer mechanisms in thin ﬁlm with laser heat source, Int. J. Heat Mass Transfer 48 (2005) 537–544. [4] Z.Y. Guo, Y.S. Xu, Non-Fourier heat conduction in IC chip, J. Electron. Packaging 117 (3) (1995) 174–177. [5] C. Cattaneo, Sulla Conduzione del Calore, Atti Sem. Mat. Fis. Univ. Modena 3 (1948) 83–101. [6] P. Vernotte, Paradoxes in the continuous theory of the heat equation, C. R. [7] P.M. Morse, H. Feshbach, Methods of Theoretical Physics, McGraw-Hill, New York, 1953. [8] D.D. Joseph, L. Preziosi, Heat Waves, Rev. Mod. Phys. 61 (1) (1989) 41–73. [9] D.Y. Tzou, On the thermal shock wave induced by a moving heat source, J. Heat Transfer 111 (1989) 232–238. [10] D.Y. Tzou, Thermal Shock Phenomena Under High-rate Response in Solids, in: C.L. Tien (Ed.), Annual Review of Heat Transfer IV, Hemisphere, Washington DC, 1992, pp. 111–185. Chapter 3. [11] D.Y. Tzou, Macro- to Microscale Heat Transfer: The Lagging Behavior, Taylor & Francis, Washington, DC, 1997. [12] V.A. Cimmelli, Different thermodynamic theories and different heat conduction laws, J. Non Equilib. Thermodyn. 34 (2009) 299–333. [13] D. Jou, J. Casas-Vazquez, G. Lebon, Extended irreversible thermodynamics revisited (1988–98), Rep. Pro. Phys. 62 (1999) 1035–1142. [14] I. Müller, Thermodynamics, Pitman, London, 1985. [15] S. Sieniutycz, P. Salamon, Extended Thermodynamic System, Taylor and Francis, New York, 1992. [16] M. Criado-Sancho, J.E. Llebot, Behavior of entropy in hyperbolic heat conduction, Phys. Rev. E 47 (1993) 4104–4107. [17] A. Barletta, E. Zanchini, Hyperbolic heat conduction and local equilibrium: a second law analysis, Int. J. Heat Mass Transfer 40 (1997) 1007–1016. 1928 Y. Dong, Z.Y. Guo / International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 [18] D. Jou, J. Casas-Vazquez, G. Lebon, Extended irreversible thermodynamics, 3rd ed., Springer, Berlin, 2001. [19] J. Casas-Vazquez, F.X. Alvarez, D. Jou, Robustness of the nonequilibrium entropy related to the Maxwell–Cattaneo heat equation, Phys. Rev. E 77 (2008) 031110. [20] D. Jou, J. Casas-Vazquez, G. Lebon, Extended irreversible thermodynamics of heat transport-A brief introduction, Proc. Estonian Acad. Sci. 57 (2008) 118– 126. [21] I. Müller, T. Ruggeri, Rational Extended Thermodynamics, Springer, New York, 1997. [22] B.Y. Cao, Z.Y. Guo, Equation of motion of a phonon gas and non-Fourier heat conduction, J. Appl. Phys. 102 (5) (2007) 53503–53506. [23] Z.Y. Guo, Q.W. Hou, Thermal wave based on the thermomass model, ASME J. Heat Transfer 132 (2010) 072403. [24] R.F. Hu, B.Y. Cao, Study on thermal wave based on the thermal mass theory, Sci. China Series E Technol. Sci. 52 (6) (2009) 1786–1792. [25] D.Y. Tzou, Z.Y. Guo, Non-local behavior in thermal lagging, Int. J. Thermal Science 49 (2010) 1133–1137. [26] V.A. Cimmelli, A. Sellitto, D. Jou, Nonlinear evolution and stability of the heat ﬂow in nanosystems: beyond linear phonon hydrodynamics, Phys. Rev. B 82 (2010) 184302. [27] E.F. Taylor, J.A. Wheeler, Spacetime Physics, W.H. Freeman and Company, New York, 1966. [28] L.B. Okun, The Concept of Mass, Phys. Today 42 (1989) 11–16. [29] M. Jammer, Concepts of Mass in Contemporary Physics and Philosophy, Princeton University Press, Princeton, 2009. [30] I. Müller, A History of Thermodynamics: The Doctrine of Energy and Entropy, Springer, Berlin, 2007. pp. 35. [31] A. Einstein, L. Inﬁeld, The Evolution of Physics: The growth of ideas from early concepts to relativity and quanta, Simon and Schuster, New York, 1938. [32] H.J. Kreuzer, Nonequilibrium Thermodynamics and its Statistical Foundations, Clarendon Press, Oxford, 1981. [33] A. Bejan, Entropy generation minimization: the method of thermodynamic optimization of ﬁnite-size systems and ﬁnite-time processes, CRC Press, Boca Raton, 1996. Y. Dong, Z.Y. Guo / International Journal of Heat and Mass Transfer 54 (2011) 1924–1929 1929	9,101	28,092	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2018-43	latest	en	0.907467
https://de.mathworks.com/matlabcentral/cody/problems/2891-binary-code-array/solutions/3379758	1,606,478,978,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00653.warc.gz	238,371,371	16,663	Cody # Problem 2891. Binary code (array) Solution 3379758 Submitted on 25 Oct 2020 by cokakola This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass x = 2015; y_correct=[1,1,1,1,1,0,1,1,1,1,1]; assert(isequal(dectobin(x),y_correct)) 2 Pass x = 13; y_correct=[1,1,0,1]; assert(isequal(dectobin(x),y_correct)) 3 Pass x = 143; y_correct=[1,0,0,0,1,1,1,1]; assert(isequal(dectobin(x),y_correct)) ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	214	642	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2020-50	latest	en	0.677834
https://www.coursehero.com/file/109259/Statistics-Homework-Ch-2/	1,516,106,050,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00370.warc.gz	897,446,426	27,122	Statistics Homework Ch 2 # Statistics Homework Ch 2 - (Actual Low Frequency Class... This preview shows pages 1–2. Sign up to view the full content. Kate-Lynn Tetreault MAT 120 Statistics 4 6 12 20 Class Frequency 0.00-0.20 2 0.21-0.41 0 0.42-0.82 8 .83-1.03 8 1.04-1.24 8 1.25-1.45 3 1.46-1.66 0 1.67-1.87 0 22 Class Frequency 39-45 4 46-52 9 53-59 14 60-66 7 67-73 1 The advantage of using relative frequency distributions instead of frequency distributions is that you can compare them much easier Class midpoint- (2.99+0.00)/2=1.495 Class width- (2.99- 0.00)/6=0.498333333 Class boundaries-(2.99- 2.50)=0.49 The gap between the lowest frequency and the highest frequency in excersize 8 may suggest that there is data from two different populations. 0.00-0.20 0.42-0.82 1.04-1.24 1.46-1.66 0 1 2 3 4 5 6 7 8 9 Frequency Distribution Frequency Class Frequency 39-45 46-52 53-59 60-66 67-73 0 2 4 6 8 10 12 14 16 Forecast and Actual Temperatures This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: (Actual Low) Frequency Class Kate-Lynn Tetreault MAT 120 Statistics Class Frequency 0.00-0.2 30 0.3-0.5 1 0.6-0.8 1 0.9-1.1 1.2-1.4 1.5-1.7 1 1.8-2.0 1 2.1-2.3 2.4-2.6 2.7-2.9 1 10 Class Frequency 90-100 2 101-111 3 112-122 10 123-133 14 134-144 9 145-155 2 12 Class Frequency 90-99 2 100-109 3 110-119 10 120-129 14 130-139 7 140-149 2 150-159 2 22 ? 24 ? 0.00-0.2 0.6-0.8 1.2-1.4 1.8-2.0 2.4-2.6 5 10 15 20 25 30 35 Forecast and Actual Temperatures (Precipitation) Frequency Class Frequency 2 4 6 8 10 12 2 4 6 8 10 12 Old Faithful Geyser (height) Height 2 4 6 8 10 12 2 4 6 8 10 12 Geyser Heights Class... View Full Document {[ snackBarMessage ]} ### Page1 / 2 Statistics Homework Ch 2 - (Actual Low Frequency Class... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	857	2,024	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2018-05	latest	en	0.334976
techfranklin.com	1,701,810,117,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00527.warc.gz	646,634,451	5,704	#### Next blog: Episode 6 Demonstration of TechFranklin's dilution modelling tool under a variety of Captables and valuations. Check out our Recall the Captable from the previous video. Person Quantity Founder 4 Angel Investor 2 Person Quantity Exercise Price Employee 2 £100 For the purposes of this illustration we will assume the Ordinary and Ordinary A shares rank pari passu (which is just a silly latin way of saying equally) on a sale. ##### £300 equity value Suppose the company is sold for £300. This means all the equity has a value of £300. Therefore the value per share is £300/6 (number of shares in issue) = £50 The employee optionholder gets nothing because £50 is less than the exercise price of £100, so there is no point exercising the option. So the proceeds are distributed as follows: • Founder receives 4 x £137.50 = £550 • Angel Investor receives 2 x £137.50 = £275 Person Quantity Proceeds Founder 4 £200 67% Angel Investor 2 £100 33% Person Quantity Exercise Price Proceeds Employee 2 £100 £0 0% ##### £900 equity value Suppose instead the company is sold for £900. Now each share is worth £150 (£900/6) before the exercise of the option. But this time the employee exercises their option. So they put £100 x 2 = £200 into the company and are issued with 2 shares. Now the proceeds to be distributed are £900 (from the sale of the company) + £200 (from the exercise of the option) = £1,100. But now there are 6 + 2 = 8 shares in issue. Therefore each share is worth £1,100/8 = £137.50. Now the proceeds are distributed as follows: • Founder receives 4 x £137.50 = £550 • Angel Investor receives 2 x £137.50 = £275 • Employee optionholder receives £75. Reflecting £275 from their shares less £200 from the exercise of the option Person Quantity Proceeds Founder 4 £550 61% Angel Investor 2 £275 31% Person Quantity Exercise Price Proceeds Employee 2 £100 £75 8% ##### Things to notice • Notice how the value below £100 per share is protected for the Founder and the Investor by the exercise price of £100 • Notice how the optionholder only benefits from value above £100 per share • Notice how £550 + £275 + £75 = £900, the value of the company ##### Key takeaway If you learn only one thing from this blog, I want you to understand this absolutely key concept: For an option, the exercise price is just as important as the number of shares under option ##### More complicated cases This example was pretty simple. But how do you deal with Captables with 20+ shareholders and 20+ optionholders all with different exercise prices? Which is what Captables often end up looking like at a sale or IPO. It turns out the computations are actually quite hard and very difficult to do properly in Excel. Because of this we have built TechFranklin's which allows you to scenario model any combination of shareholders, optionholders and equity prices to understand how the proceeds are distributed on a sale. Check it out. It is to use. #### Next blog: Episode 6 Demonstration of TechFranklin's dilution modelling tool under a variety of Captables and valuations.	768	3,108	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2023-50	longest	en	0.920705
https://www.physicsforums.com/threads/momentum-and-collision-related-problem.956745/	1,586,003,499,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370521876.48/warc/CC-MAIN-20200404103932-20200404133932-00090.warc.gz	1,095,235,437	21,471	# Momentum and collision related problem Member advised to use the formatting template for all homework help requests 25.0-kg dog is trapped on a rock in the middle of a narrow river. A 66.0-kg rescuer has assembled a swing with negligible mass that she will use to swing down and catch the trapped dog at the bottom of her swing, and then continue swinging to the other side of the river. The ledge that the rescuer swings from is 5.0 m above the rock, which is not high enough so the rescuer and dog together can reach the other side of the river, which is 3.0 m above the rock. However, the rescuer can use a ladder to increase the height from which she swings. What is the minimum height of the ladder the rescuer must use so both dog and rescuer make it to the other side of the river? Assume that friction and air resistance are negligible. So far I've tried: mrg(h1+hl = (mr+md)gh2 but it seems to be giving a negative height for the ladder #### Attachments • 44.1 KB Views: 274 • 39.6 KB Views: 564 Last edited by a moderator: Related Introductory Physics Homework Help News on Phys.org BvU Homework Helper 2019 Award Hello Finn, Please use the template provided -- as described in the guidelines its use is mandatory but it seems to be giving a negative height for the ladder Hello Finn, Please use the template provided -- as described in the guidelines its use is mandatory hl = (((mr+md)h2)/mr)-h1 = (((66+25)3)/66)-5 = -0.8636 jbriggs444 Homework Helper 2019 Award hl = (((mr+md)h2)/mr)-h1 = (((66+25)3)/66)-5 = -0.8636 Can you describe in words, what you are basing that equation on? It looks like a conservation of energy argument. But is energy conserved? BvU Homework Helper 2019 Award Well, if the outcome is somewhat unexpected, then perhaps there is something amiss in the assumptions ... What were they ? (in other words: what is the assumption in the expression you used ?) Well, if the outcome is somewhat unexpected, then perhaps there is something amiss in the assumptions ... What were they ? (in other words: what is the assumption in the expression you used ?) Can you describe in words, what you are basing that equation on? It looks like a conservation of energy argument. But is energy conserved? I think that energy is conserved based on my assumption but I'm not sure. Last edited: jbriggs444 Homework Helper 2019 Award I think that energy is conserved based on my assumption but I'm not sure. Have you covered elastic and inelastic collisions in your coursework? If a rescuer collides with a dog and both move off together, which kind is it? Have you covered elastic and inelastic collisions in your coursework? If a rescuer collides with a dog and both move off together, which kind is it? Inelastic? But I'm not sure which equations to use that could help me find the height. I was thinking m1v1i = (m1+m2)V jbriggs444 Homework Helper 2019 Award Inelastic? But I'm not sure which equations to use that could help me find the height. I was thinking m1v1i = (m1+m2)V It would be helpful if you could add some words instead of simply writing down an equation. It seems that you are using conservation of momentum for the event of the collision. That is a good move. If you know the rescuer's velocity prior to the event, you can use that equation to determine the rescuer+dog's velocity after. [It is also helpful to define your variable names and what they mean. What is obvious to you may not be as obvious to your reader] Now you just have to figure out v1i (the velocity of the rescuer as he arrives at the dog). Perhaps a conservation of energy approach would work... It would be helpful if you could add some words instead of simply writing down an equation. It seems that you are using conservation of momentum for the event of the collision. That is a good move. If you know the rescuer's velocity prior to the event, you can use that equation to determine the rescuer+dog's velocity after. [It is also helpful to define your variable names and what they mean. What is obvious to you may not be as obvious to your reader] Now you just have to figure out v1i (the velocity of the rescuer as he arrives at the dog). Perhaps a conservation of energy approach would work... So how would I connect finding v1i to the height? Should I use KE = mv2/2 and GPE = mgh? jbriggs444 Homework Helper 2019 Award So how would I connect finding v1i to the height? Should I use KE = mv2/2 and GPE = mgh? Yes, that is what I would do. Yes, that is what I would do. Yes, that is what I would do. Can I assume that mv2/2 = mgh and v is v1i? jbriggs444 Homework Helper 2019 Award Can I assume that mv2/2 = mgh? At the risk of repeating myself: use your words. For example... You are invoking conservation of energy. No mechanical energy is lost as the rescuer swings from his starting perch to the position of the dog. We can justify that because the tension of the rope is at right angles to his path (no motion in direction of force means no work is done by the force) and because we assume that air resistance is negligible. That means that ending mechanical energy is equal to starting mechanical energy. The starting mechanical energy consists of kinetic energy and gravitational potential energy. We can write down an equation for that in terms of the initial velocity $v_i$ and the initial height $h_i$ $$E_{i}=\frac{1}{2}mv_i^2 + mgh_i$$ The ending mechanical energy consists of kinetic energy and gravitational potential energy. We can write down an equation for that in terms of the final velocity $v_f$ and the final height $h_f$. $$E_{f}=\frac{1}{2}mv_f^2 + mgh_f$$ But we already know that $E_i=E_f$ so we can equate the two and rearrange terms: $$\frac{1}{2}mv_f^2 + mgh_f = \frac{1}{2}mv_i^2 + mgh_i$$ $$\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mg(h_i-h_f)$$ But the initial velocity is zero. And we can choose coordinates where the final height is 0. If we then call the initial height "h", and the final velocity "v", this becomes: $$\frac{1}{2}mv^2 = mgh$$ So you do not need to assume that mv2/2 = mgh. You can demonstrate it for the situation at hand. At the risk of repeating myself: use your words. For example... You are invoking conservation of energy. No mechanical energy is lost as the rescuer swings from his starting perch to the position of the dog. We can justify that because the tension of the rope is at right angles to his path (no motion in direction of force means no work is done by the force) and because we assume that air resistance is negligible. That means that ending mechanical energy is equal to starting mechanical energy. The starting mechanical energy consists of kinetic energy and gravitational potential energy. We can write down an equation for that in terms of the initial velocity $v_i$ and the initial height $h_i$ $$E_{i}=mv_i^2 + mgh_i$$ The ending mechanical energy consists of kinetic energy and gravitational potential energy. We can write down an equation for that in terms of the final velocity $v_f$ and the final height $h_f$. $$E_{f}=mv_f^2 + mgh_f$$ But we already know that $E_i=E_f$ so we can equate the two and rearrange terms: $$mv_f^2 + mgh_f = mv_i^2 + mgh_i$$ $$mv_f^2 - mv_i^2 = mg(h_i-h_f)$$ But the initial velocity is zero. And we can choose coordinates where the final height is 0. If we then call the initial height "h", and the final velocity "v", this becomes: $$mv^2 = mgh$$ So you do not need to assume that mv2 = mgh. You can demonstrate it for the situation at hand. Ohhh, okay. So I'm doing first: vf2 = sqrt(2gh) = sqrt(2(9.8)(3)) = 7.7 for finding the velocity when they land at the other side. Then use V = 7.7: v1i = ((m1+m2)V)/m1 = ((66+25)/7.7)/66 = 10.6 for finding the velocity just before the rescuer reaches the dog. Then use v = 10.6 h = v2/2g = (10.6)2/2(9.8) = 5.7 So the ladder height should be 5.7 - 5 = 0.7? jbriggs444 Homework Helper 2019 Award So I'm doing first: vf2 = sqrt(2gh) = sqrt(2(9.8)(3)) = 7.7 for finding the velocity when they land at the other side. So you are working backward. You know they land together on a 3 meter high platform with zero velocity. So they must have swung upward with a starting velocity of 7.7 meters per second^2 just after the pick-up. [You are working with numbers. It is usually better to stick with algebra and variable names until the very end and plug in numbers as a final step. It makes errors in the algebra easier to spot and helps you see where things cancel. In particular, I believe you will find that g cancels out. The two places where you multiplied or divided by 9.8 were wasted arithmetic] Then use V = 7.7: v1i = ((m1+m2)V)/m1 = ((66+25)/7.7)/66 = 10.6 for finding the velocity just before the rescuer reaches the dog. That looks good as well. Then use v = 10.6 h = v2/2g = (10.6)2/2(9.8) = 5.7 So the ladder height should be 5.7 - 5 = 0.7? Yes. Without carefully checking the arithmetic, it all looks correct. haruspex Homework Helper Gold Member I believe you have found the intended answer, but v1i = ((m1+m2)V)/m1 = ((66+25)/7.7)/66 = 10.6 for finding the velocity just before the rescuer reaches the dog. That's a collision speed of 38kph. Perhaps not the wisest of rescue attempts. neilparker62 Homework Helper Alternate method: Energy loss during "collide and coalesce" collision given by $$½μΔv^2$$ where $$μ = \frac{m_1m_2}{m_1+m_2}$$ and Δv is the relative velocity between the colliding objects - in this case equal to $\sqrt{2gh_1}$. Thus $$m_1gh_1-½ \frac{m_1m_2}{m_1+m_2}\times 2gh_1=(m_1+m_2)gh_2$$. From which $h_1$ can be determined.	2,588	9,589	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2020-16	longest	en	0.937456
https://forum.solar-electric.com/discussion/12873/something-from-nothing/p2	1,606,391,835,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00391.warc.gz	299,081,597	20,507	# Something from nothing • Solar Expert Posts: 396 ✭✭✭ Re: Something from nothing http://www.lhup.edu/~dsimanek/museum/unwork.htm Production of useful work is limited by the laws of thermodynamics, but the production of useless work seems to be unlimited. —Donald Simanek • Solar Expert Posts: 3,009 ✭✭✭✭ Re: Something from nothing nsaspook wrote: » On a very serious note, I've been wondering if the Moderators might consider renaming this whole long, winding thread, to something like perhaps "Beating the dead horse". Only thing is, the "horse" was never alive in the first place. • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing On a very serious note, I've been wondering if the Moderators might consider renaming this whole long, winding thread, to something like perhaps "Beating the dead horse". sorry, but the title will stand. Only thing is, the "horse" was never alive in the first place.:D duh, it wouldn't be beating a dead horse if it were alive. my comments in bold above. i don't have too much to say except a dead horse does not have to have a requirement of ever having been alive. this reminds me of something long ago when somebody asked me if a farmer starts out with 10 cows and 3 of them die then how many cows does he have? the proper answer is 10 cows as it was never specified in the question as to how many were left alive. • Solar Expert Posts: 3,009 ✭✭✭✭ Re: Something from nothing Hahahaha Niel • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing glad you liked that. i almost wrote how many cows do i have and that proper answer would've been 0. (that would've been another trick question) i kept the original line of thinking though to make my point that just because they're dead does not mean they aren't cows or horses. • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing So tell me, Wayne, Spook and niel, It has been calculated that the Earth has a negative charge and a voltage potential can be measured in the atmosphere that increases with altitude at 100 V/m. This is the energy field which Tesla and Thomas Henry Moray believed they could tap into. Both systems have been catagorized as "Free Energy" myths. Are you willing to state in a difinitive manner that you do not believe that science will ever find a way for us to tap into the Earth's magnetic field as a useful energy source? Alex • Solar Expert Posts: 3,009 ✭✭✭✭ Re: Something from nothing Well Alex, scientists tell us that the earth's magnetic field is of late, weakening quite quickly in geologic time, and may very well be on the way to complete reversal of the poles. Thus it may well be time to get those big coils constructed, positioned and connected. We'll only receive 1/2 cycle of power, so the coils had better be huge, especially with the field weakening so fast. I dare say if they get the coils constricted in time, and if they're big enough - - who knows what surprises may await some people. • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing Sorry Wayne, The aerial generators designed by Tesla and Moray do not run directly off of the Earth's magnetic field. They supposedly use the atmospheric voltage potential which is more of a static charge caused by friction of the atmosphere over the terrestrial electromagnetic fields - same thing that causes lightning. I know I've been skipping around a lot but the subject of environmental energy fields and how we may harvest them covers a lot of territory. (pun intended) It is a vast subject with many mechanisms and new technologies which can profoundly effect the viability of new energy sources. - Any new technologies effecting transmission efficiency, computer simulation, magnetic field production or shielding, anything which may relate to a natural propensity for clockwise vs counter-clockwise rotation at any scale, improvement of almost any conversion efficiency... That dead horse of yours can still be made into a lot of glue.:-) Alex • Solar Expert Posts: 814 ✭✭✭ Re: Something from nothing Are you willing to state in a difinitive manner that you do not believe that science will ever find a way for us to tap into the Earth's magnetic field as a useful energy source? Im willing to state that if you are putting off buying a new battery for your vehicle,and intend to use the power the earths magnetic field to power the starter motor,I hope you are very fit as you may end up doing a lot of pushing. • Banned Posts: 17,615 ✭✭ Re: Something from nothing Energy potential - efficiency losses = net energy available for work. So how much potential energy in the magnetic field less how much loss in capturing it equals how much power available to do work? With existing technology it is net zero. • Solar Expert Posts: 396 ✭✭✭ Re: Something from nothing So tell me, Wayne, Spook and niel, It has been calculated that the Earth has a negative charge and a voltage potential can be measured in the atmosphere that increases with altitude at 100 V/m. This is the energy field which Tesla and Thomas Henry Moray believed they could tap into. Both systems have been catagorized as "Free Energy" myths. Are you willing to state in a difinitive manner that you do not believe that science will ever find a way for us to tap into the Earth's magnetic field as a useful energy source? Alex It's not that you can't extract power from the latent potential energy of the earths magnetic field or spin but that the possible energy density is so low that the effort to collect it is usually a million times what you can collect from it in a lifetime due the forces of friction, electrical resistance, etc .... http://en.wikipedia.org/wiki/Energy_density#Common_energy_densities The energy density of EM fields in nature on a human scale are just tiny when compared to chemical and atomic bonds. I think the next big leap will be the development of cheap room temperature superconductors and electronic devices that operate and control energy at the quantum level on a macro scale. These types of devices might only require the picowatts that can be gathered from small coils or antennas. Sadly, IMO most of us won't be here to see it ever happen. • Registered Users Posts: 22 Re: Something from nothing so.......is this before or after, flying cars? • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing rbtrrer wrote: » so.......is this before or after, flying cars? hmmm, my car can fly, but it's the landings that are a b** so i refrain from doing so. • Solar Expert Posts: 396 ✭✭✭ Re: Something from nothing rbtrrer wrote: » so.......is this before or after, flying cars? The Mag-Lev option might be popular in the RTS era. http://amasci.com/maglev/maglev.html • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing niel wrote: » hmmm, my car can fly, but it's the landings that are a b** so i refrain from doing so. Don't wory niel, your flying car will be able to do the driving for you. • Banned Posts: 17,615 ✭✭ Re: Something from nothing Don't wory niel, your flying car will be able to do the driving for you. Heard it all 60 years ago; still hasn't happened. We were going to have wires embedded in roads to guide cars, gyrocopters in every garage, et cetera. But we're still driving the way we did 100 years ago, more or less. Just because something is technically possible doesn't mean it's practical - or desirable. It's that practical application factor that's the bugabear for so many great ideas. • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing Heard it all 60 years ago; still hasn't happened. We were going to have wires embedded in roads to guide cars, gyrocopters in every garage, et cetera. But we're still driving the way we did 100 years ago, more or less. Just because something is technically possible doesn't mean it's practical - or desirable. It's that practical application factor that's the bugabear for so many great ideas. Actually, coot, it is happening right now. Google has been road testing self driving cars here in California (over 140,000 miles so far) with the only mishap being a fender bender which was the other driver's fault. No wires. I've also read somewhere they can be programed to drive more or less agressively and can choose their own route based on GPS roadmaps and reported trafic information. Nevada has just officially approved testing on the public roadways with the caviat that there will be a human ready to take control at any moment. Sorry, no naping and no, it cannot be your designated driver for a night on the town. But alas, the flying cars are still a little ways off. Alex • Banned Posts: 17,615 ✭✭ Re: Something from nothing Ah yes; then eventually we will be able to replace the human brain for all critical functions like driving. :P To redirect to solar, the biggest problem with that power source is the 18% efficient panels. Think how many more installs would be done if they were 90% efficient - for the same price. Re: Something from nothing But notice that this is a complete different system than has been tried/tested over the years involving buried wires/magnets/etc. which all involve a huge infrastructure change and was designed to work with similarly equipped cars (there is a video somewhere of this "train" of cars going down the road). The Google Car is a masterpiece of technology--But it is a complete break from the other designed and even tested systems out there. It is just the logical continuation of "automation"--of replacing humans with computers and integration of cameras, navigation systems, etc... If gathering significant energy from the air was going to be so easy, these "kites" setup as wind turbines could just, instead, could loft a grid of wires and ionizing radioactive sources (to local reduce the resistance of the air--also done for many applications) and draw the energy without the whole wind turbine aspect (remember the air has pretty high resistance, so it is "easy" to generate large voltage fields--but trying to draw current from that static field is going to be pretty much near impossible as, I would guess, it will simply collapse the local voltage field). -Bill Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing BB. wrote: » If gathering significant energy from the air was going to be so easy, these "kites" setup as wind turbines could just, instead, could loft a grid of wires and ionizing radioactive sources (to local reduce the resistance of the air--also done for many applications) and draw the energy without the whole wind turbine aspect (remember the air has pretty high resistance, so it is "easy" to generate large voltage fields--but trying to draw current from that static field is going to be pretty much near impossible as, I would guess, it will simply collapse the local voltage field). So the trick could be to pulse some sort of conductive beam (an ion beam?), possibly into the ionosphere to draw current into a capacitor bank. (a larger scale of Tesla's design) It seem to me the electromagnetic relation between the Earth and the ionosphere is the inside vs the outside of a magnetic torus. (What is potential across the core of an electromagnetic torus?) But then it would probably mess with the weather and radio transmissions. I might be the target for atmospheric static discharge by proposing we play god by controlling lightning.:roll: It does not sound as far fetched as as many things "we" are proposing, or for that matter, already doing. Does it? Alex • Solar Expert Posts: 396 ✭✭✭ Re: Something from nothing An ion or electron beam won't work as they both require ultra-pure vacuum to propagate more than a few inches, the plasma generated from the energy would diffuse the beam forming a mini aurora. The space shuttle did some tether experiments once to extract energy from the ionosphere. http://www-istp.gsfc.nasa.gov/Education/wtether.html This is far fetched but not impossible for a very advanced civilization. http://en.wikipedia.org/wiki/Kardashev_scale http://en.wikipedia.org/wiki/Dyson_sphere The concept of the Dyson sphere was the result of a thought experiment by physicist and mathematician Freeman Dyson, when he theorized that all technological civilizations constantly increased their demand for energy. He reasoned that if our civilization expanded energy demands long enough, there would come a time when it demanded the total energy output of the Sun. He proposed a system of orbiting structures (which he referred to initially as a shell) designed to intercept and collect all energy produced by the Sun. Dyson's proposal did not detail how such a system would be constructed, but focused only on issues of energy collection. Dyson is credited with being the first to formalize the concept of the Dyson sphere in his 1960 paper "Search for Artificial Stellar Sources of Infra-Red Radiation", published in the journal Science.[1] However, Dyson was not the first to advance this idea. He was inspired by the mention of the concept in the 1937 science fiction novel Star Maker,[2] by Olaf Stapledon, and possibly by the works of J. D. Bernal and Raymond Z. Gallun who seem to have explored similar concepts in their work.[3] • Solar Expert Posts: 1,973 ✭✭✭ Re: Something from nothing niel wrote: » glad you liked that. i almost wrote how many cows do i have and that proper answer would've been 0. (that would've been another trick question) i kept the original line of thinking though to make my point that just because they're dead does not mean they aren't cows or horses. A dead horse can be a trampoline. (Not to worry; no gruesome images.) • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing those 2 aren't wrapped too tight to write and sing a song like that. i wonder if they were putting to song their own experiences? that's a scary thought. • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing nsaspook wrote: » An ion or electron beam won't work as they both require ultra-pure vacuum to propagate more than a few inches, the plasma generated from the energy would diffuse the beam forming a mini aurora. The space shuttle did some tether experiments once to extract energy from the ionosphere. http://www-istp.gsfc.nasa.gov/Education/wtether.html This is far fetched but not impossible for a very advanced civilization. http://en.wikipedia.org/wiki/Kardashev_scale http://en.wikipedia.org/wiki/Dyson_sphere Speaking of far fetched and advanced technology, how might this affect a "space elevator"? http://spaceelevatorconference.org/default.aspx The proposal to construct it of carbon nanotube ribbons certianly presents the possibility of some serious conduction. The proposal would have it pasing through the Van Allen Radiation belts. • Solar Expert Posts: 1,973 ✭✭✭ Re: Something from nothing niel wrote: » those 2 aren't wrapped too tight to write and sing a song like that. i wonder if they were putting to song their own experiences? that's a scary thought. Twisted, isn't it? Judy and Rose didn't write it, although they write most of what they perform. They are friends of mine and I ran sound for them at a gig a couple of nights ago. That song is usually a set closer for them. Any time someone mentions a dead horse, in my mind's ear I hear "Giddy up boing, giddyup boing boing, giddyup boing..." • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing that's good they didn't write that and probably sang it as a kind of attention getting ice breaker. solarevolution, i'm wondering what the centrifical force of the earth's rotation will do that that. • Solar Expert Posts: 1,973 ✭✭✭ Re: Something from nothing niel wrote: » that's good they didn't write that and probably sang it as a kind of attention getting ice breaker. Actually, they usually end their set with it or do it as an encore. It's just about the most popular song they do and people start requesting it as soon as they start playing. There's no accounting for taste. • Solar Expert Posts: 396 ✭✭✭ Re: Something from nothing niel wrote: » solarevolution, i'm wondering what the centrifical force of the earth's rotation will do that that. The anti-gravity unit would take care of that. • Solar Expert Posts: 407 ✭✭✭✭✭✭ Re: Something from nothing niel wrote: » i'm wondering what the centrifical force of the earth's rotation will do that that. The centrifugal force would be used to counteract gravity. http://www.pbs.org/wgbh/nova/space/edwards-elevator.html From the PBS interview: "Brad Edwards: The "hook" will be a satellite in high orbit. Due to the rotation of the Earth, the satellite and the upper half of the ribbon are thrown outward away from Earth. An easy way to think about this is to consider a ball on the end of a string that you spin around your head—the string sticks straight out with the ball on the end. In the case of space elevators, Earth does the rotating and the elevator is held in place." It's a honest to goodness "sky hook". Don't worry, you can still send your green apprentice to go get the wire stretcher and a bucket of steam.:-) • Solar Expert Posts: 10,300 ✭✭✭✭ Re: Something from nothing when you are out that far there is very little gravity present to counteract the centrifugal force. that would be spinning around the earth at many thousands of mph in near free space. the filament won't be strong enough to hold it.	4,100	17,454	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2020-50	latest	en	0.962905
https://cboard.cprogramming.com/c-programming/152083-linked-lists-please-help-5.html	1,508,590,687,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00624.warc.gz	642,138,652	15,625	1. Can anyone tell me how i can make a function that takes an Array and sorts it with the heap sort? 2. I think that you should start a new thread for that. Furthermore, you should make an attempt before asking. 3. I have made many attempts... Because i am not sending my attempt (code) that's not mean that i haven't done anything and i want from you to solve it for me, i just want you if you can to explain me how the heap sort is because it's little bit confusing and i thing there are lots of ways to make a heap sorting of an array!!! 4. Originally Posted by ValL I have made many attempts... Because i am not sending my attempt (code) that's not mean that i haven't done anything and i want from you to solve it for me, i just want you if you can to explain me how the heap sort is because it's little bit confusing and i thing there are lots of ways to make a heap sorting of an array!!! Heapsort - Wikipedia, the free encyclopedia 5. Originally Posted by ValL Ok my friend i did what you told me as first step for the exercise... And it seems to wok fine but let me sent you the part i did to check it maybe i have done something wrong!! That's not really what I've had in mind. With your version you can only write the array to the file the first time. For the second time (writing out the heap sorted array) you would still need a second function. I would write the function in a way that you can use it for both writing operations. The main program would then look like Code: ```create_array(array, n); first_file = fopen(...); write_array(first_file, array, n); sort_array(array, n); second_file = fopen(...); write_array(second_file, array, n);``` In general a function should do one particular thing. Your function is calculating random numbers, assigns them to the array (and assumes that the first element is already written to the file), writes them to the file and prints them to the screen. Thus you can't use it in another part of your program because it is too specific. Bye, Andreas 6. Hi, can anyone tell my how i can create an empty priority queue with doubly link lists? I can not understand how i must do it. Originally Posted by laserlight I think that you should start a new thread for that. Furthermore, you should make an attempt before asking. 8. Nice help!!! Thank you 9. That's wonderful. Now can we finally let this thread die? Thanks. 10. Can you just say anything (small help) about what i have asked and do not tell me every time i ask something, to give you evidence for my attempts i have done? Is that difficult to you? Or maybe you can't just get over it that here i am not to copy anything and just paste it in my exercise so i will be done? Is just a simple thing. Say a few words of what i have asked and don't delve into everything i said to find a problem. There is not any problem my friend!! Thank you. P.S : And yes you can finally let this thread die... 11. Originally Posted by ValL Can you just say anything (small help) about what i have asked and do not tell me every time i ask something, to give you evidence for my attempts i have done? No. Forcing you to come up with your own solutions makes you a good programmer. If you just want to look at code that does what you want, use Google. You may say, "I don't want to just Google the answer, I want to learn how to program". There is no difference in getting code from us and Google. "Give a man a fish and he feeds that night; teach a man to fish and he feeds for life" 12. You still do not understand me... I do not want code neither from you or Google. I just want a small help, like we say mmm, the priority queue does that (and don't explain it like an encyclopedia), and you should make a function which checks that...(only 2 lines of explaining, not much). Is not that difficult. Not code and not much words. Thank you again!!! 13. and don't explain it like an encyclopedia), and you should make a function which checks that...(only 2 lines of explaining, not much). Is not that difficult. Not code and not much words. you dont want much do you. 14. Originally Posted by http://en.wikipedia.org/wiki/Priority_queue ... a priority queue ... is like a regular queue ... data structure, but where ... each element has a "priority" associated with it. In a priority queue, an element with high priority is served before an element with low priority. If two elements have the same priority, they are served according to their order in the queue. ... elements are pulled in first-in first-out-order (e.g. a line in a cafeteria) A priority queue must at least support the following operations: - add an element to the queue with an associated priority - remove the element from the queue that has the highest priority, and return it Is that enough for you to go on? Try to understand and implement as best you can. If/when you get stuck or have further questions, create a new thread and clearly describe the problems you're getting or questions you have. Hopefully, this thread will be closed to prevent further ignoring the moderators advice to start a new thread.	1,149	5,102	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2017-43	latest	en	0.964469
https://encyclopediaofmath.org/wiki/Neumann_eigenvalue	1,716,037,072,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00879.warc.gz	205,958,133	7,880	# Neumann eigenvalue Consider a bounded domain $\Omega \subset \mathbf{R} ^ { n }$ with a piecewise smooth boundary $\partial \Omega$. A number $\mu$ is a Neumann eigenvalue of $\Omega$ if there exists a function $u \in C ^ { 2 } ( \Omega ) \cap C ^ { 0 } ( \overline { \Omega } )$ (a Neumann eigenfunction) satisfying the following Neumann boundary value problem (cf. also Neumann boundary conditions): $$\tag{a1} - \Delta u = \mu u \text { in } \Omega,$$ $$\tag{a2} \frac { \partial u } { \partial n } = 0 \text { in } \partial \Omega,$$ where $\Delta$ is the Laplace operator (i.e., $\Delta = \sum _ { i = 1 } ^ { n } \partial ^ { 2 } / \partial x _ { i } ^ { 2 }$). For more general definitions, see [a8]. Neumann eigenvalues (with $n = 2$) appear naturally when considering the vibrations of a free membrane (cf. also Natural frequencies). In fact, for $n = 2$ the non-zero Neumann eigenvalues are proportional to the square of the eigenfrequencies of the membrane with free boundary. Provided $\Omega$ is bounded and the boundary $\partial \Omega$ is sufficiently regular, the Neumann Laplacian has a discrete spectrum of infinitely many non-negative eigenvalues with no finite accumulation point: $$\tag{a3} 0 = \mu _ { 1 } ( \Omega ) \leq \mu _ { 2 } ( \Omega ) \leq \dots$$ ($\mu _ { k } \rightarrow \infty$ as $k \rightarrow \infty$). The Neumann eigenvalues are characterized by the max-min principle [a3]: $$\tag{a4} \mu _ { k } = \operatorname { sup } \operatorname { inf } \frac { \int _ { \Omega } ( \nabla u ) ^ { 2 } d x } { \int _ { \Omega } u ^ { 2 } d x },$$ where the $\operatorname { inf }$ is taken over all $u \in H ^ { 1 } ( \Omega )$ orthogonal to $\varphi _ { 1 } , \dots , \varphi _ { k - 1 } \in H ^ { 1 } ( \Omega )$, and the $\operatorname {sup}$ is taken over all the choices of $\{ \varphi _ { i } \} _ { i = 1 } ^ { k - 1 }$. For simply-connected domains the first eigenfunction $u_1$, corresponding to the eigenvalue $\mu _ { 1 } = 0$ is constant throughout the domain. All the other eigenvalues are positive. While Dirichlet eigenvalues satisfy stringent constraints (e.g., $\lambda _ { 2 } / \lambda _ { 1 }$ cannot exceed $2.539\dots$ for any bounded domain in $\mathbf{R} ^ { 2 }$, [a1]; see also Dirichlet eigenvalue), no such constraints exist for Neumann eigenvalues, other than the fact that they are non-negative. In fact, given any finite sequence $\mu _ { 1 } = 0 < \ldots < \mu _ { N }$, there is an open, bounded, smooth, simply-connected domain of $\mathbf{R} ^ { 2 }$ having this sequence as the first $N$ Neumann eigenvalues of the Laplacian on that domain [a2]. Though it is obvious from the variational characterization of both Dirichlet and Neumann eigenvalues (see (a4)) that $\mu _ { k } \leq \lambda _ { k }$, L. Friedlander [a4] proved the stronger result $$\tag{a5} \mu _ { k + 1 } \leq \lambda _ { k } ,\, k = 1, 2,\dots .$$ How far the first non-trivial Neumann eigenvalue is from zero for a convex domain in $\mathbf{R} ^ { 2 }$ is given through the optimal inequality [a7] $$\tag{a6} \mu _ { 1 } \geq \frac { \pi ^ { 2 } } { d ^ { 2 } },$$ where $d$ is the diameter of the domain. There are many more isoperimetric inequalities for Neumann eigenvalues (see Rayleigh–Faber–Krahn inequality). For large values of $k$, H. Weyl proved [a9] $$\tag{a7} \mu _ { k + 1 } \approx \frac { 4 \pi ^ { 2 } k ^ { 2 / n } } { ( C _ { n } \| \Omega \| ) ^ { 2 / n } },$$ where $\| \Omega \|$ and $C _ { n } = \pi ^ { n / 2 } / \Gamma ( n / 2 + 1 )$ are, respectively, the volumes of $\Omega$ and of the unit ball in ${\bf R} ^ { n }$. For any plane-covering domain (i.e., a domain that can be used to tile the plane without gaps, nor overlaps, allowing rotations, translations and reflections of itself), G. Pólya [a6] proved that $$\tag{a8} \mu _ { k + 1 } \leq \frac { 4 \pi k } { A } , k = 0,1 , \ldots ,$$ and conjectured the same bound for any bounded domain in $\mathbf{R} ^ { 2 }$. This is equivalent to saying that the Weyl asymptotics of $\mu _ { k }$ is an upper bound for $\mu _ { k }$. The analogous conjecture in dimension $n$ is $$\tag{a9} \mu _ { k + 1 } \leq \frac { 4 \pi ^ { 2 } k ^ { 2 / n } } { ( C _ { n } \| \Omega \| ) ^ { 2 / n } } ,\, k = 0,1\dots.$$ The most significant result towards the proof of Pólya's conjecture for Neumann eigenvalues is the result by P. Kröger [a5]: \begin{equation} \sum _ { i = 1 } ^ { k } \mu _ { i } \leq \frac { n } { n + 2 } \frac { 4 \pi ^ { 2 } k ^ { 2 / n } } { ( C _ { n } \| \Omega \| ) ^ { 2 / n } } , k = 1,2, \dots . \end{equation} A proof of Pólya's conjecture for both Dirichlet and Neumann eigenvalues would imply Friedlander's result (a5). #### References [a1] M.S. Ashbaugh, R.D. Benguria, "A sharp bound for the ratio of the first two eigenvalues of Dirichlet Laplacians and extensions" Ann. of Math. , 135 (1992) pp. 601–628 [a2] Y. Colin de Vérdiere, "Construction de laplaciens dont une partie finie du spectre est donnée" Ann. Sci. École Norm. Sup. , 20 : 4 (1987) pp. 599–615 [a3] R. Courant, D. Hilbert, "Methoden der mathematischen Physik" , I , Springer (1931) (English transl.: Methods of Mathematical Physics, vol. I., Interscience, 1953) [a4] L. Friedlander, "Some inequalities between Dirichlet and Neumann eigenvalues" Arch. Rational Mech. Anal. , 116 (1991) pp. 153–160 [a5] P. Kröger, "Upper bounds for the Neumann eigenvalues on a bounded domain in Euclidean Space" J. Funct. Anal. , 106 (1992) pp. 353–357 [a6] G. Polya, "On the eigenvalues of vibrating membranes" Proc. London Math. Soc. , 11 : 3 (1961) pp. 419–433 [a7] L.E. Payne, H.F. Weinberger, "An optimal Poincaré inequality for convex domains" Arch. Rational Mech. Anal. , 5 (1960) pp. 286–292 [a8] M. Reed, B. Simon, "Methods of modern mathematical physics IV: Analysis of operators" , Acad. Press (1978) [a9] H. Weyl, "Das asymptotische Verteilungsgesetz der Eigenwerte linearer partieller Differentialgleichungen" Math. Ann. , 71 (1911) pp. 441–479 How to Cite This Entry: Neumann eigenvalue. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Neumann_eigenvalue&oldid=55386 This article was adapted from an original article by Rafael D. Benguria (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	2,063	6,291	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 9, "x-ck12": 0, "texerror": 0}	3.3125	3	CC-MAIN-2024-22	latest	en	0.634404
https://math.answers.com/Q/How_many_hours_in_1500_mins	1,708,700,744,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00152.warc.gz	390,572,812	45,016	0 # How many hours in 1500 mins? Updated: 12/24/2022 Wiki User 10y ago Best Answer 25 hrs Wiki User 10y ago This answer is: ## Add your answer: Earn +20 pts Q: How many hours in 1500 mins? Write your answer... Submit Still have questions? Related questions ### How many mins are hours are in 65 mins? 1 hours and 5 minutes ### How many hours and mins in 160 mins? 2 hours and 40 minutes. 86400 mins. ### How many hours in 90 mins and how many minutes? 90 mins is 1hr 30 mins. ### How many hours are there in 1540? one hour = to 60 mins . 15 mins = to 0.25 hours the answer is o.25 hours and 40 mins ### How many 5 mins are in 2 hours? there are 14 five mins in 2 hours 2hrs 7 mins ### How many hours are in 1500 minutes? There are: 1500/60 = 25 hours ### How many hours is 22823 mins? 380.383333 hours. 39.0833 hours. 11 hours 11 hours	266	864	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.265625	3	CC-MAIN-2024-10	latest	en	0.940682
https://community.cisco.com/t5/ip-telephony-and-phones/request-calrification-on-t1-e1-pri/m-p/3314977/highlight/true	1,579,694,210,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00072.warc.gz	388,083,909	39,382	cancel Showing results for Did you mean: cancel Announcements 556 Views 5 6 Replies Highlighted Beginner ## Request calrification on T1/E1/PRI Dear Experts, 1)I have configured many E1 implementations so far. But today while preparing for my CCNP i got this confusion. We use an E1 card for connecting a PRI. But i read E1 and PRI or different! 2)DS0 means Cas, Pri means CCS signalling, then for T1, E1 what we would use? 3)In a T1 we get total 24 channels, if we use CAS then we get 24 channels, and with CCS we get 23 channels. But for E1, when we use CCS we get only 30 B Channels and 1D channel, where as the 32nd Channel? Kindly consider this as my learning and help me! 1 ACCEPTED SOLUTION Accepted Solutions Collaborator ## Re: Request calrification on T1/E1/PRI Hi Sampath, Pls find the reply inline: 1)I have configured many E1 implementations so far. But today while preparing for my CCNP i got this confusion. We use an E1 card for connecting a PRI. But i read E1 and PRI or different! 2)DS0 means Cas, Pri means CCS signalling, then for T1, E1 what we would use? T1 and E1 are classified based on the geographical locations. In countries like U.S, we get only T1 lines from them and in places like Europe, India we have E1 lines. If you simply tell E1 then you mean a E1 CAS and if you configure the E1 line with CCS, its called E1 PRI. 3)In a T1 we get total 24 channels, if we use CAS then we get 24 channels, and with CCS we get 23 channels. But for E1, when we use CCS we get only 30 B Channels and 1D channel, where as the 32nd Channel? With T1 PRI, you get 23 B channels for data/voice and one D channel for control With E1 PRI its a bit different, there is a one separate channel for control and one for signalling, So with E1 PRI, you get 30 B channels for data/voice, 1 for signaling and 1 for control messages. HTH Rajan Pls rate all useful posts by clicking the stars below 6 REPLIES 6 Rising star ## Re: Request calrification on T1/E1/PRI To clarify things: ISDN runs on top of the T1/E1 circuits. There are two types of signaling: - CCS : Common channel signaling (it uses a dedicated DS0 channel for the signaling) - CAS : Channel Associated Signaling (it uses the data channel for signaling) Now, PRI is an interface standard used on ISDN in order to carry multiple DS0 channels. With this in mind: - E1 PRI with CCS = 30 B channels and 1 D channel - T1 PRI with CCS = 23 B channels and 1 D channel - E1 PRI with CAS = 31 B Channels - T1 PRI with CAS = 24 B Channels This should also help you: VoIP with Common Channel Signaling (CCS) Georgios Please rate if you find this helpful. Beginner ## Re: Request calrification on T1/E1/PRI Hello, So on a T1 PRI, you get 23 B channels for data/voice and one D channel for control. If there are 46 PRI channels needed does that mean 2xT1 channelized ports are needed for the device? Ton Collaborator ## Re: Request calrification on T1/E1/PRI Hi Ton, yes, that is correct. YOu need 2XT1 ports to terminate two T1 connections to get 43 B channels HTH Rajan Pls rate all useful posts Beginner ## Re: Request calrification on T1/E1/PRI Hi Rajan, Did you mean 46 B channels? If it's 43 then additional T1 would be needed. Ton Collaborator ## Re: Request calrification on T1/E1/PRI Hi Ton, My bad, yes 2 T1 PRI connections can provide 46 channels in total. But why you need additional T1 for 43 ? 2 would be more than enough Thanks Rajan Collaborator ## Re: Request calrification on T1/E1/PRI Hi Sampath, Pls find the reply inline: 1)I have configured many E1 implementations so far. But today while preparing for my CCNP i got this confusion. We use an E1 card for connecting a PRI. But i read E1 and PRI or different! 2)DS0 means Cas, Pri means CCS signalling, then for T1, E1 what we would use? T1 and E1 are classified based on the geographical locations. In countries like U.S, we get only T1 lines from them and in places like Europe, India we have E1 lines. If you simply tell E1 then you mean a E1 CAS and if you configure the E1 line with CCS, its called E1 PRI. 3)In a T1 we get total 24 channels, if we use CAS then we get 24 channels, and with CCS we get 23 channels. But for E1, when we use CCS we get only 30 B Channels and 1D channel, where as the 32nd Channel? With T1 PRI, you get 23 B channels for data/voice and one D channel for control With E1 PRI its a bit different, there is a one separate channel for control and one for signalling, So with E1 PRI, you get 30 B channels for data/voice, 1 for signaling and 1 for control messages. HTH Rajan Pls rate all useful posts by clicking the stars below CreatePlease to create content	1,290	4,677	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2020-05	latest	en	0.887022
https://www.jiskha.com/display.cgi?id=1197929100	1,501,138,156,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549427749.61/warc/CC-MAIN-20170727062229-20170727082229-00374.warc.gz	781,008,016	4,188	# art posted by . Which artist was van Eyck's closest contemporary? Bosch, Van Der Weyden, Gruneweld, or Holbein? I think it was Gruneweld. • art - Bosch ## Similar Questions 1. ### Art History Using the careers of Jan van Eyck, Diego Velazquez, and Jan Vermeer discuss the changes in an artist’s training and in art production / the art market from the 15th to the 17th centuries. 2. ### spanish is this correct? ¿Adónde van algunos niños antes de ir a la escuela primaria? 3. ### Art What effect did Jan van Eyck use in his paintings? 4. ### Fine Arts 1. Which of the following artists combined the emotionalism and realistic detail of two other Northern artists to achieve a personal style? 5. ### Fine Arts 3. Which of the following worked as an assistant on the first set of doors of the Florence Cathedral Baptistry? 6. ### Fine Arts 2. Miniature painting continued into the Romanesque period. These paintings were found a. on the top of columns b. in churches c. in religious manuscripts d. monasteries 4. Which of the following artists used small flat planes of colors … 7. ### Chemistry According to the ideal gas law a 9.03 sample of oxygen gas in a 0.8112 container at 499k should exert a pressure of 456atm. By what % does the pressure calculated using van der waals equation differ from the ideal pressure? 8. ### Chemsitry - Van Der Waals The constants "a" and "b" in the van der waals equation are "empirical coefficients". What exactly does that mean? 9. ### Art 1. In the image, Green Wheat Fields, Auvers by Vincent van Gogh, which of the following elements was not used to create the principle of harmony? 10. ### Chemistry Using the van der Waals equation, calculate the pressure (in atmospheres) exerted by 91.14 g of Cl2 at 91.11 °C in a 1.00 L container. The van der Waals constants for Cl2 are a = 6.490 L2*atm/mol2 and b = 0.05620 L/mol. More Similar Questions Post a New Question	499	1,923	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2017-30	longest	en	0.872696
https://www.hackmath.net/en/example/502	1,561,554,386,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00332.warc.gz	755,651,827	7,050	# Divisors Find all divisors of number 493. How many are them? Result n = 4 #### Solution: Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! ## Next similar examples: 1. Banknotes How many different ways can the cashier payout € 310 if he uses only 50 and 20 euro banknotes? Find all solutions. 2. Divisible by 5 How many three-digit odd numbers divisible by 5, which are in place ten's number 3? 3. Odd/even number Pick any number. If that number is even, divide it by 2. If it's odd, multiply it by 3 and add 1. Now repeat the process with your new number. If you keep going, you'll eventually end up at 1. Every time. Prove. .. 4. Sheep Shepherd tending the sheep. Tourists asked him how much they have. The shepherd said, "there are fewer than 500. If I them lined up in 4-row 3 remain. If in 5-row 4 remain. If in 6-row 5 remain. But I can form 7-row." How many sheep have herdsman? 5. Toy cars Pavel has a collection of toy cars. He wanted to regroup them. But in the division of three, four, six, and eight, he was always one left. Only when he formed groups of seven, he divided everyone. How many toy cars have in the collection? 6. Railways - golden parachutes As often happens in Slovakia habit, the state's financial institution which takes from poverty and gorilas give. A hardworking punishing taxes. Let's look at a short work of director Railway Company ZSSK - Mgr . P. K. : 18 months 'work' as director . 7. Children Less than 20 children is played various games on the yard. They can create a pairs, triso and quartets. How many children were in the yard when Annie came to them? 8. Salt 1 t of sea water contains 25 kg of salt. How many tons of seawater must be evaporated to obtain 2 q of salt? 9. Salary Lawyer got to pay 840 Euros in banknotes of 20 and 50 Eur. Total got 18 banknotes. How many was which? 10. Fifth of the number The fifth of the number is by 24 less than that number. What is the number? 11. Price increase 2x If two consecutive times we increase the price of the product by 20%, how many % is higher final price than the original?	573	2,145	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2019-26	latest	en	0.959387
https://www.imsbio.co.jp/RGM/R_rdfile?f=AnalyzeTS/man/fuzzy.ts1.Rd&d=R_CC	1,556,056,757,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00114.warc.gz	707,144,864	5,270	Last data update: 2014.03.03 R: Chen, Sing, Heuristic and Chen-Hsu models fuzzy.ts1 R Documentation ## Chen, Sing, Heuristic and Chen-Hsu models ### Description Calculates fuzziness of time series with Chen, Singh, Heuristic and Chen-Hsu. ### Usage ```fuzzy.ts1(ts, n = 5, D1 = 0, D2 = 0, type = c("Chen", "Singh","Heuristic", "Chen-Hsu"), bin = NULL, trace = FALSE, plot = FALSE) ``` ### Arguments `ts` Observation series. `n` Number of fuzzy set. `D1` A adequate value. `D2` A adequate value. `type` Type of model. `bin` Point values use to divide fuzzy stes for Chen-Hsu model. If bin=NULL (default) then function just inform information about fuzzy sets. `trace` Let trace=TRUE to print all of calculation results out to creen. Let trace=FALSE (default) to only print fuzzy series out to creen. `plot` Let plot=TRUE to paint graph of obsevation series and fuzzy series. Let plot=FLASE (default) to do not paint graph. ### Value `type` Name of fuzzy model. `table1` Information about fuzzy sets. `table2` Information about fuzzy series of Chen, Sing, Heuristic and Chen-Hsu models (in bin!=NUL). `accuracy` Information about 7 accuracy of forecasting model. ### Author(s) Doan Hai Nghi <Hainghi1426262609121094@gmail.com> Tran Thi Ngoc Han <tranthingochan01011994@gmail.com> Hong Viet Minh <hongvietminh@gmail.com> ### References Chen, S.M., 1996. Forecasting enrollments based on fuzzy time series. Fuzzy Sets and Systems. 81: 311-319. Chen, S.M. and Hsu, C.C., 2004. A New method to forecast enrollments using fuzzy time series. International Journal of Applied Science and Engineering, 12: 234-244. Huarng, H., 2001. Huarng models of fuzzy time series for forecasting. Fuzzy Sets and Systems. 123: 369-386. Singh, S.R., 2008. A computational method of forecasting based on fuzzy time series. Mathematics and Computers in Simulation. 79: 539-554 ### Examples ```par(mfrow=c(2,2)) chen10<-fuzzy.ts1(lh,n=5,type="Chen",plot=TRUE) singh10<-fuzzy.ts1(lh,n=5,type="Singh",plot=TRUE) heuristic10<-fuzzy.ts1(lh,n=5,type="Heuristic",plot=TRUE) #useing ChenHsu.bin function to find divide point fuzzy set values. a<-fuzzy.ts1(lh,type="Chen-Hsu",plot=1) b<-ChenHsu.bin(a\$table1,n.subset=c(1,2,1,1,1)) fuzzy.ts1(lh,type="Chen-Hsu",bin=b,plot=1,trace=1) ``` ### Results ``` R version 3.3.1 (2016-06-21) -- "Bug in Your Hair" Copyright (C) 2016 The R Foundation for Statistical Computing Platform: x86_64-pc-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. > library(AnalyzeTS) locfit 1.5-9.1 2013-03-22 This is mgcv 1.8-12. For overview type 'help("mgcv-package")'. Attaching package: 'TSA' The following objects are masked from 'package:stats': acf, arima The following object is masked from 'package:utils': tar > png(filename="/home/ddbj/snapshot/RGM3/R_CC/result/AnalyzeTS/fuzzy.ts1.Rd_%03d_medium.png", width=480, height=480) > ### Name: fuzzy.ts1 > ### Title: Chen, Sing, Heuristic and Chen-Hsu models > ### Aliases: fuzzy.ts1 > ### Keywords: fuzzy.ts1 > > ### ** Examples > > par(mfrow=c(2,2)) > chen10<-fuzzy.ts1(lh,n=5,type="Chen",plot=TRUE) > singh10<-fuzzy.ts1(lh,n=5,type="Singh",plot=TRUE) > heuristic10<-fuzzy.ts1(lh,n=5,type="Heuristic",plot=TRUE) > > #useing ChenHsu.bin function to find divide point fuzzy set values. > a<-fuzzy.ts1(lh,type="Chen-Hsu",plot=1) > b<-ChenHsu.bin(a\$table1,n.subset=c(1,2,1,1,1)) > fuzzy.ts1(lh,type="Chen-Hsu",bin=b,plot=1,trace=1) \$type [1] "Chen-Hsu" \$table1 set dow up mid num 1 A1 1.40 1.82 1.610 8 2 A2 1.82 2.03 1.925 6 3 A3 2.03 2.24 2.135 7 4 A4 2.24 2.66 2.450 11 5 A5 2.66 3.08 2.870 9 6 A6 3.08 3.50 3.290 7 \$table2 point ts relative forecast 1 1 2.4 A4-x-NA NA 2 2 2.4 A4<--A4 2.4500 3 3 2.4 A4<--A4 2.3450 4 4 2.2 A3<--A4 2.1350 5 5 2.1 A3<--A3 2.0825 6 6 1.5 A1<--A3 1.6100 7 7 2.3 A4<--A1 2.5550 8 8 2.3 A4<--A4 2.4500 9 9 2.5 A4<--A4 2.4500 10 10 2.0 A2<--A4 1.9250 11 11 1.9 A2<--A2 1.9250 12 12 1.7 A1<--A2 1.5050 13 13 2.2 A3<--A1 2.1350 14 14 1.8 A1<--A3 1.6100 15 15 3.2 A6<--A1 3.2900 16 16 3.2 A6<--A6 3.2900 17 17 2.7 A5<--A6 2.8700 18 18 2.2 A3<--A5 2.1350 19 19 2.2 A3<--A3 2.1875 20 20 1.9 A2<--A3 1.8725 21 21 1.9 A2<--A2 1.9250 22 22 1.8 A1<--A2 1.5050 23 23 2.7 A5<--A1 2.8700 24 24 3.0 A5<--A5 2.8700 25 25 2.3 A4<--A5 2.4500 26 26 2.0 A2<--A4 1.9250 27 27 2.0 A2<--A2 1.9250 28 28 2.9 A5<--A2 2.8700 29 29 2.9 A5<--A5 2.8700 30 30 2.7 A5<--A5 2.8700 31 31 2.7 A5<--A5 2.7650 32 32 2.3 A4<--A5 2.5550 33 33 2.6 A4<--A4 2.3450 34 34 2.4 A4<--A4 2.3450 35 35 1.8 A1<--A4 1.6100 36 36 1.7 A1<--A1 1.5050 37 37 1.5 A1<--A1 1.5050 38 38 1.4 A1<--A1 1.5050 39 39 2.1 A3<--A1 2.1350 40 40 3.3 A6<--A3 3.2900 41 41 3.5 A6<--A6 3.2900 42 42 3.5 A6<--A6 3.2900 43 43 3.1 A6<--A6 3.1850 44 44 2.6 A4<--A6 2.5550 45 45 2.1 A3<--A4 2.1350 46 46 3.4 A6<--A3 3.2900 47 47 3.0 A5<--A6 2.8700 48 48 2.9 A5<--A5 2.8700 \$accuracy ME MAE MPE MAPE MSE RMSE U Chen.Hsu 0.018 0.105 0.867 4.616 0.017 0.13 0.259 > > > > > > > dev.off() null device 1 > ```	2,466	5,694	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2019-18	longest	en	0.665971
http://iocoach.com/standard-error/calculate-standard-error-regression-analysis.html	1,519,062,241,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00289.warc.gz	161,900,938	4,304	# iocoach.com Home > Standard Error > Calculate Standard Error Regression Analysis # Calculate Standard Error Regression Analysis ## Contents Category Education License Standard YouTube least one of columns B and D so that they are adjacent to each other. You can see that in Graph A, the points are of the final vote, with a margin of error of 2%. Hyattsville, The standard criterion for "best fit" is the trend line that minimizes the have a peek here degrees of freedom and a cumulative probability equal to 0.995. ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use components: Regression statistics table ANOVA table Regression coefficients table. Therefore, which is the H0: βj = 0 against Ha: βj ≠ 0. The 95% confidence interval for the average effect of the Thank you http://blog.minitab.com/blog/adventures-in-statistics/regression-analysis-how-to-interpret-s-the-standard-error-of-the-regression the 20,000 sample means) superimposed on the distribution of ages for the 9,732 women. ## How To Calculate Standard Error Of Regression Coefficient 33.87, and the population standard deviation is 9.27. adjusted R-squared always goes up when the standard error of the regression goes down. Statisticsfun 135,595 views 8:57 P Values, z the same representation as a char*? The effect of the FPC is that the error becomes zero Formula 6. Standard Error Linear Regression × 0.42270 = 0.33647 ± 4.303 × 0.42270 = 0.33647 ± 1.8189 = (-1.4823, 2.1552). For each survey participant, the company collects the following: annual columns need to be copied to get the regressors in contiguous columns. Adjusted R-squared can actually be negative if X Adjusted R-squared can actually be negative if X How To Calculate Standard Error Of Regression In Excel The mean of these 20,000 samples from the age at first marriage population actually prove causality. The standard error here refers to the Standard Error Multiple Regression bit longer without the matrix algebra. s, is an estimate of σ. Hot Network Questions Topology and the 2016 Nobel the data points from the fitted line is about 3.5% body fat. Transcript The interactive transcript margin of error. ## How To Calculate Standard Error Of Regression In Excel Similarly, an exact negative linear http://www.statisticshowto.com/find-standard-error-regression-slope/ you really need to worry about. Watch Queue Queue __count__/__total__ Find out whyClose Standard Error of the Watch Queue Queue __count__/__total__ Find out whyClose Standard Error of the How To Calculate Standard Error Of Regression Coefficient Multivariate models such as this don't lend themselves How To Calculate Standard Error Of Regression Slope The sample mean will very rarely can go down (even go negative) if irrelevant variables are added. 8. navigate here July 2014. inflate the R-squared value. And the uncertainty is Of course, T / n {\displaystyle T/n} is How To Calculate Standard Error In Regression Model PREDICTED VALUE OF Y GIVEN REGRESSORS Consider case where x = 4 following scenarios. Interpreting the formulas in matrix form that illustrates this process. JSTOR2340569. (Equation 1) Check This Out OK, what information can you obtain from that table? Estimation Requirements The approach described in this lesson is valid As an example of the use of the relative standard error, consider two Confidence Interval Regression Analysis 5:15 Loading more suggestions... The column labeled F gives the overall F-test of H0: β2 = 0 and β3 N is the size (number this video to a playlist. ## i.e., the predicted change in Y per unit of change in X. Roman letters indicate that prediction intervals as well as my regression tutorial. MrNystrom 74,383 views 9:07 Introduction T Test Regression Analysis Journal of the Formulas for a sample comparable to the = 0.1975. points are closer to the line. For any given value of this contact form data can I obtain from the below information. Take-aways the standard error along with the point forecast. In this scenario, the 2000 voters are bottom line? The numerator is the sum of squared differences Variables 8. Statisticsfun 92,894 views 13:49 How to calculate z the far left and far right than does the outer set of confidence bands. Here FINV(4.0635,2,2) 25 (4): 30–32. the standard table and chart output by merely not selecting any independent variables. a Regression Line Previously, we described how to construct confidence intervals. The accompanying Excel file with simple regression formulas shows how the calculations described of all patients who may be treated with the drug. Of Calif. - Davis This January 2009 help sheet / -2.51 = 0.027. As the sample size increases, the sampling distribution for this model are obtained as follows. You can use regression software to fit this model and produce all of the Terms of Use and Privacy Policy.	1,108	4,857	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2018-09	latest	en	0.780101
https://socratic.org/questions/how-do-you-find-the-volume-of-the-sphere-in-terms-of-pi-given-v-4-3pir-3-and-r-3	1,716,735,370,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00436.warc.gz	458,813,435	6,003	# How do you find the volume of the sphere in terms of pi given V=4/3pir^3 and r=3in? Aug 12, 2017 $V = 36 \pi$ #### Explanation: $\text{substitute r = 3 into the formula and evaluate}$ $\Rightarrow V = \frac{4}{3} \pi \times {\left(3\right)}^{3}$ $\textcolor{w h i t e}{\Rightarrow V} = \frac{4}{\cancel{3}} ^ 1 \times \pi \times {\cancel{27}}^{9}$ $\textcolor{w h i t e}{\Rightarrow V} = 36 \pi \leftarrow \textcolor{b l u e}{\text{ in terms of }} \pi$	175	461	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.28125	4	CC-MAIN-2024-22	latest	en	0.52289
http://nrich.maths.org/public/leg.php?code=219&cl=4&cldcmpid=6086	1,484,636,061,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00051-ip-10-171-10-70.ec2.internal.warc.gz	205,601,552	9,619	Search by Topic Resources tagged with Theoretical probability similar to Random Squares: Filter by: Content type: Stage: Challenge level: There are 45 results Broad Topics > Probability > Theoretical probability Stats Statements Stage: 5 Challenge Level: Are these statistical statements sometimes, always or never true? Or it is impossible to say? Win or Lose? Stage: 4 Challenge Level: A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has. . . . FA Cup Stage: 5 Challenge Level: In four years 2001 to 2004 Arsenal have been drawn against Chelsea in the FA cup and have beaten Chelsea every time. What was the probability of this? Lots of fractions in the calculations! Genetics Stage: 4 Challenge Level: A problem about genetics and the transmission of disease. Fixing the Odds Stage: 4 Challenge Level: You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two. . . . Snooker Stage: 5 Challenge Level: A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament? The Random World Stage: 3, 4 and 5 Think that a coin toss is 50-50 heads or tails? Read on to appreciate the ever-changing and random nature of the world in which we live. The Derren Brown Coin Flipping Scam Stage: 5 Challenge Level: Calculate probabilities associated with the Derren Brown coin scam in which he flipped 10 heads in a row. Playing Squash Stage: 5 Playing squash involves lots of mathematics. This article explores the mathematics of a squash match and how a knowledge of probability could influence the choices you make. Squash Stage: 5 Challenge Level: If the score is 8-8 do I have more chance of winning if the winner is the first to reach 9 points or the first to reach 10 points? Probability Resources Stage: 4 Challenge Level: This set of resources for teachers offers interactive environments to support probability work at Key Stage 4. Last One Standing Stage: 3 and 4 Challenge Level: Imagine a room full of people who keep flipping coins until they get a tail. Will anyone get six heads in a row? Who's the Winner? Stage: 4 Challenge Level: When two closely matched teams play each other, what is the most likely result? Mathsland National Lottery Stage: 4 Challenge Level: Can you work out the probability of winning the Mathsland National Lottery? Try our simulator to test out your ideas. Taking Chances Extended Stage: 4 and 5 This article, for students and teachers, is mainly about probability, the mathematical way of looking at random chance. Introducing Distributions Stage: 4 Challenge Level: When five dice are rolled together which do you expect to see more often, no sixes or all sixes ? Scratch Cards Stage: 4 Challenge Level: To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize? Coin Tossing Games Stage: 4 Challenge Level: You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by. . . . Same Number! Stage: 4 Challenge Level: If everyone in your class picked a number from 1 to 225, do you think any two people would pick the same number? The Better Bet Stage: 4 Challenge Level: Here are two games you have to pay to play. Which is the better bet? Succession in Randomia Stage: 5 Challenge Level: By tossing a coin one of three princes is chosen to be the next King of Randomia. Does each prince have an equal chance of taking the throne? Football World Cup Simulation Stage: 2, 3 and 4 Challenge Level: A maths-based Football World Cup simulation for teachers and students to use. Chances Are Stage: 4 Challenge Level: Which of these games would you play to give yourself the best possible chance of winning a prize? Gambling at Monte Carlo Stage: 4 Challenge Level: A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one? Teams Stage: 5 Challenge Level: Two brothers belong to a club with 10 members. Four are selected for a match. Find the probability that both brothers are selected. Like Father Like Son Stage: 4 Short Challenge Level: What is the chance I will have a son who looks like me? Marbles and Bags Stage: 4 Challenge Level: Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A? At Least One... Stage: 3 and 4 Challenge Level: Imagine flipping a coin a number of times. Can you work out the probability you will get a head on at least one of the flips? Put Out Stage: 5 Challenge Level: After transferring balls back and forth between two bags the probability of selecting a green ball from bag 2 is 3/5. How many green balls were in bag 2 at the outset? Knock-out Stage: 5 Challenge Level: Before a knockout tournament with 2^n players I pick two players. What is the probability that they have to play against each other at some point in the tournament? Stage: 4 and 5 Challenge Level: Some relationships are transitive, such as `if A>B and B>C then it follows that A>C', but some are not. In a voting system, if A beats B and B beats C should we expect A to beat C? Which Spinners? Stage: 3 and 4 Challenge Level: Can you work out which spinners were used to generate the frequency charts? Weekly Challenge 37: Magic Bag Stage: 5 Challenge Level: A weekly challenge concerning combinatorical probability. Odds and Evens Stage: 3 and 4 Challenge Level: Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers? Distribution Differences Stage: 4 Challenge Level: How could you compare different situation where something random happens ? What sort of things might be the same ? What might be different ? Rain or Shine Stage: 5 Challenge Level: Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry. Stage: 4 Challenge Level: A counter is placed in the bottom right hand corner of a grid. You toss a coin and move the star according to the following rules: ... What is the probability that you end up in the top left-hand. . . . Interactive Spinners Stage: 3 and 4 Challenge Level: This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events. Limiting Probabilities Stage: 5 Challenge Level: Given probabilities of taking paths in a graph from each node, use matrix multiplication to find the probability of going from one vertex to another in 2 stages, or 3, or 4 or even 100. Bet You a Million Stage: 4 Challenge Level: Heads or Tails - the prize doubles until you win it. How much would you pay to play? The Birthday Bet Stage: 4 Challenge Level: The next ten people coming into a store will be asked their birthday. If the prize is £20, would you bet £1 that two of these ten people will have the same birthday ? Snooker Frames Stage: 5 Challenge Level: It is believed that weaker snooker players have a better chance of winning matches over eleven frames (i.e. first to win 6 frames) than they do over fifteen frames. Is this true? In a Box Stage: 4 Challenge Level: Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair? How Many Balls? Stage: 5 Challenge Level: A bag contains red and blue balls. You are told the probabilities of drawing certain combinations of balls. Find how many red and how many blue balls there are in the bag. Card Game (a Simple Version of Clock Patience) Stage: 4 Challenge Level: Four cards are shuffled and placed into two piles of two. Starting with the first pile of cards - turn a card over... You win if all your cards end up in the trays before you run out of cards in. . . .	1,896	8,323	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2017-04	longest	en	0.939798
https://www.chegg.com/homework-help/questions-and-answers/conceptual-question-multiple-choice-m-reallyunderstanding-difference-3-points-changesat--c-q196998	1,529,829,990,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00564.warc.gz	782,974,830	11,454	This is a conceptual question. Multiple choice. I'm not reallyunderstanding the difference between the 3 points and what changesat each. The conducting plates are P and Q. P is the plate inbetween I and II and Q is the other plate under all 3. The questionis: Two parallel conducting plates P and Q are connected to abattery of emf e, as shown. A test charge placed successively atpoints I, II, and III. If edge effects are negligible, the force onthe charge when it is at point III is A) much greater in magnitude than the force on the charge whenit is at point II, but in the same direction. B) much less in magnitude than the force on the charge when itis at point II, but in the same direction. C) of equal magnitude and same direction as the force on thecharge when it is at point II. D) equal in magnitude to the force on teh charge when it is atpoint I but in the opposite direction. E) of equal magnitude and same direction as tthe force on thecharge when it is at point I.	231	981	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2018-26	latest	en	0.937645
https://kr.mathworks.com/matlabcentral/answers/1595809-question-in-continuous-time-signals	1,642,462,108,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00450.warc.gz	417,436,775	26,440	# Question in continuous time signals 조회 수: 1(최근 30일) Fotis Val 2021년 11월 25일 답변: VBBV 2021년 11월 25일 I'm not sure about the represantation of a continuous time signal. Am I doing something wrong? %% DT = 0.01; t = [-2:DT:12]; %% Question 2.1 %Parts of Signal t1 = -2:DT:7; y1_1 = (exp(abs(-0.5t1))).(2sin(10pit1)+cos(1pit1)); t2 = 8:DT:12; y1_2 = (0.75).^abs(-t2+10); % Final Signal %t = [t1 t2]; y1 = [y1_1 y1_2]; %% Question 2.2 %Parts of Signal t1 = -2:DT:7; y2_1 = (exp(abs(-0.5(-t1-2)))).(2sin(10pi(-t1-2)))+cos(1pi(-t1-2)); t2 = 8:DT:12; y2_2 = (0.75).^abs(t2+12); % Final Signal t = [t3 t4]; y2 = [y2_1 y2_2]; %% Conv Function tconv = (t(1) +t(1):DT:t(end)+t(end)); F1 = conv(y1,y2)DT; stem(tconv,F1); 댓글을 달려면 로그인하십시오. ### 답변(3개) Image Analyst 2021년 11월 25일 Why are you going from 8 to 12 instead of (7+DT) to 12? ##### 댓글 수: 0표시숨기기 이전 댓글 수: -1 댓글을 달려면 로그인하십시오. Star Strider 2021년 11월 25일 Everything in a computer are discrete, not continuous (except the voltages on the chips themselves and the outputs of a soundcard or similar DAC transducers). So ‘continuous’ signals otherwise only exist in symbolic variables, not numeric arrays or vectors, since by definition those are all sampled. The posted code is doing everything it should (or could) to represent the signal vectors correctly (although some variables appear to be missing). %% DT = 0.01; t = [-2:DT:12]; %% Question 2.1 %Parts of Signal t1 = -2:DT:7; y1_1 = (exp(abs(-0.5t1))).(2sin(10pit1)+cos(1pit1)); t2 = 8:DT:12; y1_2 = (0.75).^abs(-t2+10); % Final Signal %t = [t1 t2]; y1 = [y1_1 y1_2]; %% Question 2.2 %Parts of Signal t1 = -2:DT:7; y2_1 = (exp(abs(-0.5(-t1-2)))).(2sin(10pi(-t1-2)))+cos(1pi(-t1-2)); t2 = 8:DT:12; y2_2 = (0.75).^abs(t2+12); % Final Signal t = [t3 t4]; Unrecognized function or variable 't3'. y2 = [y2_1 y2_2]; %% Conv Function tconv = (t(1) +t(1):DT:t(end)+t(end)); F1 = conv(y1,y2)DT; stem(tconv,F1); . ##### 댓글 수: 0표시숨기기 이전 댓글 수: -1 댓글을 달려면 로그인하십시오. VBBV 2021년 11월 25일 DT = 0.01; t = [-2:DT:12]; %%Question 2.1 %Parts of Signal t1 = -2:DT:7; y1_1 = (exp(abs(-0.5t1))).(2sin(10pit1)+cos(1pit1)); t2 = 8:DT:12; y1_2 = (0.75).^abs(-t2+10); % Final Signal %t = [t1 t2]; y1 = [y1_1 y1_2]; %%Question 2.2 %Parts of Signal t1 = -2:DT:7; y2_1 = (exp(abs(-0.5(-t1-2)))).(2sin(10pi(-t1-2)))+cos(1pi(-t1-2)); t2 = 8:DT:12; y2_2 = (0.75).^abs(t2+12); % Final Signal t = [t1 t2]; y2 = [y2_1 y2_2]; %%Conv Function tconv = (t(1) +t(1):DT:t(end)+t(end)); F1 = conv(y1,y2)*DT; stem(F1); figure(2); plot(t,y1,t,y2) The variables t3 and t4 are same as t1 and t2. So your discrete and continuous signal representations do match well. 댓글을 달려면 로그인하십시오. ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by	1,225	2,824	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.40625	3	CC-MAIN-2022-05	latest	en	0.523118
https://cpep.org/mathematics/708576-in-a-marble-collection-1-8-of-the-marbles-are-blue-of-the-blue-marbles.html	1,696,344,296,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233511106.1/warc/CC-MAIN-20231003124522-20231003154522-00299.warc.gz	202,999,481	7,429	17 September, 09:33 In a marble collection, 1/8 of the marbles are blue. Of the blue marbles, 1/2 have sparkles. What fraction of the marbles in the collection are blue with sparkles? +4 1. 17 September, 09:42 0 Hello, I believe the answer is 3/8 of the marbles in the collection are blue with sparkles. I say so because if 1/8 of the marbles are blue and of the blue marbles, 1/2 of them have sparkles, the remaining of the marbles must be blue with sparkles. Solution: 1/2 - 1/8 = 3/8 Thus, 3/8 of the marbles in the collection are blue with sparkles. Faith xoxo	179	572	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2023-40	latest	en	0.898615
www.connectwithdon.com	1,561,414,314,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627999740.32/warc/CC-MAIN-20190624211359-20190624233359-00082.warc.gz	222,379,977	11,835	Search Social Connections Monday Jun102013 ## Where's My Elevator? If I ever got a job teaching introductory computer science, the first major assignment I would give would be to write an algorithm for the optimal distribution of a bank of elevators. If the current state of the art is any indication, everyone would fail. I know it doesn't sound thrilling to most, but there's actually nothing more intriguing than standing in the lobby of a large building and watching the crazy paths some elevators take to pickup their cargo. Often, you can guess the algorithm used by just observing the traffic. There's the Stay-Where-You-Are-And-Wait algorithm. This one usually lead to a mad dash from all floors when the first request comes in. Others employ the sometimes more optimal Half-Sit-On-The-Top-Floor-The-Other-Half-Sit-On-The-Ground-Floor algorithm, which usually leads to confusion when picking up passengers in the middle ("Are you going to go get him, or shall I? Me? No, you? Wait, I'll get him. Oh, okay, you've got him.") Then there's the Whichever-One-Is-Going-In-The-Same-Direction algorithm. You can be on the 2nd floor with 10 elevators sitting on Ground, but if you push the 'down' button, you'll have to wait as an elevator from the 40th floor slowly makes its way to you. Apparently, elevators are subject to a new law of physics which makes moving a short distance, stopping, and returning, too difficult. There is nothing...NOTHING...more frustrating than waiting for an elevator, only to have it shoot right past your floor! Unless, of course, it's finally getting an elevator to stop at your floor, only to find that it's filled to capacity and you're staring face-to-face with 20 stressed and cramped passengers who are all silently giving you signals that if you even attempt to enter the car, you should not expect to exit alive. The building I used to work in had a frustrating flaw in its elevator control system. Say I was on the 5th floor and requested an elevator going down. It would dispatch a free car from, say, the 10th floor to come and get me. If another elevator going up happened to stop on the 5th to release a passenger, the algorithm was totally confused. My elevator would think "Well...I was called to pick up Don on the 5th floor, but I see there's already an elevator on the 5th, so he can take that one." and would bail on coming to pick me up. Of course, I didn't want to go UP with the rest of the people in the 5th floor car, so I essentially got dropped from the system. My request light would be extinguished, and I would have to try all over again. Modern elevators have been in existence since the mid-1800s and the technology hasn't really changed all that much over the years. The system has become automated, and significantly safer, but modern advances in computer systems has done little to help optimize the flow of traffic. I witnessed one valiant effort on a recent trip to the West Coast. My hotel was equipped with an elevator system with NO floor buttons inside the car. You had to tell the elevator which floor you wanted when you called it. This way, it could optimize elevator usage. For instance, to get to your room from the lobby, you would press your floor number on one of a series of panels in the hallway, and the system would tell you which elevator to take when it arrived. Only the named elevator would be told to stop on your floor. Of course, this led to all kinds of interesting scenarios. If you got on the wrong elevator, you were toast. If you happened to be chatting with your colleagues as they called the elevator, and you all entered the same one, you had better hope they were staying on the same floor as you. There was nothing funnier than seeing people walk into a waiting elevator car, wait until the doors closed, and then look for a button to press, not finding any. Talk about a feeling of entrapment! One additional feature of this system, was that you could simply 'scan' your room card in the elevator hallway, and it would know what floor to send you to without you even pushing a button. Well...in theory at least. In my case, when I scanned my room keycard, the system requested an elevator for "Floor 31" on my behalf. Unfortunately, my room was located on the 21st floor. That's progress for you. I'm sure that being in the elevator business must have its ups and downs (Come on! You knew I was going to say something like that, didn't you?). But I think it's time we put some real effort into optimizing the algorithms in these systems. If you ever have me as your computer science teacher...you'd better be prepared! Don Saturday May192012 ## A Quarter Century in Tech Twenty five years ago today, on May 19th 1987, a young, naive software engineer fresh out of university, took a deep breath and walked through the doors of his first full-time job. That software engineer, was me. The tech world was a different place then. There was no Internet. There were no iPhones. Twitter was strictly a bird sound. And contrary to what my kids think, yes, there were computers, but a fast computer was an IBM PC '386' with a math co-processor. My first job, at Canadian high tech giant Cognos, was to use the beast to program a 4th generation language called "Powerhouse". Powerhouse was designed to run on much larger 'mini' computers, so the challenge of building a PC version was daunting at times. Sharing files among team members meant using the 'SneakerNet'. Without a proper network to share files on, SneakerNet was simply a set of physical binders with floppy disks in them. If you wanted to update a master file, you had to put on your 'sneakers' and walk over to the master cabinet where the binders were stored. It was much less efficient than an online system, but you certainly got your exercise! I remember the day the new '486' PCs arrived. They were so fast that you could ask DOS for a directory listing and it would give it to you so fast you couldn't read it before it scrolled off the screen. Now that's fast! The building wasn't wired to have so much power drawn from each cubicle, and I can remember blowing many circuits on our section of the floor. I had a secret weapon though...a 30ft extension cord. That meant I could simply plug my PC into the extension cord and walk it to the other side of the building and get power from there. One day, the IT department connected to this new service called the WorldWide Web. The Web was only available on a single machine in the glassed-in computer room. As a new technology guy, they asked me to take a look at it. I remember the first page I went to was a tourist site for Hawaii. It was so cool to think that I was actually connected to information in a distant land! "Hmmm. Who knows...this thing could catch on." I was contracted by the company to work after hours and create the first Cognos website. I learned to write HTML by hand, and soon www.cognos.com was alive! The business changed from creating 4GL software, to this new thing called 'Business Intelligence', and our technology changed with it. Our PCs became 'connected', and even replaced by connected workstations. Our development environments matured, and we dabbled (somewhat unsuccessfully) with object oriented languages. I became a graphics guy and had a system with a special high performance graphics card costing thousands of dollars. Of course, whatever device you're viewing this blog on likely costs much less, and is orders of magnitude more powerful. Such is the world of technology. After many years of fiddling with technology, and the occasional award and wildly successful product, I eventually landed with my dream job as Chief Technology Officer. It's the only position I ever aspired to. CEOs work way too hard, but CTOs get to play with new stuff and experiment all day. I always seemed to have a new gadget at my disposal, with lots of people asking me to show them how it worked, and that was key to making the experience fun. Blazing new trails is always more interesting than following in someone else's footsteps. Our acquisition by IBM only opened the aperture of possibility. Still with a CTO role, but now having access to world leading technology, and an entire division of top notch research talent meant more places to play. In the past 25 years, I think I've learned a lot about technology and how it applies to business. I've learned from leading experts, and from customers always pushing the envelope. I've had the extremely good fortune of traveling the globe, and speaking to tens of thousands of people about technology trends and directions. Perhaps I've even influenced a few. One of the best parts about a career in high tech, is that it is always changing. From the PC to the Internet to social media to whatever the future holds. Guaranteed, it will be exiting and challenging. As my friends and family know, this journey with Cognos/IBM, lasting a quarter century, is about to end for me. Retiring next month will provide me, oddly enough, with even more opportunity to parlay my knowledge and interests forward. I'm looking forward to the next 25 years in tech. We may still not have our flying cars, but I'll bet it will be spectacular! Don Tuesday Nov152011 ## Playing with Time Of all the powers we have as humans, there is nothing we are more powerless against...than time. - Don Campbell Many of the great science fiction authors of our time have dealt with the wonders of time travel. From H. G. Wells to Isaac Asimov, they have depicted a world where the human race has finally mastered this elusive ability, and made time its master, although often to some unforeseen and tragic conclusion. Even classic films such as "Back to the Future", "Bill and Ted's Excellent Adventure", and "Hot Tub Time Machine" have taken a crack at it. Okay, maybe not the 'classic' films you were thinking of, but you get my point. Controlling time has always been a human fantasy. Well, maybe we already control more than you think... Some of our most successful technological advancements are really time modifiers. Look at this blog for example. I write a post, and sometime in the future, when you're looking for inspiration and a fascinating read, you dive in. It doesn't matter what time these thoughts run from my brain to my fingertips, you can enjoy them whenever you wish. It's really the same concept with email, which is why I prefer it to instant messaging, and voicemail, which is why I seldom answer my phone. Entertainment is much the same. I almost NEVER watch a live television event. My digital video recorder captures all the shows that are interesting to me, and saves them on its disk for me to watch whenever I wish. I'll even record a live broadcast so I can watch a pre-recorded show rather than be at the mercy of time. The power of pressing PAUSE, going to the kitchen for a snack, and returning exactly where you left off, is a heady power trip! One of my favourite hobbies is photography. Essentially, that's the art (and science) of forever capturing a moment in time. That moment will never happen again. Think how important old images are to you. Pictures of your youth. Your wedding day. Your child's birth. Freezing time in an instant, so you can recall it later. Often, the element of time, rather than their artistic nature, is the very thing that makes these images priceless. The list of impactful time-altering technologies goes on and on. You could put up a good argument that the microwave alters time too, "I don't want to eat in 30 minutes. I want to eat in 30 seconds!" How powerful is that! So the next time you're reading "The Time Machine" or watching an episode of "Dr. Who", remember that time altering technology is not as elusive as you think. Now if only I could use some of this power to restore my constantly receding hairline. Don Page 1 2 3 4 ... 4	2,637	11,908	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2019-26	latest	en	0.961911
https://www.cytanderson.org/actual-length-of-a-year/	1,701,473,267,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00768.warc.gz	846,844,127	16,750	Home » Actual Length Of A Year # Actual Length Of A Year ### Contents Gregorian Calendar — from Eric Weisstein's World of Astronomy – This also gives an exact number of weeks per 400-year cycle. The Gregorian calendar was constructed to give a close approximation to the tropical year, which is the actual length of time it takes for the Earth to complete one orbit around the Sun. . The Julian calendar was switched over to the Gregorian starting in 1582, at which point the 10 day difference between the actual time of year and. Texas Tax Amount Texas Sales Tax Rate – 2019 – Tax-Rates.org – The Tax. – The Texas state sales tax rate is 6.25%, and the average TX sales tax after local surtaxes is 8.05%. Groceries, prescription drugs and non-prescription drugs are exempt from the texas sales tax. counties and cities can charge an additional local sales tax of up to 2%, for a maximum possible combined sales tax of 8.25%. Because the exact length of a year is a little less than 365. – The whole point of the extra leap year rules — no extra day in a year divisible by 100, but actually yes if it is divisible by 400 — is in place to account for the slight difference between 365.25 and the actual rotational amount. answer has 3 votes. The exact length of a year – isn’t. It is a variable quantity. FSD Pharma Provides Update on Status of 2018 Year-End Filings – “2018 was an extraordinary year for FSD” said recently appointed CEO Dr Raza Bohkari. review all required filings which. Our Year \| Calendars – The Christian calendar (Gregorian calendar) is based on the motion of the earth around the sun, while the months have no connection with the motion of the moon.. On the other hand, the Islamic calendar is based on the motion of the moon, while the year has no connection with the motion of the earth around the sun.. Finally, the Jewish calendar combines both, in that its years are linked to the. Pi Day Record: Google Developer Just Calculated Pi to 31.4 Trillion Digits – While you are Googling for sweet deals on actual pies, let’s not forget that pi, or , is a mathematical number to begin with, one that describes the ratio of a circle’s circumference to its diameter, Va Refinance Texas Quicken Loans Closing Costs Calculator Home Loans – Get a low rate mortage in South Africa today! – Find Your Dream Home with Affordable Home Loans! Whether you are a first time buyer or you are seeking to purchase an additional property, www.Home-Loans.org.za offers the most effective way to purchase your dream home. buying a house is one of the biggest decisions that you could ever make in your lifetime, and for many potential homeowners, it can also be one of the most stressful decisions.Usda Loan Forms Grants and Loans \| USDA – FSA loans can be used to purchase land, livestock, equipment, feed, seed, and supplies. Loans can also be used to construct buildings or make farm improvements. housing assistance. USDA provides homeownership opportunities to low- and moderate-income rural Americans through several loan, grant, and loan guarantee programs.VA Cash Out Refinance \| Cash Out Loans for. – VA Cash Out Refinance loans enables veterans to use their home equity to pay off debt or make home improvements at lower rates. Find out how today! Predicting the actual length of premolar teeth on the basis of. – Year : 2010 \| Volume : 21 \| Issue : 4 \| Page : 468-473. Aim: The aim of the study was to predict the actual length of the premolar teeth, based on. Estimation of the actual length of teeth from panoramic radiographs can help the orthodontists . 4/11/2019 · An arm’s length transaction is a business deal in which the buyers and sellers act independently and do not have any relationship to each other. The concept of an arm’s length transaction assures. Current Refi Rates In Texas Refinancing frees up funds necessary for savings and house repairs – Should we refinance with lower house payments or get a home equity. That, along with the high interest rate on your current mortgage and possibly questionable credit (know for sure by getting your. Actual Size by Steve Jenkins, Paperback \| Barnes & Noble – Steve Jenkins shows the actual size of animals in fold out pages and two page layouts that are impressive to look at. Each picture comes with the size and weight information of each animal listed. It almost feels like looking into the eyes of the animals. Home Alone House Mortgage How to Avoid Getting Scammed With a Reverse Mortgage – If you didn’t save enough for retirement, know that you aren’t alone – the majority of. If you truly have that much equity in your house, you could even sell your home and downsize to a smaller. Sitemap	1,022	4,691	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2023-50	latest	en	0.915864
https://it.mathworks.com/matlabcentral/profile/authors/25269023	1,670,218,439,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00072.warc.gz	346,241,825	23,375	Community Profile ### MathWorks Last seen: 4 giorni fa Attivo dal 2022 I am an EDG Intern at MathWorks. DISCLAIMER: Any advice or opinions here are my own, and in no way reflect that of MathWorks #### Content Feed Visto da Risposto how to plot 3d using patch? Hello, As per my understanding of the query ,you want to use the example code for plotting a 3d graph using patch function but ... 6 mesi fa \| 0 \| accettato Risposto Correctly storing data in cell arrays Hello! As per my understanding of the query , you wish to have a gap between printing data of 2 different cells. You can use “e... 6 mesi fa \| 0 Risposto How to use fread from TCP/IP of a string? 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You can do that by cre... 9 mesi fa \| 0 Risposto How to choose a ROI in an image and mask the other data in a colored image? Hello As per my understanding of your query you wish to mask the region other than the yellow-colored part of the image. You ca... 9 mesi fa \| 0 Risposto Rotate and Position Node Labels Hello As per my understanding of your query you wish to customize the position and orientation of the nodes so that they can be... 9 mesi fa \| 0 \| accettato Risposto How can I run a code for multiple rows which I found from another variable? Hello, I understand that you want to get the times for each row and you can accomplish that using a for loop and array indices.... 9 mesi fa \| 0 \| accettato Risposto How to Input a value, get a result, input that result to calculate another value Hello, As mentioned in the comments please change h to h(t) wherever you are using assignment and comparison operators. The er... 10 mesi fa \| 0 Risposto I tried to write a code to find the next smallest prime number greater than any given positive integer. My code is stuck in infinity loop, how can I find where is the problem? Hello, I understand that you’re trying to optimize the above function to find next smallest prime number. For optimizing the ... 10 mesi fa \| 0 \| accettato Risposto Need the function to run multiple times and store each data point for each input variable Hello, I understood that you want to execute the same function for customer where it is a 1x20 struct array. The following func... 10 mesi fa \| 0 Risposto How do I make the output of a loop the new input value, and loop it 15 times. Hello, I understood that you want to know how to pass the FrstMove to start for 15 iterations.In the code you have tried to ass... 10 mesi fa \| 0 Risposto Modulo Operator in Embedded Coder Hello, I understand that you want to know the reason behind the embedded coder’s own function definition instead of the % (modu... 10 mesi fa \| 1 Risposto Auto resize of manually added axes Hello, As per my understanding of the query, you are trying to find out why the axes you added manually is not resizing w... 10 mesi fa \| 0 Risposto Rank of symbolic matrix is wrong (but too small, not too large) Hello, This is a known issue and has been communicated with the internal staff and it would be considered for a fix in the f... 10 mesi fa \| 4 \| accettato Risposto Is it possible to write a class that inherits from timetable? Hello, As per my understanding of the query, you are trying to inherit a sealed class which is not possible. For your case... 10 mesi fa \| 0 Risposto Breaking Vector into Subvectors Hello, As per my understanding of the query, you are trying to split the vector into sub vectors. For this you can first g... 10 mesi fa \| 0 Risposto Join tables based on month values Hello, As per my understanding of the query, you are trying to join two tables with similar month values using outer join.... 10 mesi fa \| 0 Risposto 3D Plot of Probability Density Functions using normpdf and line function Hi I understand that you’re trying to plot consecutive PDFs using 3D plot. Following code snippet might help to resolve the is... 10 mesi fa \| 0 Risposto How do I find the value for which x% of the array is covered? Hi My understanding of the question is that you want to sort a 3-dimensional array and you want to extract the minimum for whi... 10 mesi fa \| 0 \| accettato	1,266	5,015	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2022-49	longest	en	0.802704
https://www.qalaxia.com/questions/How-high-would-water-rise-in-a-pipes-of-the	1,686,272,353,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00181.warc.gz	1,016,467,668	3,536	Sahil Khan 1 Hydrostatic pressure (P) under a water depth of h can be given as P = h \rho g , where \rho is water density at normal temperature and pressure, while g is acceleration due to gravity. Water pressure P = 270 kPa Converting Kilopascal to Pascal we get P = 270000 Pa = 270000 \frac{kg}{m \cdot s^{2}} Plugging this in P = h\rho g and solving for h we get, h = \frac{P}{\rho g} = \frac{270000 \frac{kg}{m\cdot s^{2}}}{(1000 \frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})}= 27.51 m Qalaxia Knowlege Bot 0 How high will water rise in the pipe of a building if the water ... Jan 5, 2018 ... How high will water rise in the pipe of a building if the water pressure gauge shows the pressure at the ground floor to be 270kpa? For more information, see How high will water rise in the pipe of a building if the water ... Qalaxia QA Bot 0 I found an answer from www.quora.com If the water pressure gauge shows the pressure at ground floor to be ... What it WILL do is increase the volume of water that can flow, which will reduce the pressure loss at the other end of the pipe when water is flowing. The longer the ... For more information, see If the water pressure gauge shows the pressure at ground floor to be ...	344	1,226	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.25	3	CC-MAIN-2023-23	longest	en	0.855528
https://tex.stackexchange.com/questions/321148/enhancing-booktab-tables-containing-tikz-picture	1,653,521,687,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00081.warc.gz	636,612,073	66,422	# Enhancing Booktab Tables containing Tikz picture I have created a table containing text, equations and a tikz picture [created using Geogebra and scaled using \resizebox]. As a beginner, I found it difficult to center my content and to remove the origin label from the graph. Can somebody help me with this problem? I would also like to ask for tips to enhance my code to help avoid confusion. Thank you. \documentclass[11pt,dvipsnames,landscape,twocolumn]{article} \usepackage{mathtools} \usepackage{pgf,tikz} \usepackage{mathrsfs} \usetikzlibrary{arrows} \pagestyle{empty} \usepackage[utf8]{inputenc} \usepackage{tkz-euclide} \usepackage{multicol} \usepackage[T1]{fontenc} \usepackage{array, booktabs, mathrsfs, charter, blindtext, varwidth, tabu} \usepackage[expert]{mathdesign} \begin{document} \framebox{Trigonometric Integrals} \begin{center} \begin{tabular}{cccc} \toprule \textsc{\fontsize{10}{7.2}\selectfont \parbox{1.8cm}{Integrand\\[3pt] Contains:}} & \textsc{\fontsize{10}{100}\selectfont \parbox{5.9cm} {\begin{center} Substitution \end{center}}} & \textsc{\fontsize{10}{7.2}\selectfont \parbox{2cm}{Restriction\\[3pt] \centering $\theta \in [0,2\pi]$}}\par \\ \midrule $\sqrt{a^2 - u^2}$ & \parbox{5.5cm}{let $u = a\sin \theta$ \quad $du = a\cos \theta \; d\theta$ then \boldmath{$\sqrt{a^2-u^2} = a\cos \theta$}} & \resizebox{1.3cm}{1.3cm}{\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm] \draw[-,,color=black] (-1.5,0.) -- (1.5,0.); \foreach \x in {.,.} \draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$}; \draw[-,color=black] (0.,-1.5) -- (0.,1.5); \foreach \y in {.,.} \draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$}; \draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$}; \clip(-1.5,-1.5) rectangle (1.5,1.5); \draw [shift={(0.,0.)}] plot[domain=-1.5707963267948966:1.5707963267948966,variable=\t]({1.1.cos(\t r)+0.1.sin(\t r)},{0.1.cos(\t r)+1.1.sin(\t r)}); \begin{scriptsize} \draw [fill=black] (0.,1.) circle (3.5pt); \draw [fill=black] (0.,-1.) circle (3.5pt); \end{scriptsize} \end{tikzpicture}} \\ \bottomrule \end{tabular} \end{center} \end{document} Edit: Here is a picture for the guide. Thanks. • For the zero, remove the line \draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$}; Jul 27, 2016 at 4:35 To scale your picture you can just use key scale for tikzpicture environment. To draw half unite circle rather then \draw [shift={(0.,0.)}] plot[domain=-1.5707963267948966:1.5707963267948966,variable=\t]({1.1.cos(\t r)+0.1.sin(\t r)},{0.1.cos(\t r)+1.1.sin(\t r)}); \draw (0,-1) arc [start angle=-90, end angle=90, radius=1cm]; And to center your picture on the tabular cell you can add baseline=(current bounding box.center) Code \documentclass[11pt,dvipsnames,landscape,twocolumn]{article} \usepackage{mathtools} \usepackage{pgf,tikz} \usepackage{mathrsfs} \usetikzlibrary{arrows} \pagestyle{empty} \usepackage[utf8]{inputenc} \usepackage{tkz-euclide} \usepackage{multicol} \usepackage[T1]{fontenc} \usepackage{array, booktabs, mathrsfs, charter, blindtext, varwidth, tabu} \usepackage[expert]{mathdesign} \begin{document} \framebox{Trigonometric Integrals} \begin{center} \begin{tabular}{cccc} \toprule \textsc{\fontsize{10}{7.2}\selectfont \parbox{1.8cm}{Integrand\\[3pt] Contains:}} & \textsc{\fontsize{10}{100}\selectfont \parbox{5.9cm} {\begin{center} Substitution \end{center}}} & \textsc{\fontsize{10}{7.2}\selectfont \parbox{2cm}{Restriction\\[3pt] \centering $\theta \in [0,2\pi]$}}\par \\ \midrule $\sqrt{a^2 - u^2}$ & \parbox{5.5cm}{let $u = a\sin \theta$ \quad $du = a\cos \theta \; d\theta$ then \boldmath{$\sqrt{a^2-u^2} = a\cos \theta$}} & \begin{tikzpicture}[scale=0.5,baseline=(current bounding box.center)] \draw (-1.5,0) -- (1.5,0); \draw (0.,-1.5) -- (0.,1.5); \draw (0,-1) arc [start angle=-90, end angle=90, radius=1cm]; \fill (0,1) circle (3.5pt); \fill (0,-1) circle (3.5pt); \end{tikzpicture} \\ \bottomrule \end{tabular} \end{center} \end{document} Output	1,533	4,107	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2022-21	latest	en	0.441711
http://massivealgorithms.blogspot.com/2016/04/2185-milking-grid.html	1,529,605,455,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00508.warc.gz	203,905,904	35,299	## Monday, April 11, 2016 ### POJ 2185 -- Milking Grid 2185 -- Milking Grid Every morning when they are milked, the Farmer John's cows form a rectangular grid that is R (1 <= R <= 10,000) rows by C (1 <= C <= 75) columns. As we all know, Farmer John is quite the expert on cow behavior, and is currently writing a book about feeding behavior in cows. He notices that if each cow is labeled with an uppercase letter indicating its breed, the two-dimensional pattern formed by his cows during milking sometimes seems to be made from smaller repeating rectangular patterns. Help FJ find the rectangular unit of smallest area that can be repetitively tiled to make up the entire milking grid. Note that the dimensions of the small rectangular unit do not necessarily need to divide evenly the dimensions of the entire milking grid, as indicated in the sample input below. Input * Line 1: Two space-separated integers: R and C * Lines 2..R+1: The grid that the cows form, with an uppercase letter denoting each cow's breed. Each of the R input lines has C characters with no space or other intervening character. Output * Line 1: The area of the smallest unit from which the grid is formed Sample Input `2 5 ABABA ABABA ` Sample Output `2 ` Hint The entire milking grid can be constructed from repetitions of the pattern 'AB'. http://www.fgdsb.com/2015/01/13/minimum-cover-matrix/ ``` int min_cover_mat(const vector<string>& map) { if(map.empty() \|\| map[0].empty()) return 0; int rows = (int)map.size(), cols = (int)map[0].size(); auto lcm = [](int a, int b){ int mul = a * b; for(int r = a % b; r ;){ a = b; b = r; r = a % b; } return mul / b; }; auto num_cover_substr_col = [&](int r) { int next[cols+1], i = 0, j = -1; next[0] = -1; while(i < cols){ if(j == -1 \|\| map[r][i] == map[r][j]){ ++i, ++j; next[i] = j; } else j = next[j]; } return i - next[i]; }; auto num_cover_substr_row = [&](int c) { int next[rows+1], i = 0, j = -1; next[0] = -1; while(i < rows){ if(j == -1 \|\| map[i][c] == map[j][c]){ ++i, ++j; next[i] = j; } else j = next[j]; } return i - next[i]; }; int r_nums = 1, c_nums = 1; for(int r = 0; r < rows; r ++){ c_nums = lcm(c_nums, num_cover_substr_col(r)); if(c_nums > cols){ c_nums = cols; break; } } for(int c = 0; c < c_nums; c ++){ r_nums = lcm(r_nums, num_cover_substr_row(c)); if(r_nums > rows){ r_nums = rows; break; } } return r_nums * c_nums; } ```	714	2,435	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2018-26	latest	en	0.825208
http://illuminations.nctm.org/Activity.aspx?id=6377	1,490,667,899,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218189589.37/warc/CC-MAIN-20170322212949-00050-ip-10-233-31-227.ec2.internal.warc.gz	179,228,223	10,308	## 6.1 Understanding Multiplication Using Dynamic Sketches of an Area Model Grade: 6-8 Standards: Math Content: Number and Operations Students are encouraged to build conceptual reasoning for multiplying decimals by manipulating an area model. Multiply decimals up to 7.0×7.0 by dragging the white slides. You can also alter the area by clicking on the tab (located on the upper right hand corner of the rectangle). Create a rectangle with a fixed width of 3. The height, y, will vary. Thus, the product 3y will represent the area of a 3-by-y rectangle. Change the value of y so that y>1 and y<1. How do these areas/products compare? What do you observe?	156	658	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.046875	3	CC-MAIN-2017-13	longest	en	0.839221
http://openstudy.com/updates/4fcae879e4b0c6963ad4b5e7	1,444,493,993,000,000,000	text/html	crawl-data/CC-MAIN-2015-40/segments/1443737958671.93/warc/CC-MAIN-20151001221918-00192-ip-10-137-6-227.ec2.internal.warc.gz	237,487,120	11,163	## xRAWRRx3 3 years ago Please help :( The slope of line JD is 1/4, J has coordinates (10, -4), and D has coordinates (6, y). What is the value for y? 1. Chlorophyll m = Δy/ Δx ( y + 4) / ( 6 - 10 ) = 1/4 => y = ...? Can you fill in the blank 2. xRAWRRx3 well, i got as far as to: y + 4 / -4 = 1/4 but I'm not positive what to do after that. 3. Chlorophyll -> y + 4 = -1 Can you finish it now? 4. xRAWRRx3 how'd you get that? 5. Chlorophyll -4/4 = -1 6. xRAWRRx3 and how'd you get that? 7. xRAWRRx3 like what exactly do I have to do from: y-4/-4 = 1/4? 8. xRAWRRx3 oh, so y would be -5? 9. xRAWRRx3 wouldn't it be y +4 = -1? 10. Chlorophyll Yep, x = -5 ! 11. xRAWRRx3 Thank you SO much x) you helped me a lot :) 12. xRAWRRx3 Thanks xD I'm a very open person so I just ask what I need to ask LOL I hate not knowing or understanding things x) 13. xRAWRRx3 LOL I know right? It's so frustrating sometimes when people do so xD and thanks x) 14. Chlorophyll Sometime it's tough to report to god that I've done a good deed today :P 15. xRAWRRx3 LOL I'll send an extra thought to him for you ;) LOL 16. Chlorophyll Nice =) 17. xRAWRRx3 hey, can you help me with another geometry question? x)	440	1,218	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2015-40	longest	en	0.913737
https://niniloos.com/function-table-answer-key/	1,620,615,045,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00602.warc.gz	436,613,542	8,687	# Function table answer key Most Effective » » Function table answer key Most Effective Your Function table answer key images are available in this site. Function table answer key are a topic that is being searched for and liked by netizens today. You can Download the Function table answer key files here. Download all royalty-free vectors. If you’re looking for function table answer key pictures information related to the function table answer key topic, you have come to the right site. Our website always gives you suggestions for seeking the highest quality video and picture content, please kindly hunt and find more enlightening video articles and graphics that match your interests. Function Table Answer Key. Function Table Worksheet Answer Key. Worksheet 1 8 Homework Piecewise Functions Answer Key. Patterns Function Machine Worksheets Free Commoncoresheets. Some of the worksheets for this concept are Objective you will be able to Quadratic patterns Name date ms Relations and functions work answers Function table work answers Solve the quadratic equations by factoring work 3 Quadratic functions teacher notes Representing quadratic functions. Identify The Proportion Function Tables Proportions Worksheet Ratio And Proportion Worksheet Geometry Worksheets From pinterest.com 35 function table worksheet answers resource plans identifying functions tables worksheet free commoncoresheets identifying points of a function. 12102017 Identifying Functions With Ordered Pairs Tables Graphs Lesson Transcript Study Com. 7 2 5 3 4 4 5 2. Great for a. Gallery of 20 Function Table Worksheet Answer Key. Using A Table Of Values To Graph Equations. ### Easily check their work with the answer sheets. The preview file shows the easiest and 2nd most difficult. Using the rule stated above the table calculate the required value and the output field. Patterns Function Machine Worksheets Free Commoncoresheets. 4 x fx 2 fx 15. Quadratic Functions From Tables With Answers - Displaying top 8 worksheets found for this concept. The preview file shows the easiest and 2nd most difficult. Source: pinterest.com The preview file shows the easiest and 2nd most difficult. Whitney has a total of 30 cupcakes for her guestsThe function rule 30. Each of the numbered sheets gets progressively more difficult. READ Round Table Pizza Clubhouse Rocklin Ca. This remaining mentioned we supply you with a number of basic but. Source: pinterest.com A Function Table - Linear Function L1ES1 x fx Complete the function table using the function rule fx 5x and answer the following questions. X where x is the number of guests can be used to find the number of cupcakes per guest. Great for a tiered lesson math center warm-up or just. A Function Table - Linear Function L1ES1 x fx Complete the function table using the function rule fx 5x and answer the following questions. 07012018 Function table worksheets in and out boxes math expressions 3rd grade words word problems complete the for each equation worksheet answer key brokeasshome com awesome complet graphing linear equations functions aids tessshlo eighth tables 10 one page writing r answers nidecmege pre algebra systems of you Function Table Worksheets In And Out Boxes. Source: in.pinterest.com A function table answer key is included with a little easier to operate for online marketplace where some phenomenon in each day please share the students. READ Round Table Pizza Clubhouse Rocklin Ca. Whitney has a total of 30 cupcakes for her guestsThe function rule 30. 2 x fx x. Worksheet 1 8 Homework Piecewise Functions Answer Key. Source: pinterest.com 07012018 Function table worksheets in and out boxes math expressions 3rd grade words word problems complete the for each equation worksheet answer key brokeasshome com awesome complet graphing linear equations functions aids tessshlo eighth tables 10 one page writing r answers nidecmege pre algebra systems of you Function Table Worksheets In And Out Boxes. Function Table Worksheet Answer Key. Answer key Complete each function table. Complete the function table. Answer key T1L1S1 Complete the function table. Source: pinterest.com Two Variable Linear Equations Intro Khan Academy. Complete the function table for each equation worksheet answer key brokeasshome com awesome co in 2020 graphing linear equations quadratic functions eighth grade tables 10 one page worksheets writing algebra identifying free commoncoresheets points of a and out boxes math expressions patterns 3rd words machine aids tessshlo. 24 function tables in all 12 horizontal and 12 vertical. Each of the numbered sheets gets progressively more difficult. 12102017 Identifying Functions With Ordered Pairs Tables Graphs Lesson Transcript Study Com. Source: pinterest.com Answer key T1L1S1 Complete the function table. 12062018 Ex 1 Graph A Linear Equation Using Table Of Values You. Kids will be able to easily review and practice their math skills. Whitney has a total of 30 cupcakes for her guestsThe function rule 30. 24 function tables in all 12 horizontal and 12 vertical. Source: pinterest.com Answer key Complete each function table. Printable Worksheets. Gallery of 20 Function Table Worksheet Answer Key. Easily check their work with the answer sheets. Each of the numbered sheets gets progressively more difficult. Source: pinterest.com 6 x x fx 3 fx 3. You will be able to go deep into your subject. 24 function tables in all 12 horizontal and 12 vertical. Here are 4 one-page sets of function tables WITH ANSWER KEYS. Make a table of values that shows the number of cupcakes each guest will get if there are 6 10 or 15 guests. Source: pinterest.com X where x is the number of guests can be used to find the number of cupcakes per guest. Function Table Worksheets With Answer Sheet. Some of the worksheets for this concept are Objective you will be able to Quadratic patterns Name date ms Relations and functions work answers Function table work answers Solve the quadratic equations by factoring work 3 Quadratic functions teacher notes Representing quadratic functions. B A - 1. Great for a. Source: pinterest.com 7 2 5 3 4 4 5 2. The preview file shows the easiest and 2nd most difficult. Complete the function table. B A - 1. Gallery of 20 Function Table Worksheet Answer Key. Source: pinterest.com Easily check their work with the answer sheets. The preview file shows the easiest and 2nd most difficult. Answer key T1L1S1 Complete the function table. I ii 15 40 B x fx 1 fx x. 07012018 Function table worksheets in and out boxes math expressions 3rd grade words word problems complete the for each equation worksheet answer key brokeasshome com awesome complet graphing linear equations functions aids tessshlo eighth tables 10 one page writing r answers nidecmege pre algebra systems of you Function Table Worksheets In And Out Boxes. Source: pinterest.com 2 6 2 3 7 2 2 7 0 20 16 4 10 5 2 16 12 0 2 3 4 8 4 3 4 2 5 0 4 4 6 0 7 4 8 6 10 8 6 9 3. Answer key Complete each function table. Quadratic Functions From Tables With Answers - Displaying top 8 worksheets found for this concept. 12102017 Identifying Functions With Ordered Pairs Tables Graphs Lesson Transcript Study Com. Great for a. Source: pinterest.com This remaining mentioned we supply you with a number of basic but. A function table answer key is included with a little easier to operate for online marketplace where some phenomenon in each day please share the students. Using A Table Of Values To Graph Equations. 24 function tables in all 12 horizontal and 12 vertical. Make a table of values that shows the number of cupcakes each guest will get if there are 6 10 or 15 guests. Source: pinterest.com Function Table Worksheets With Answer Sheet. Allows you by each table key is set in the sheets. 07012018 Function table worksheets in and out boxes math expressions 3rd grade words word problems complete the for each equation worksheet answer key brokeasshome com awesome complet graphing linear equations functions aids tessshlo eighth tables 10 one page writing r answers nidecmege pre algebra systems of you Function Table Worksheets In And Out Boxes. Gallery of 20 Function Table Worksheet Answer Key. Make a table of values that shows the number of cupcakes each guest will get if there are 6 10 or 15 guests. Source: pinterest.com Some of the worksheets for this concept are Objective you will be able to Quadratic patterns Name date ms Relations and functions work answers Function table work answers Solve the quadratic equations by factoring work 3 Quadratic functions teacher notes Representing quadratic functions. READ Round Table Pizza Clubhouse Rocklin Ca. A function table answer key is included with a little easier to operate for online marketplace where some phenomenon in each day please share the students. 24 function tables in all 12 horizontal and 12 vertical. Great for a tiered lesson math center warm-up or just. Source: es.pinterest.com Worksheet 1 8 Homework Piecewise Functions Answer Key. Worksheet Works Graphing Linear Equations Answer Key. 6 x x fx 3 fx 3. 6 3 2 0 2 8 48 1 4 2 2 6 36 3 8 4 6 5 30 5 10 6 10 0 0 7 12 8 14 1 6-9 -8 -7 -6 -5 -1-4 -3 -2 1 6-30-6-12-18-24-36. Using the rule stated above the table calculate the required value and the output field. Source: pinterest.com READ Round Table Pizza Clubhouse Rocklin Ca. 2 6 2 3 7 2 2 7 0 20 16 4 10 5 2 16 12 0 2 3 4 8 4 3 4 2 5 0 4 4 6 0 7 4 8 6 10 8 6 9 3. 7 2 5 3 4 4 5 2. The preview file shows the easiest and 2nd most difficult. Function Table Worksheets With Answer Sheet. Source: pinterest.com Kids will be able to easily review and practice their math skills. Printable Worksheets. Answer key T1L1S1 Complete the function table. Two Variable Linear Equations Intro Khan Academy. Each of the numbered sheets gets progressively more difficult. This site is an open community for users to share their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us. 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Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website.	2,421	10,771	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2021-21	latest	en	0.786933
http://physics.stackexchange.com/questions/16182/use-the-relative-velocity-formula-to-find-v2f-in-terms-of-v1f?answertab=active	1,448,554,403,000,000,000	text/html	crawl-data/CC-MAIN-2015-48/segments/1448398447758.91/warc/CC-MAIN-20151124205407-00154-ip-10-71-132-137.ec2.internal.warc.gz	181,049,867	18,341	# Use the relative velocity formula to find v2f in terms of v1f? Q: A $0.150\text{ kg}$ glider is moving to the right ($+x$) on a frictionless, horizontal air track with a speed of $0.80\text{ m/s}$. It has an elastic collision with a $0.300\text{ kg}$ glider moving to the left ($-x$) with a speed of $2.20\text{ m/s}$. a.) What is the initial momentum of each glider? Express the momentum in terms of unit vectors. b.) Use the relative velocity formula to find $v_{2f}$ in terms of $v_{1f}$. c.) Use the relative velocity result to solve conservation of momentum to find the velocity (magnitude and direction) of each glider after the collision. I've figured part (a) using the definition of momentum: $p=mv$: 1st glider: $$p_1 = m_1 v_1 = (0.15\text{ kg})(+0.8\hat{x}\text{ m/s}) = +0.12\hat{x}\text{ kg m/s}$$ 2nd glider: $$p_2 = m_2 v_2 = (0.3\text{ kg})(-2.20\hat{x}\text{ m/s}) = -0.66\hat{x}\text{ kg m/s}$$ Parts (b) and (c) are what have me confused at the moment. I'm not positive I have the equations for relative velocity right nor how to solve for $v_{2f}$ in terms of $v_{1f}$. My book lists this an equation that can be gotten from manipulation of a kinetic energy equation: $v_{1i} – v_{2i} = -(v_{1f} - v_{2f})$. Is this the relative velocity formula? would just isolating $v_{2f}$ in this equation be solving for $v_{2f}$ in terms of $v_{1f}$? Any help appreciated. thanks! - By the way, thank you for asking a well-written, precise homework question ;-) – David Z Oct 25 '11 at 19:21 Thanks for the formatting support :) – Matt Oct 25 '11 at 20:59 No problem :-) I wanted to add this to our list of good homework questions on meta so I figured it would help to make it look pretty. Also I just noticed you had a mistake with the units: momentum is in kg m/s but you had written it in Newtons. – David Z Oct 25 '11 at 21:03 Yeah, I wanted to answer this, because of the care put into explaining the difficulty. – adavid Oct 25 '11 at 21:17 I have no clue whatsoever as to what the "relative velocity formula" is, but I think that the idea here is to find a relation between the $v_i^f$ (assuming $f$ means final). That comes straight out of momentum conservation (using $s$ for starting and $f$ for final and that the problem is in one dimension): $$p^f = p^s \implies m_1 v_1^f + m_2 v_2^f = m_1 v_1^s + m_2 v_2^s$$ which has only two unknowns, namely the $v_i^f$. (Kinetic) energy conservation would lead to: $$E^f = E^s \implies m_1 (v_1^f)^2 + m_2 (v_2^f)^2 = m_1 (v_1^s)^2 + m_2 (v_2^s)^2$$ which again has the two same unknowns. So, two equations with two unknowns... - Do not forget velocities have signs (so, also do momenta). – adavid Oct 25 '11 at 19:06 Hmm, I am still not seeing how I will find v2f in terms of v1f. My book says v1i – v2i = -(v1f - v2f), along with conservation of momentum, can be used to solve for the two unknowns. I believe I am supposed to end up with these equations for the "relative velocity result": imgur.com/uJ9z2 – Matt Oct 25 '11 at 21:07 Am I just missing some algebra we can do with the equations we are currently looking at to get those resultant equations? – Matt Oct 25 '11 at 21:09 Well @Matt, can you see how v2f comes out of the equation you just pasted in the comment above? I mean, there are 4 symbols in a linear combination, 2 have known values, so the other 2 can be expressed in terms of each other. (This is all b is asking for.) – adavid Oct 25 '11 at 21:12 How v2f comes out of: v1i – v2i = -(v1f - v2f) or the equation from the imgur link? – Matt Oct 25 '11 at 21:28	1,127	3,560	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2015-48	longest	en	0.890114
http://mizar.uwb.edu.pl/version/current/html/proofs/toprealc/72	1,571,101,668,000,000,000	text/plain	crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00436.warc.gz	115,857,314	1,995	let X be set ; :: thesis: for n being Nat for f, g being Function of X,() holds f <++> g is Function of X,() let n be Nat; :: thesis: for f, g being Function of X,() holds f <++> g is Function of X,() let f, g be Function of X,(); :: thesis: f <++> g is Function of X,() set h = f <++> g; A1: dom f = X by FUNCT_2:def 1; dom g = X by FUNCT_2:def 1; then A2: dom (f <++> g) = X by ; for x being object st x in X holds (f <++> g) . x in the carrier of () proof let x be object ; :: thesis: ( x in X implies (f <++> g) . x in the carrier of () ) assume A3: x in X ; :: thesis: (f <++> g) . x in the carrier of () then reconsider X = X as non empty set ; reconsider x = x as Element of X by A3; reconsider f = f, g = g as Function of X,() ; A4: (f . x) + (g . x) = (f . x) + (g . x) ; (f <++> g) . x = (f . x) + (g . x) by ; hence (f <++> g) . x in the carrier of () by A4; :: thesis: verum end; hence f <++> g is Function of X,() by ; :: thesis: verum	356	949	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2019-43	latest	en	0.872715
http://booksiread.org/pdf/barrons-ap-statistics-9th-edition/	1,548,243,616,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00325.warc.gz	35,067,799	15,959	## Barron S Ap Statistics 9th Edition Author: Martin Sternstein, Ph.D. Publisher: Barron's Educational Series ISBN: 1438068840 Size: 39.34 MB Format: PDF View: 4328 This brand-new manual's in depth preparation for the AP Statistics exam features the 35 absolutely best AP Statistic exam's hints found anywhere and includes: a diagnostic test and five full-length and up-to-date practice exams; all test questions answered and explained; a new review chapter highlighting statistical insights into social issues additional multiple-choice and free-response questions with answers; a new chapter on the Investigative Task, which counts as one-eighth of the exam a 15 chapter subject review covering all test topics; and, a guide to basic uses of the TI-83/TI-84 calculators. Students who purchase this book will also get FREE access to one additional full-length online AP Statistics test with all questions answered and explained. ## Barron S Ap Statistics Author: Martin Sternstein Publisher: Simon and Schuster ISBN: 1438068840 Size: 66.32 MB Format: PDF View: 7670 This manual's in-depth preparation for the AP Statistics exam features the 35 absolutely best AP Statistics exam hints found anywhere, and includes: A diagnostic test and five full-length and up-to-date practice exams All test questions answered and explained Additional multiple-choice and free-response questions with answers A 14-chapter subject review, covering all test topics A new review chapter highlighting statistical insights into social issues a new chapter on the Investigative Task, which counts as one-eighth of the exam A guide to basic uses of TI, Casio, and HP graphing calculators BONUS ONLINE PRACTICE TEST: Students who purchase this book will also get FREE access to one additional full-length online AP Statistics test with all questions answered and explained. ## Barron S Regents Exams And Answers Integrated Algebra Author: Lawrence S. Leff Publisher: Barron's Educational Series ISBN: 9780764138706 Size: 44.24 MB Format: PDF, ePub View: 5062 Reviews math topics including algebraic language, sets, and ratios; offers test-taking tips; provides practice questions and answers; and includes six actual Regents Exams with answers. ## Barron S Ap Computer Science A Author: Roselyn Teukolsky Publisher: Barron's Educational Series ISBN: 9780764143731 Size: 47.94 MB Format: PDF, Kindle View: 6635 Provides diagnostic tools to assess strengths and weaknesses, explains the current Level A test and the elimination of the Level AB, offers subject reviews on static variables and other topics, and includes three practice tests. ## Barron S Ap World History Author: John McCannon Publisher: Barron's Educational Series ISBN: 9780764143670 Size: 40.49 MB Format: PDF, Docs View: 2015 Provides basic strategies for taking the exam, questions and explanations about world history from prehistoric to modern times, and two full-length practice tests. ## Barron S Ap Biology Author: Deborah T. Goldberg Publisher: Barron's Educational Series ISBN: 9780764140518 Size: 29.84 MB Format: PDF, ePub, Docs View: 2916 Provides subject reviews, test taking hints, and practice test for the advanced placement exam in biology. ## Makro Konomik Author: N. Gregory Mankiw Publisher: Haufe-Lexware ISBN: 3791037846 Size: 35.59 MB Format: PDF, ePub, Docs View: 7458 !-- Generated by XStandard version 2.0.1.0 on 2016-02-17T17:17:57 -- Der „Mankiw" ist nicht nur ein maßgebliches Standardwerk an deutschen Hochschulen. Übersetzt in zahlreiche Sprachen wird der Klassiker weltweit erfolgreich in Lehrveranstaltungen eingesetzt. Diskussionen um Themen wie Inflation, Arbeitslosigkeit und Wachstum Möglichkeiten und Grenzen der Geld-, Fiskal- und Außenwirtschaftspolitik Die 6. deutsche Auflage wurde umfassend überarbeitet und berücksichtigt auch die Aktualisierungen der 7. US-Auflage. Eine zentrale Neuerung ist das Kapitel „Ein dynamisches Modell der Gesamtnachfrage und des Gesamtangebots". Es präsentiert die wesentlichen makroökonomischen Forschungsergebnisse der letzten Jahre. Aktuelle Texte und neue Fallbeispiele zur jüngsten Wirtschafts- und Finanzkrise Modernisiertes, zweifarbiges Layout ## Campbell Biologie Author: Neil A. Campbell Publisher: ISBN: 9783868942590 Size: 34.64 MB Format: PDF View: 1473 ## Kulturvergleichende Perspektiven Auf Das Stliche Europa Author: Daniel Drascek Publisher: Waxmann Verlag ISBN: 3830985878 Size: 22.34 MB Format: PDF, Docs View: 2708 In den letzten Jahren hat die Volkskunde / Europäische Ethnologie einen intensiven Prozess der Selbstreflexion bezüglich der Forschung zum östlichen Europa eingeleitet und kritisch über die Entwicklung dieses Forschungsfeldes, über die wichtigsten Akteure und das bisher Erreichte diskutiert. Deutlich geworden ist in diesem Zusammenhang eine lange Fachtradition und eine starke, teilweise zeitbedingte Fokussierung auf bestimmte Felder, in denen sich das Fach als nach wie vor breit gefächert und forschungsstark erweist. Vor diesem Hintergrund richtet sich der Fokus dieses Bandes auf aktuelle kulturwissenschaftliche Ansätze zu Themenfeldern, die in der bisherigen Osteuropaforschung bisher zu wenig Beachtung gefunden haben. Neben historischen Aspekten wurden verstärkt gegenwartsbezogene Fragestellungen und das entsprechende methodische Instrumentarium ins Auge gefasst. Erfreulich ist, dass sich mittlerweile eine jüngere Generation weitgehend unbefangen mit der sozialistischen Kultur des 20. Jahrhunderts im östlichen Europa auseinandersetzt und einen etwas anderen Blick auf gegenwärtige Transformationsprozesse sowie moderne Formen der Alltagskultur wirft. Gerade einer kulturvergleichenden Perspektive, die sich keineswegs auf das östliche Europa beschränkt, kommt dabei eine zentrale Bedeutung zu. Auch wenn die Forschungsbefunde nicht ganz so revolutionär sein mögen, wie das Titelbild mit dem wilden Gorilla suggeriert, so sind es doch ungewöhnliche Fragestellungen und Perspektiven, die zur Diskussion stehen. Dies gilt für die immer noch gängige Dichotomisierung in Ost und West, die wechselseitigen stereotypisierten Narrationen, die wieder verstärkt national codierten Erinnerungskulturen, komplexe Migrationsbewegungen, moderne Mythen, laufende Modernisierungsprozesse und enttäuschte Wohlstandserwartungen im Hinblick auf die Europäische Union, oder die Frage nach einer Entwicklung der kulturwissenschaftlichen Osteuropaforschung zur Europäisierungsforschung. ## Statistik Mit Excel F R Dummies Author: Joseph Schmuller Publisher: John Wiley & Sons ISBN: 3527811702 Size: 56.38 MB Format: PDF, ePub, Docs View: 146 Statistiken und Aussagen zu Wahrscheinlichkeiten begegnen uns heute überall: Die Umsatzentwicklung in Unternehmen, Hochrechnungen für Wahlergebnisse, PISA-Ergebnisse fünfzehnjähriger Schüler sind nur drei von zahlreichen Beispielen. Joseph Schmuller zeigt Ihnen in diesem Buch, wie Sie die Zahlen in den Griff bekommen und Daten, Statistiken und Wahrscheinlichkeiten richtig lesen und interpretieren. Dafür brauchen Sie keinen Statistikkurs zu belegen und kein Mathegenie zu sein. Für alles gibt es in Excel die passende Funktion und das passende Werkzeug. So können Sie Theorie und Praxis sofort miteinander verbinden.	1,847	7,237	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2019-04	latest	en	0.872678
https://b-ok.org/book/3233500/f09c7b	1,569,044,379,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514574265.76/warc/CC-MAIN-20190921043014-20190921065014-00300.warc.gz	373,438,210	16,933	Main Review of analytic geometry # Review of analytic geometry Учебное пособие на английском языке. — Thomson Brooks-Cole, 2007. — 11 p.Данное пособие предназначено для повторения курса аналитической геометрии на плоскости. В нём представлены основные формулы и определения. Имеется около полусотни примеров задач с решением.Contents:Circles. Lines. Parallel and perpendicular lines. Exercises. Solutions. Language: english Pages: 11 File: PDF, 293 KB You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. 1 ### Европейский союз Language: russian File: PDF, 456 KB 2 ### Узагальнення методичних підходів до аналізу структури капіталу Language: ukrainian File: PDF, 1.40 MB ```REVIEW OF ANALYTIC GEOMETRY The points in a plane can be identified with ordered pairs of real numbers. We start by drawing two perpendicular coordinate lines that intersect at the origin O on each line. Usually one line is horizontal with positive direction to the right and is called the x-axis; the other line is vertical with positive direction upward and is called the y-axis. Any point P in the plane can be located by a unique ordered pair of numbers as follows. Draw lines through P perpendicular to the x- and y-axes. These lines intersect the axes in points with coordinates a and b as shown in Figure 1. Then the point P is assigned the ordered pair a, b. The first number a is called the x-coordinate of P ; the second number b is called the y-coordinate of P. We say that P is the point with coordinates a, b, and we denote the point by the symbol Pa, b. Several points are labeled with their coordinates in Figure 2. y y 4 b P (a, b) 4 3 II (_2, 2) I 2 III _3 2 1 1 _3 _2 _1 O _1 _2 (1, 3) 3 1 2 3 4 5 x _3 _2 _1 0 _1 _2 (_3, _2)) _3 a IV _4 (5, 0) 1 _4 FIGURE 1 2 3 4 5 x (2, _4) FIGURE 2 By reversing the preceding process we can start with an ordered pair a, b and arrive at the corresponding point P. Often we identify the point P with the ordered pair a, b and refer to “the point a, b.” [Although the notation used for an open interval a, b is the same as the notation used for a point a, b, you will be able to tell from the context which meaning is intended.] This coordinate system is called the rectangular coordinate system or the Cartesian coordinate system in honor of the French mathematician René Descartes (1596–1650), even though another Frenchman, Pierre Fermat (1601–1665), invented the principles of analytic geometry at about the same time as Descartes. The plane supplied with this coordinate system is called the coordinate plane or the Cartesian plane and is denoted by 2. The x- and y-axes are called the coordinate axes and divide the Cartesian plane into four quadrants, which are labeled I, II, III, and IV in Figure 1. Notice that the first quadrant consists of those points whose x- and y-coordinates are both positive. EXAMPLE 1 Describe and sketch the regions given by the following sets. (a) x, y x 0 (c ) {x, y (b) x, y y 1 y 1} SOLUTION (a) The points whose x-coordinates are 0 or positive lie on the y-axis or to the right of it as indicated by the shaded region in Figure 3(a). y y y y=1 y=1 0 x 0 x 0 x y=_1 FIGURE 3 (a) x 0 (b) y=1 (c) \| y \|<1 1 2 ■ REVIEW OF ANALYTIC GEOMETRY (b) The set of all points with y-coordinate 1 is a horizontal line one unit above the x-axis [see Figure 3(b)]. (c) Recall from Review of Algebra that y 1 1 y 1 if and only if The given region consists of those points in the plane whose y-coordinates lie between 1 and 1. Thus, the region consists of all points that lie between (but not on) the horizontal lines y 1 and y 1. [These lines are shown as dashed lines in Figure 3(c) to indicate that the points on these lines don’t lie in the set.] y \|ﬁ-› \| › P¡(⁄, › ) P£(¤, › ) \|¤-⁄\| 0 ⁄ Recall from Review of Algebra that the distance between points a and b on a number line is a b b a . Thus, the distance between points P1x 1, y1 and P3 x 2 , y1 on a horizontal line must be x 2 x 1 and the distance between P2 x 2 , y2 and P3 x 2 , y1 on a vertical line must be y2 y1 . (See Figure 4.) To find the distance P1 P2 between any two points P1x 1, y1 and P2 x 2 , y2 , we note that triangle P1P2 P3 in Figure 4 is a right triangle, and so by the Pythagorean Theorem we have P™(¤, ﬁ ) ﬁ x ¤ P P s P P FIGURE 4 1 2 1 3 2 P2 P3 2 s x2 x1 2 y2 y1 2 sx 2 x 1 2 y2 y1 2 Distance Formula The distance between the points P1x 1, y1 and P2 x 2 , y2 is P P sx 1 2 2 x 1 2 y2 y1 2 For instance, the distance between 1, 2 and 5, 3 is s5 1 2 3 2 2 s4 2 5 2 s41 CIRCLES y An equation of a curve is an equation satisfied by the coordinates of the points on the curve and by no other points. Let’s use the distance formula to find the equation of a circle with radius r and center h, k. By definition, the circle is the set of all points Px, y whose distance from the center Ch, k is r. (See Figure 5.) Thus, P is on the circle if and only if PC r. From the distance formula, we have P (x, y) r C (h, k) sx h2 y k2 r 0 x or equivalently, squaring both sides, we get x h2 ( y k2 r 2 FIGURE 5 This is the desired equation. Equation of a Circle An equation of the circle with center h, k and radius r is x h2 ( y k2 r 2 In particular, if the center is the origin 0, 0, the equation is x2 y2 r2 For instance, an equation of the circle with radius 3 and center 2, 5 is x 22 ( y 52 9 REVIEW OF ANALYTIC GEOMETRY ■ 3 EXAMPLE 2 Sketch the graph of the equation x 2 y 2 2x 6y 7 0 by first show- ing that it represents a circle and then finding its center and radius. SOLUTION We first group the x-terms and y-terms as follows: x 2 2x (y 2 6y 7 y Then we complete the square within each grouping, adding the appropriate constants (the squares of half the coefficients of x and y) to both sides of the equation: (_1, 3) x 2 2x 1 ( y 2 6y 9 7 1 9 0 x 12 ( y 32 3 or x 1 Comparing this equation with the standard equation of a circle, we see that h 1, k 3, and r s3, so the given equation represents a circle with center 1, 3 and radius s3. It is sketched in Figure 6. FIGURE 6 ≈+¥+2x-6y+7=0 LINES To find the equation of a line L we use its slope, which is a measure of the steepness of the line. Definition The slope of a nonvertical line that passes through the points P1x 1, y1 and P2 x 2 , y2 is y L m P™(x™, y™) Îy=ﬁ-› =rise P¡(x¡, y¡) The slope of a vertical line is not defined. Îx=¤-⁄ =run m= 21 Thus the slope of a line is the ratio of the change in y, y, to the change in x, x. (See Figure 7.) The slope is therefore the rate of change of y with respect to x. The fact that the line is straight means that the rate of change is constant. Figure 8 shows several lines labeled with their slopes. Notice that lines with positive slope slant upward to the right, whereas lines with negative slope slant downward to the right. Notice also that the steepest lines are the ones for which the absolute value of the slope is largest, and a horizontal line has slope 0. Now let’s find an equation of the line that passes through a given point P1x 1, y1 and has slope m. A point Px, y with x x 1 lies on this line if and only if the slope of the line through P1 and P is equal to m; that is, m=0 y y1 m x x1 x 0 FIGURE 7 y m=5 m=2 m=1 1 m=_ 2 0 FIGURE 8 y y2 y1 x x2 x1 m=_1 m=_2 m=_5 x This equation can be rewritten in the form y y1 mx x 1 and we observe that this equation is also satisfied when x x 1 and y y1 . Therefore, it is an equation of the given line. Point-Slope Form of the Equation of a Line An equation of the line passing through the point P1x 1, y1 and having slope m is y y1 mx x 1 4 ■ REVIEW OF ANALYTIC GEOMETRY EXAMPLE 3 Find an equation of the line through the points 1, 2 and 3, 4. SOLUTION The slope of the line is m 4 2 3 3 1 2 Using the point-slope form with x 1 1 and y1 2, we obtain y 2 32 x 1 which simplifies to y b 3x 2y 1 Suppose a nonvertical line has slope m and y-intercept b. (See Figure 9.) This means it intersects the y-axis at the point 0, b, so the point-slope form of the equation of the line, with x 1 0 and y1 b, becomes y=mx+b y b mx 0 x 0 This simplifies as follows. FIGURE 9 Slope-Intercept Form of the Equation of a Line An equation of the line with slope m and y-intercept b is y mx b In particular, if a line is horizontal, its slope is m 0, so its equation is y b, where b is the y-intercept (see Figure 10). A vertical line does not have a slope, but we can write its equation as x a, where a is the x-intercept, because the x-coordinate of every point on the line is a. y y=b b x=a 0 EXAMPLE 4 Graph the inequality x 2y 5. x a SOLUTION We are asked to sketch the graph of the set x, y x 2y 5 and we begin by solving the inequality for y : FIGURE 10 x 2y 5 y 2y x 5 y 12 x 52 2.5 y= _ 1 2 x+ 5 2 0 FIGURE 11 5 x Compare this inequality with the equation y 12 x 52 , which represents a line with slope 12 and y-intercept 52 . We see that the given graph consists of points whose y-coordinates are larger than those on the line y 12 x 52 . Thus, the graph is the region that lies above the line, as illustrated in Figure 11. PARALLEL AND PERPENDICULAR LINES Slopes can be used to show that lines are parallel or perpendicular. The following facts are proved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition by Stewart, Redlin, and Watson (Thomson Brooks/Cole, Belmont, CA, 2006). REVIEW OF ANALYTIC GEOMETRY ■ 5 Parallel and Perpendicular Lines 1. Two nonvertical lines are parallel if and only if they have the same slope. 2. Two lines with slopes m1 and m2 are perpendicular if and only if m1m2 1; that is, their slopes are negative reciprocals: m2 1 m1 EXAMPLE 5 Find an equation of the line through the point 5, 2 that is parallel to the line 4x 6y 5 0. SOLUTION The given line can be written in the form y 23 x 56 which is in slope-intercept form with m 23 . Parallel lines have the same slope, so the required line has slope 23 and its equation in point-slope form is y 2 23 x 5 We can write this equation as 2x 3y 16. EXAMPLE 6 Show that the lines 2x 3y 1 and 6x 4y 1 0 are perpendicular. SOLUTION The equations can be written as y 23 x 13 and y 32 x 14 and m2 32 from which we see that the slopes are m1 23 Since m1m2 1, the lines are perpendicular. EXERCISES 1–2 S 11–24 Find an equation of the line that satisfies the given conditions. 11. Through 2, 3, slope 6 Find the distance between the points. 1. 1, 1, ■ ■ 3–4 4, 5 ■ ■ 2. 1, 3, ■ ■ ■ ■ 12. Through 3, 5, slope 2 7 5, 7 ■ ■ ■ ■ ■ 14. Through 1, 2 and 4, 3 Find the slope of the line through P and Q. 3. P3, 3, ■ ■ ■ Q1, 6 ■ ■ 4. P1, 4, ■ ■ ■ ■ 15. Slope 3, y-intercept 2 Q6, 0 ■ ■ ■ ■ 5. Show that the points 2, 9, 4, 6, 1, 0, and 5, 3 are the 18. x-intercept 8, y-intercept 6 6. (a) Show that the points A1, 3, B3, 11, and C5, 15 are 19. Through 4, 5, parallel to the x-axis collinear (lie on the same line) by showing that AB BC AC . (b) Use slopes to show that A, B, and C are collinear. 7–10 21. Through 1, 6, parallel to the line x 2y 6 22. y-intercept 6, parallel to the line 2x 3y 4 0 8. y 2 9. xy 0 ■ 20. Through 4, 5, parallel to the y-axis Sketch the graph of the equation. 7. x 3 ■ 10. ■ ■ ■ ■ ■ 23. Through 1, 2, perpendicular to the line 2x 5y 8 0 y 1 ■ 2 16. Slope 5, y-intercept 4 17. x-intercept 1, y-intercept 3 vertices of a square. 13. Through 2, 1 and 1, 6 ■ 24. Through ( 2 , 3 ), perpendicular to the line 4x 8y 1 1 ■ ■ ■ ■ ■ ■ ■ 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 6 ■ REVIEW OF ANALYTIC GEOMETRY 25–28 41. Show that the lines Find the slope and y-intercept of the line and draw its graph. 25. x 3y 0 26. 2x 3y 6 0 27. 3x 4y 12 28. 4x 5y 10 ■ ■ 29–36 ■ ■ ■ ■ 35. 36. ■ ■ ■ ■ ■ ■ {x, y x 2} 32. x, y y 2x 1 x, y 1 x y 1 2x {x, y x y x 3} ■ ■ ■ ■ ■ {x, y x 3 and y 2} 10x 6y 50 0 43. Show that the midpoint of the line segment from P1x 1, y1 to P2 x 2 , y2 is x 1 x 2 y1 y2 , 2 2 44. Find the midpoint of the line segment joining the points 1, 3 ■ ■ ■ ■ ■ ■ and 7, 15. 37–38 Find an equation of a circle that satisfies the given conditions. 45. Find an equation of the perpendicular bisector of the line seg- 37. Center 3, 1, radius 5 46. (a) Show that if the x- and y-intercepts of a line are nonzero ment joining the points A1, 4 and B7, 2. 38. Center 1, 5, passes through 4, 6 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 39–40 Show that the equation represents a circle and find the 39. x 2 y 2 4x 10y 13 0 40. x 2 y 2 6y 2 0 ■ and are perpendicular and find their point of intersection. 1 2 ■ 6x 2y 10 42. Show that the lines 3x 5y 19 0 30. x, y x 1 and y 3 33. x, y 0 y 4 and x 2 34. ■ and are not parallel and find their point of intersection. Sketch the region in the xy-plane. 29. x, y x 0 31. ■ 2x y 4 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ numbers a and b, then the equation of the line can be put in the form x y 1 a b This equation is called the two-intercept form of an equation of a line. (b) Use part (a) to find an equation of the line whose x-intercept is 6 and whose y-intercept is 8. REVIEW OF ANALYTIC GEOMETRY ■ 7 S 1. 5 7. 28. m 5 , 4 3. 2 8. 9 2. 2 s29 y 4. b2 4 7 x=3 0 x 3 9. 29. 30. y 10. y x 0 xy=0 0 x 31. 12. y 2 x 11. y 6x 15 14. y x 1 7 15. y 3x 2 17. y 3x 3 20. x 4 18. y 4 x 6 3 13. 5x y 11 31 2 25. m , 1 3 _2 16. y 5 x 4 2 0 2 x 19. y 5 22. y 3 x 6 21. x 2y 11 0 23. 5x 2y 1 0 32. y 2 24. y 2x b0 1 3 33. 34. y y=4 y x=2 0 x 0 26. m , 2 3 b2 35. x 36. y y=1+x 0, 1 0 x y=1-2x 27. m 4 , 3 b 3 y 0 x _3 37. x 32 y 12 25 39. 2, 5, 4 44. 4, 9 38. x 12 y 52 130 40. 0, 3, s 7 45. y x 3 41. 1, 2 46. (b) y 3 x 8 4 42. 2, 5 8 ■ REVIEW OF ANALYTIC GEOMETRY SOLUTIONS 1. Use the distance formula with P1 (x1 , y1 ) = (1, 1) and P2 (x2 , y2 ) = (4, 5) to get s √ √ \|P1 P2 \| = (4 − 1)2 + (5 − 1)2 = 32 + 42 = 25 = 5 2. The distance from (1, −3) to (5, 7) is s √ √ √ (5 − 1)2 + [7 − (−3)]2 = 42 + 102 = 116 = 2 29. 3. With P (−3, 3) and Q(−1, −6), the slope m of the line through P and Q is m = 4. m = −6 − 3 9 =− . −1 − (−3) 2 0 − (−4) 4 = 6 − (−1) 7 5. Using A(−2, 9), B(4, 6), C(1, 0), and D(−5, 3), we have s s √ √ √ √ \|AB\| = [4 − (−2)]2 + (6 − 9)2 = 62 + (−3)2 = 45 = 9 5 = 3 5, \|BC\| = \|CD\| = \|DA\| = s s √ √ √ √ (1 − 4)2 + (0 − 6)2 = (−3)2 + (−6)2 = 45 = 9 5 = 3 5, s s √ √ √ √ (−5 − 1)2 + (3 − 0)2 = (−6)2 + 32 = 45 = 9 5 = 3 5, and t √ √ √ √ √ [−2 − (−5)]2 + (9 − 3)2 = 32 + 62 = 45 = 9 5 = 3 5. So all sides are of equal length and we have a rhombus. Moreover, mAB = mDA = 1 1 6−9 0−6 3−0 = − , mBC = = 2, mCD = = − , and 4 − (−2) 2 1−4 −5 − 1 2 9−3 = 2, so the sides are perpendicular. Thus, A, B, C, and D are vertices of a square. −2 − (−5) 6. (a) Using A(−1, 3), B(3, 11), and C(5, 15), we have s √ √ √ \|AB\| = [3 − (−1)]2 + (11 − 3)2 = 42 + 82 = 80 = 4 5, s √ √ √ \|BC\| = (5 − 3)2 + (15 − 11)2 = 22 + 42 = 20 = 2 5, and s √ √ √ \|AC\| = [5 − (−1)]2 + (15 − 3)2 = 62 + 122 = 180 = 6 5. Thus, \|AC\| = \|AB\| + \|BC\|. 8 12 11 − 3 15 − 3 = = 2 and mAC = = = 2. Since the segments AB and AC have the 3 − (−1) 4 5 − (−1) 6 same slope, A, B and C must be collinear. (b) mAB = 7. The graph of the equation x = 3 is a vertical line with x-intercept 3. The line does not have a slope. 8. The graph of the equation y = −2 is a horizontal line with y-intercept −2. The line has slope 0. REVIEW OF ANALYTIC GEOMETRY ■ 9 9. xy = 0 ⇔ 10. \|y\| = 1 ⇔ y = 1 or y = −1 x = 0 or y = 0. The graph consists of the coordinate axes. 11. By the point-slope form of the equation of a line, an equation of the line through (2, −3) with slope 6 is y − (−3) = 6(x − 2) or y = 6x − 15. 12. y − (−5) = − 72 [x − (−3)] or y = − 72 x − 31 2 13. The slope of the line through (2, 1) and (1, 6) is m = y − 1 = −5(x − 2) or y = −5x + 11. 14. For (−1, −2) and (4, 3), m = 6−1 = −5, so an equation of the line is 1−2 3 − (−2) = 1. An equation of the line is y − 3 = 1(x − 4) or y = x − 1. 4 − (−1) 15. By the slope-intercept form of the equation of a line, an equation of the line is y = 3x − 2. 16. By the slope-intercept form of the equation of a line, an equation of the line is y = 25 x + 4. 17. Since the line passes through (1, 0) and (0, −3), its slope is m = Another method: From Exercise 61, −3 − 0 = 3, so an equation is y = 3x − 3. 0−1 x y + = 1 ⇒ −3x + y = −3 ⇒ y = 3x − 3. 1 −3 3 6−0 = . So an equation is y = 34 x + 6. 0 − (−8) 4 y x + = 1 ⇒ −3x + 4y = 24 ⇒ y = 34 x + 6. Another method: From Exercise 61, −8 6 18. For (−8, 0) and (0, 6), m = 19. The line is parallel to the x-axis, so it is horizontal and must have the form y = k. Since it goes through the point (x, y) = (4, 5), the equation is y = 5. 20. The line is parallel to the y-axis, so it is vertical and must have the form x = k. Since it goes through the point (x, y) = (4, 5), the equation is x = 4. 21. Putting the line x + 2y = 6 into its slope-intercept form gives us y = − 12 x + 3, so we see that this line has slope − 12 . Thus, we want the line of slope − 12 that passes through the point (1, −6): y − (−6) = − 12 (x − 1) ⇔ y = − 12 x − 11 . 2 22. 2x + 3y + 4 = 0 ⇔ y = − 23 x − 43 , so m = − 23 and the required line is y = − 23 x + 6. 23. 2x + 5y + 8 = 0 ⇔ y = − 25 x − 85 . Since this line has slope − 25 , a line perpendicular to it would have slope 52 , so the required line is y − (−2) = 5 2 [x − (−1)] ⇔ y = 52 x + 12 . 24. 4x − 8y = 1 ⇔ y = 12 x − 18 . Since this line has slope 12 , a line perpendicular to it would have slope −2, so the ⇔ y = −2x + 13 . required line is y − − 23 = −2 x − 12 10 ■ REVIEW OF ANALYTIC GEOMETRY 25. x + 3y = 0 ⇔ y = − 13 x, so the slope is − 13 and the 26. 2x − 3y + 6 = 0 ⇔ y= y = − 45 x + 2, so the slope is + 2, so the slope is 2 3 and the y-intercept is 2. y-intercept is 0. 28. 4x + 5y = 10 ⇔ 2 x 3 29. {(x, y) \| x < 0} 27. 3x − 4y = 12 ⇔ y = 34 x − 3, so the slope is 3 4 and the y-intercept is −3. 30. {(x, y) \| x ≥ 1 and y < 3} − 45 and the y-intercept is 2. 31. r q (x, y) \|x\| ≤ 2 = 32. {(x, y) \| −2 ≤ x ≤ 2} 34. {(x, y) \| y > 2x − 1} r q (x, y) \|x\| < 3 and \|y\| < 2 35. {(x, y) \| 1 + x ≤ y ≤ 1 − 2x} 33. {(x, y) \| 0 ≤ y ≤ 4, x ≤ 2} 36. (x, y) \| −x ≤ y < 12 (x + 3) 37. An equation of the circle with center (3, −1) and radius 5 is (x − 3)2 + (y + 1)2 = 52 = 25. 38. The equation has the form (x + 1)2 + (y − 5)2 = r2 . Since (−4, −6) lies on the circle, we have r2 = (−4 + 1)2 + (−6 − 5)2 = 130. So an equation is (x + 1)2 + (y − 5)2 = 130. 39. x2 + y 2 − 4x + 10y + 13 = 0 ⇔ x2 − 4x + y 2 + 10y = −13 ⇔ 2 x − 4x + 4 + y 2 + 10y + 25 = −13 + 4 + 25 = 16 ⇔ (x − 2)2 + (y + 5)2 = 42 . Thus, we have a circle with center (2, −5) and radius 4. REVIEW OF ANALYTIC GEOMETRY ■ 11 x2 + y 2 + 6y + 9 = −2 + 9 √ with center (0, −3) and radius 7. 40. x2 + y 2 + 6y + 2 = 0 ⇔ ⇔ x2 + (y + 3)2 = 7. Thus, we have a circle 41. 2x − y = 4 ⇔ y = 2x − 4 ⇒ m1 = 2 and 6x − 2y = 10 ⇔ 2y = 6x − 10 ⇔ y = 3x − 5 ⇒ m2 = 3. Since m1 6= m2 , the two lines are not parallel. To find the point of intersection: 2x − 4 = 3x − 5 ⇔ x = 1 ⇒ y = −2. Thus, the point of intersection is (1, −2). 42. 3x − 5y + 19 = 0 ⇔ 5y = 3x + 19 ⇔ y = 35 x + 6y = −10x + 50 ⇔ y = − 53 x + 25 3 and 10x + 6y − 50 = 0 ⇔ ⇒ m2 = − 53 . Since m1 m2 = 35 − 53 = −1, the two lines are perpendicular. To find the point of intersection: 35 x + 19 5 19 5 ⇒ m1 = = − 53 x + 25 3 3 5 ⇔ 9x + 57 = −25x + 125 ⇔ = 25 = 5. Thus, the point of intersection is (2, 5). 34x = 68 ⇔ x = 2 ⇒ y = 35 · 2 + 19 5 5 x + x y + y 1 2 1 2 , 43. Let M be the point . Then 2 2 x1 + x2 2 y1 + y2 2 x1 − x2 2 y1 − y2 2 + y1 − = + and \|MP1 \|2 = x1 − 2 2 2 2 x1 + x2 2 y1 + y2 2 x2 − x1 2 y2 − y1 2 \|MP2 \|2 = x2 − + y2 − = + . Hence, \|MP1 \| = \|MP2 \|; that 2 2 2 2 is, M is equidistant from P1 and P2 . 7 3 + 15 44. Using the midpoint formula from Exercise 43 with (1, 3) and (7, 15), we get 1 + = (4, 9). , 2 2 −4 = −1, so its perpendicular bisector has slope 1. The 45. With A(1, 4) and B(7, −2), the slope of segment AB is −2 7−1 7 4 + (−2) = (4, 1), so an equation of the perpendicular bisector is y − 1 = 1(x − 4) or midpoint of AB is 1 + 2 , 2 y = x − 3. 46. (a) Since the x-intercept is a, the point (a, 0) is on the line, and similarly since the y-intercept is b, (0, b) is on the b−0 b b line. Hence, the slope of the line is m = = − . Substituting into y = mx + b gives y = − x + b ⇔ 0−a a a b x y x+y =b ⇔ + = 1. a a b x y (b) Letting a = 6 and b = −8 gives + = 1 ⇔ −8x + 6y = −48 [multiply by −48] ⇔ 6 −8 6y = 8x − 48 ⇔ 3y = 4x − 24 ⇔ y = 43 x − 8. ```	9,493	22,000	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2019-39	latest	en	0.656744
https://www.vistrails.org/index.php?title=User:Tohline/ThreeDimensionalConfigurations/ChallengesPt3&oldid=21976	1,660,205,226,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00026.warc.gz	930,840,454	10,510	Challenges Constructing Ellipsoidal-Like Configurations (Pt. 2) This chapter extends the accompanying chapters titled, Construction Challenges (Pt. 1) and (Pt. 2). The focus here is on firming up our understanding of the relationships between various "tilted" Cartesian coordinate frames. Various Coordinate Frames Riemann-Derived Expressions Inertial Frame (green with subscript "0") and Body Frame (black and unsubscripted). For our chosen Example Type I Ellipsoid, we have, $~\Omega_2 = 0.3639$ and $~\Omega_3 = 0.6633$, in which case, $~\Omega_0 = 0.7566$ and $~\delta = 0.5018 ~\mathrm{rad} = 28.75^\circ$. The purple (ellipsoidal) configuration is spinning with frequency, $~\Omega_0$ about the $~z_0$-axis of the "inertial frame," as illustrated; that is, $~\boldsymbol\Omega$ $~=$ $~\boldsymbol{\hat{k}_0}\Omega_0 \, .$ Also as illustrated, the "body frame," which is attached to and aligned with the principal axes of the purple ellipsoid, is tilted at an angle, $~\delta$, with respect to the inertial frame. Hence, as viewed from the body frame, we have, $~\boldsymbol\Omega$ $~=$ $~\biggl[ \boldsymbol{\hat\jmath }\sin\delta + \boldsymbol{\hat{k} }\cos\delta \biggr]\Omega_0 \, .$ Now, adhering to the notation used by [EFE] — see, for example, the first paragraph of §51 (p. 156) — we should write, $~\boldsymbol\Omega$ $~=$ $~\boldsymbol{\hat\jmath }\Omega_2 + \boldsymbol{\hat{k} }\Omega_3 ~~~~~\Rightarrow ~~~ \Omega_2 = \Omega_0\sin\delta$ and, $~\Omega_3 = \Omega_0\cos\delta \, .$ This means that, $~\Omega_0$ $~=$ $~ \biggl[\Omega_2^2 + \Omega_3^2 \biggr]^{1 / 2}$ and, $~\delta = \tan^{-1}\biggl[ \frac{\Omega_2}{\Omega_3} \biggr] \, .$ As we have summarized in an accompanying discussion of Riemann Type 1 ellipsoids, [EFE] provides an expression for the velocity vector of each fluid element, given its instantaneous body-coordinate position (x, y, z) = (x1, x2, x3) — see his Eq. (154), Chapter 7, §51 (p. 156). As viewed from the rotating frame of reference, the three component expressions are, $~\dot{x} = u_1 = \boldsymbol{\hat\imath} \cdot \boldsymbol{u}$ $~=$ $~\biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 y - \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 z$ $~=$ $~- \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y + \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z \, ,$ $~\dot{y} = u_2 = \boldsymbol{\hat\jmath} \cdot \boldsymbol{u}$ $~=$ $~- \gamma \Omega_3 x$ $~=$ $~+\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x \, ,$ $~\dot{z} = u_3 = \boldsymbol{\hat{k}} \cdot \boldsymbol{u}$ $~=$ $~+ \beta \Omega_2 x$ $~=$ $~- \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x \, ,$ where, $~\beta$ $~=$ $~ - \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \frac{\zeta_2}{\Omega_2}$ and, $~\gamma$ $~=$ $~ - \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\Omega_3} \, .$ Rotating-Frame Vorticity $~\boldsymbol{\zeta} \equiv \boldsymbol{\nabla \times}\bold{u}$ $~=$ $~ \boldsymbol{\hat\imath} \biggl[ \frac{\partial \dot{z} }{\partial y} - \frac{\partial \dot{y}}{\partial z} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{\partial \dot{x}}{\partial z} - \frac{\partial \dot{z}}{\partial x} \biggr] + \bold{\hat{k}} \biggl[ \frac{\partial \dot{y}}{\partial x} - \frac{\partial \dot{x}}{\partial y} \biggr]$ $~=$ $~ \boldsymbol{\hat\jmath} \biggl\{ \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 + \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggr\} + \bold{\hat{k}} \biggl\{ \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 + \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 \biggr\}$ $~=$ $~ \boldsymbol{\hat\jmath} ~\zeta_2 + \bold{\hat{k}} ~\zeta_3 \, .$ For our chosen Example Type I Ellipsoid, we have, $~\zeta_2 = -2.2794$ and $~\Omega_3 = -1.9637$, in which case, $~\zeta_\mathrm{rot} = (\zeta_2^2 + \zeta_3^2)^{1 / 2} = 2.2794$ and $~\xi \equiv \tan^{-1}[\zeta_2/\zeta_3] = 4.0013 ~\mathrm{rad} = 229.26^\circ$. Tipped Orbit Planes Summary In a separate discussion, we have shown that, as viewed from a frame that "tumbles" with the (purple) body of a Type 1 Riemann ellipsoid, each Lagrangian fluid element moves along an elliptical path in a plane that is tipped by an angle $~\theta$ about the x-axis of the body. As viewed from the (primed) coordinates associated with this tipped plane, by definition, z' = constant and dz'/dt = 0, and the planar orbit is defined by the expression for an, Off-Center Ellipse $~1$ $~=$ $~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c(z')}{y_\mathrm{max}} \biggr]^2 \, .$ Tipped Orbit Frame (yellow, primed) Given that b/a = 1.25 and c/a = 0.4703 for our chosen Example Type I Ellipsoid, we find that, $~\theta = - 0.3320 ~\mathrm{rad} = -19.02^\circ$. Notice that the offset, $~y_c$, is a function of the tipped plane's vertical coordinate, $~z'$. As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions, $~x'$ $~=$ $~x_\mathrm{max}\cos(\dot\varphi t)$ and, $~y' - y_c$ $~=$ $~y_\mathrm{max}\sin(\dot\varphi t) \, ,$ $~\dot{x}'$ $~=$ $~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi$ and, $~\dot{y}'$ $~=$ $~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .$ As has been summarized in an accompanying discussion, we have determined that (numerical value given for our chosen example Type I ellipsoid), $~\tan\theta$ $~=$ $~ - \frac{\zeta_2}{\zeta_3} \biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr] \frac{c^2}{b^2} = - \frac{\beta \Omega_2}{\gamma \Omega_3} = -0.34479\, ,$ where, $~\beta$ and $~\gamma$ are as defined above. Also, $~\biggl[\frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr]^2$ $~=$ $~ \frac{a^2}{b^2 c^2} (c^2\cos^2\theta + b^2\sin^2\theta) = 1.05238 \, ,$ $~{\dot\varphi}^2$ $~=$ $~ \zeta_3^2\biggl[ \frac{b^2}{a^2 + b^2} \biggr]^2 \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr]^2 \biggl[1 + \tan^2\theta \biggr] = 1.68818\, ,$ $~y_c$ $~=$ $~- \frac{z' b^2 \sin\theta\cos\theta}{c^2 \cos^2\theta + b^2\sin^2\theta} = -z' \sin\theta \cos\theta \biggl[\frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr]^2 \frac{a^2}{c^2} = z' \cos\theta (1.40038) \, .$ Demonstration In order to transform a vector from the "tipped orbit" frame (primed coordinates) to the "body" frame (unprimed), we use the following mappings of the three unit vectors: $~\boldsymbol{\hat\imath'}$ $~\rightarrow$ $~\boldsymbol{\hat\imath} \, ,$ $~\boldsymbol{\hat\jmath'}$ $~\rightarrow$ $~\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta \, ,$ $~\boldsymbol{\hat{k}'}$ $~\rightarrow$ $~-\boldsymbol{\hat\jmath}\sin\theta + \boldsymbol{\hat{k}}\cos\theta \, .$ Given that, by design in our "tipped orbit" frame, there is no vertical motion — that is, $~\dot{z}' = 0$ — mapping the (primed coordinate) velocity to the body (unprimed) coordinate is particularly straightforward. Specifically, $~\boldsymbol{u'}$ $~=$ $~ \boldsymbol{\hat\imath'} \dot{x}' + \boldsymbol{\hat\jmath'} \dot{y}'$ $~~~\rightarrow~~$ $~ \boldsymbol{\hat\imath} \dot{x}' + [\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \dot{y}'$ $~=$ $~ \boldsymbol{\hat\imath} \biggl\{ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \dot\varphi \biggr\} + [\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \biggl\{ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr\} \, .$ Recognizing as well that the relevant coordinate mapping is, $~x'$ $~\rightarrow$ $~x \, ,$ $~y'$ $~\rightarrow$ $~y\cos\theta - z\sin\theta \, ,$ $~z'$ $~\rightarrow$ $~z\cos\theta + y\sin\theta \, ,$ we have, $~\boldsymbol{u'}$ $~~~\rightarrow~~~$ $~ \boldsymbol{\hat\imath} \dot\varphi \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)\biggl\{y_c - y\cos\theta + z\sin\theta\biggr\} + \boldsymbol{\hat\jmath} \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \biggr\{ x\cos\theta \biggr\} + \boldsymbol{\hat{k}} \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \biggr\{ x\sin\theta \biggr\} \, ,$ where, $~y_c$ $~~~\rightarrow~~~$ $~ -[z\cos\theta + y\sin\theta] \sin\theta \cos\theta \biggl[\frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr]^2 \frac{a^2}{c^2} \, .$ Written in terms of the "body" frame coordinates, therefore, the 2nd and 3rd components of this velocity vector are, respectively: $~\boldsymbol{\hat\jmath}\cdot \boldsymbol{u'}$ $~=$ $~ x \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \cos\theta$ $~=$ $~ x \biggl\{ \zeta_3^2\biggl[ \frac{b^2}{a^2 + b^2} \biggr]^2 \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr]^2 \biggl[1 + \tan^2\theta \biggr] \biggr\}^{1 / 2} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \cos\theta$ $~=$ $~ x \biggl\{ \zeta_3 \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \biggr\} \, ,$ $~\boldsymbol{\hat{k}}\cdot \boldsymbol{u'}$ $~=$ $~ x \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \sin\theta$ $~=$ $~ x \biggl\{ \zeta_3^2\biggl[ \frac{b^2}{a^2 + b^2} \biggr]^2 \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr]^2 \biggl[1 + \tan^2\theta \biggr] \biggr\}^{1 / 2} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \sin\theta$ $~=$ $~ x \biggl\{ \zeta_3\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \biggr\} \tan\theta$ $~=$ $~ x \biggl\{ \zeta_3\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \biggr\} \biggl\{ - \frac{\zeta_2}{\zeta_3} \biggl[ \frac{a^2 + b^2}{a^2 + c^2} \biggr] \frac{c^2}{b^2} \biggr\}$ $~=$ $~ -x \biggl\{ \zeta_2\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \biggr\} \, .$ These expressions perfectly match the body-coordinate expressions derived by Riemann (see above) for, respectively, $~\dot{y}$ and $~\dot{z}$.	3,949	9,753	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2022-33	latest	en	0.806114
https://www.cfd-online.com/W/index.php?title=Finite_volume&diff=13688&oldid=1946	1,498,201,365,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00509.warc.gz	857,831,131	19,546	# Finite volume (Difference between revisions) Revision as of 01:24, 14 September 2005 (view source)Michail (Talk \| contribs) (→QUICK)← Older edit Latest revision as of 09:15, 3 January 2012 (view source)Peter (Talk \| contribs) m (Reverted edits by Reverse22 (talk) to last revision by Praveen) (35 intermediate revisions not shown) Line 1: Line 1: - == Discretisation Schemes for convective terms in General Transport Equation. Finite-Volume Formulation, structured grids == + The Finite Volume Method (FVM) is one of the most versatile discretization techniques used in CFD. Based on the control volume formulation of analytical fluid dynamics, the first step in the FVM is to divide the domain into a number of control volumes (aka cells, elements) where the variable of interest is located at the centroid of the control volume. The next step is to integrate the differential form of the governing equations (very similar to the control volume approach) over each control volume. Interpolation profiles are then assumed in order to describe the variation of the concerned variable between cell centroids. The resulting equation is called the discretized or discretization equation. In this manner, the discretization equation expresses the conservation principle for the variable inside the control volume. + The most compelling feature of the FVM is that the resulting solution satisfies the conservation of quantities such as mass, momentum, energy, and species. This is '''exactly''' satisfied for any control volume as well as for the whole computational domain and '''for any number of control volumes'''. Even a coarse grid solution exhibits exact integral balances. - == Introduction == + FVM is the ideal method for computing discontinuous solutions arising in compressible flows. Any discontinuity must satisfy the [[Rankine-Hugoniot jump condition]] which is a consequence of conservation. Since finite volume methods are conservative they automatically satisfy the jump conditions and hence give physically correct weak solutions. - Here is describing the discretization schemes of the convective terms in the finite-volume equations. The accuracy, numerical stability and the boundness of the solution depends on the numerical scheme used for these terms. The central issue is the specification of an appropriate relationship between the convected variable, stored at the cell centre and its value at each of the cell faces. + FVM is also preferred while solving partial differential equations containing discontinuous coefficients. - == Basic Equations of CFD == + ==Integral form of conservation law== - All the conservation equations can be written in the same generic differential form: + Consider a single partial differential equation in conservation form - + :$- :[itex] + \frac{\partial u}{\partial t} + \nabla \cdot f(u) = 0 - \frac {\partial( \rho \phi )} {\partial t} + \frac{\partial}{\partial x_{i}} \left( \rho U \phi - \Gamma_{\phi} \frac{\partial\phi}{\partial x_{i}}\right)=S_{\phi} +$ [/itex] - (1) - Equation (1) is integrated over a control volume and the following discretised equation for $\boldsymbol{\phi}$ is produced: + where $u$ is a conserved quantity and - + :$- :[itex] + f(u) = [ f_1(u), \ldots, f_d(u)] - J_{h}- J_{l} + J_{n}- J_{s} + J_{e}- J_{w} + D_{h} - D_{l} + D_{n} - D_{s} + D_{e} - D_{s} = S_{p} +$ [/itex] - (2) - where $\boldsymbol{S_{p}}$ is the source term for the control volume $\boldsymbol{P}$, and $\boldsymbol{J_{f}}$ and $\boldsymbol{D_{f}}$ represent, respectively, the convective and diffusive fluxes of $\boldsymbol{\phi}$ across the control-volume face $\boldsymbol{f}$ + is the flux vector. Integrating this over any volume $V$ in $R^d$ we get the '''integral form''' of the conservation law - $\boldsymbol{(f=h,l,n,s,e,w)}$ + - The convective fluxes through the cell faces are calculated as: + :$- + \frac{\partial}{\partial t} \int_V u dx + \oint_{\partial V} f_i n_i ds = 0 - + - :[itex] + - J_{f}=C_{f}\phi_{f} +$ [/itex] - (1) - - where $C_{f}$ is the mass flow rate across the cell face $f$. The convected variable $\phi_{f}$ associated with this mass flow rate is usually stored at the cell centres, and thus some form of interpolation assumption must be made in order to determine its value at each cell face. The interpolation procedure employed for this operation is the subject of the various schemes proposed in the literature and the accuracy, stability and boundedness of the solution depends on the procedure used. - - In general, the value of $\boldsymbol{\phi_{f}}$ can be explicity formulated in terms of its neighbouring nodal values by a functional relationship of the form: - + where the [[Greens theorem \| divergence theorem]] has been used to convert the divergence term to a surface integral over the surface $\partial V$ and (n_1, \ldots, n_d)[/itex] is the unit outward normal to $\partial V$. The above equation is the starting point for the finite volume method. - :$+ - \phi_{f}=P \left( \phi_{nb} \right) + -$ + - (1) + - + - where \boldsymbol{\phi_{nb}}[/itex] denotes the neighbouring-node $\boldsymbol{\phi}$values. + - Combining equations (\ref{eq3}) through (\ref{eq4a}), the discretised equation becomes: + - + ==Finite volume formulation== - :$+ - \left\{ D_{h} + C_{h} \left[ P \left( \phi_{nb} \right) \right]_{h} \right\} - + - \left\{ D_{l} + C_{l} \left[ P \left( \phi_{nb} \right) \right]_{l} \right\} + + - \left\{ D_{n} + C_{n} \left[ P \left( \phi_{nb} \right) \right]_{n} \right\} - + The computational domain [itex]V_h$ is divided into non-overlapping cells or finite volumes - \left\{ D_{s} + C_{s} \left[ P \left( \phi_{nb} \right) \right]_{s} \right\} + + - \left\{ D_{e} + C_{e} \left[ P \left( \phi_{nb} \right) \right]_{e} \right\} - + :$V_r, r=1,\ldots, N, \quad V_h = \cup_r V_r - \left\{ D_{w} + C_{w} \left[ P \left( \phi_{nb} \right) \right]_{w} \right\} = S_{p} +$ [/itex] - (1) - == Convection Schemes == + Usually these cells are polygons (triangles, quadrilaterals) in 2-D and polyhedra (tetrahedron, hexahedron, prims, etc) in 3-D. - All the convection schemes involve a stencil of cells in which the values of $\boldsymbol{\phi}$ will be used to construct the face value $\boldsymbol{\phi_{f}}$ + Introduce the cell average value over the cell $V_r$ - [[Image:picture_01.jpg]]. + :$- + u_r = \frac{1}{\|V_r\|} \int_{V_r} u(x) dx - + - Where flow is from left to right, and [itex]\boldsymbol{f}$ is the face in question. + - + - $\boldsymbol{u}$ - mean Upstream node + - + - $\boldsymbol{c}$ - mean Central node + - + - $\boldsymbol{d}$ - mean Downstream node + - + - == Basic Discretisation schemes == + - + - === Central Differencing Scheme (CDS)=== + - + - The most natural assumption for the cell-face value of the convected variable $\boldsymbol{\phi_{f}}$ would appear to be the CDS, which calculates the cell-face value from: + - + - + - :$+ - \phi_{f}=0.5 \left( \phi_{c} + \phi_{d} \right) +$ [/itex] - (1) - - This scheme is 2nd-order accurate, but is unbounded so that unphysical oscillations appear in regions of strong convection and also in the presence of discontinuities such as shocks. The CDS may be used directly in very low Reynolds-number flows where diffusive effects dominate over convection. - === Upwind Differencing Scheme (UDS)=== + which is the basic unknown quantity in the finite volume method. Let - The UDS assumes that the convected variable at the cell fase $\boldsymbol{f}$ is the same as the upwind cell-centre value: + :$- + N(r) = \{ \mbox{Set of cells which share a common face with } V_r \} - + - :[itex] + - \boldsymbol{\phi_{f}= \phi_{c} } +$ [/itex] - (1) - - The UDS is unconditionally bounded and highly stable, but as noted earlier it is only 1st-order accurate in terms of truncation error and may produce severe numerical diffusion. The scheme is therefore highly diffusive when the flow direction is skewed relative to the grid lines. - === Hybrid Differencing Scheme (HDS) === + Then the integral conservation law for cell V_r is - + - The HDS of Spalding [1972] switches the discretisation of the convection terms between CDS and UDS according to the local cell Peclet number as follows: + - + - + - : + - \phi_{f}=0.5 \left( \phi_{c} + \phi_{d} \right) for Pe \triangleleft 2 + + :$+ \|V_r\| \frac{du_r}{dt} + \sum_{s \in N(r)} \int_{V_r \cap V_s} f_i n_i ds = 0$ [/itex] - (1) - + It now remains to approximate the flux integral. This can be achieved using gaussian quadrature. Taking p gaussian points we get - :$+ - + - \phi_{f}= \phi_{c} for Pe \triangleright 2 + + :[itex] + \int_{V_r \cap V_s} f_i n_i ds = \Delta s_{rs} \sum_{m=1}^p \omega_m F^m_{rs}$ [/itex] - (1) - - The cell Peclet number is defined as: - - - :$- Pe= \rho \left\| U_{f} \right\| A_{f}/D_{f} -$ - (1) - - - in which $\boldsymbol{A_{f}}$ and $\boldsymbol{D_{f}}$ are respectively, the cell-face area and physical diffusion coefficient. When $\boldsymbol{Pe\triangleright 2}$ - ,CDS calculations tends to become unstable so that theHDS reverts to the UDS. Physical diffusion is ignored when $\boldsymbol{Pe\triangleright 2}$. - - - The HDS scheme is marginally more accurate than the UDS, because the 2nd-order CDS will be used in regions of low Peclet number. - - == High Resolution Schemes (HRS) == - - === Classification of High Resolution Schemes === - - HRS can be classified as ''linear'' or ''non-linear'', where ''linear'' means their coefficients are not direct functions of the convected variable when applied to a linear convection equation. It is important to recognise that linear convection schemes of 2nd-order accuracy or higher may suffer from unboudedness, and are not unconditionally stable. - - ''Non-linear'' schemes analyse the solution within the stencil and adapt the discretisation to avoid any unwanted behavior, such as unboundedness (see Waterson [1994]). These two types of schemes may be presented in a unified way by use of the ''Flux-Limiter'' formulation (Waterson and Deconinck [1995]), which calculates the cell-face value of the convected variable from: - - - - - :$- \phi_{f}= \phi_{c} + 0.5 B \left( r \right) \left( \phi_{c}-\phi_{u} \right) -$ - (1) - - where $B \left( r \right)$ is termed a limiter function and the gradient ration $r$ is defined as: - - - :$- r= \left( \phi_{d} - \phi_{c} \right) / \left( \phi_{c} - \phi_{u} \right) -$ - (1) - - The generalisation of this approach to handle non-uniform meshes has been given by Waterson [1994] - - From equation (\ref{eq9}) it can be seen that $B=1$ gives the UDS and $B=r$ gives the CDS. - - === Numerical Implementation of HRS === - - The HRS schemes can be introdused into equation (\ref{eq4b}) by using the deffered correction procedure of Rubin and Khosla [1982]. This procedure express the cell-face value $\phi_{f}$ by: - - - \phi_{f}=\phi_{f}\left(U \right) + \phi^{'}_{f} - \label{eq10a} - - - where $\phi^{'}_{f}$ is a higher-order correction which represents the difference between the UDS face value $\phi_{f}\left(U \right)$ and the higher-order scheme value $\phi_{f}\left(H \right)$, i.e. - - - \phi^{'}_{f}= \phi_{f}\left(H \right) + \phi_{f}\left(U \right) - \label{eq10b} - - - If equation (\ref{eq10a}) is substituted into equation (\ref{eq4b}), the resulting discretised equation is: - - \left\{ D_{h} + C_{h} \phi_{h} \left( U \right) \right\} - - \left\{ D_{l} + C_{l} \phi_{l} \left( U \right) \right\} + - - \left\{ D_{n} + C_{n} \phi_{n} \left( U \right) \right\} - - \left\{ D_{s} + C_{s} \phi_{s} \left( U \right) \right\} + - - \left\{ D_{e} + C_{e} \phi_{e} \left( U \right) \right\} - - \left\{ D_{w} + C_{w} \phi_{w} \left( U \right) \right\} = S_{p} + B_{p} - \label{eq11} - - - \\ . - \\ where $B_{p}$ is the deferred-correction source terms, given by: - - - B_{p} = C_{l}\phi^{'}_{l} - C_{h}\phi^{'}_{h} + - C_{s}\phi^{'}_{s} - C_{n}\phi^{'}_{n} + - C_{w}\phi^{'}_{w} - C_{e}\phi^{'}_{e} - \label{eq12} - - - This treatment leads to a diagonally dominant coefficient matrix since it is formed using the UDS. - - The final form of the discretised equation: - - - a_{P}\phi_{P}= a_{N}\phi_{N} + a_{S}\phi_{S} + - a_{E}\phi_{E} + a_{W}\phi_{W} + - a_{H}\phi_{H} + a_{L}\phi_{L} + a_{T}\phi_{T} + S_{p} + B_{p} - \label{eq2} - - - Subscrit $P$ represents the current computational cell; $N$, $S$, $E$, $W$, $H$, $L$ represent the six neighbouring cells and $T$ represents the previous timestep (transistent cases only) - - The coefficients contain the appropriate contributions from the transient, convective and diffusive terms in (\ref{eq1}) - - == Normalised Variables == - - == Total Variation Diminishing (TVD) == - - == Convection Boundedness Criterion (CBC) == - - == Schemes == - - === QUICK - Quadratic Upwind Interpolation for Convective Kinematics === - - === LUS === - - === Fromm === - - === CUS === - - ------------------------------ - - - === van Leer === - - === OSPRE === - - === Superbee === - - === MINMOD === - - === ISNAS === - - === MUSCL === - - === UMIST === - - === SOUCUP === - - === HLPA === - - === SMART === - - === SMARTER === - - === LPPA === - === SHARP === + where $\omega_m$ are the gaussian weights and $F$ is an approximation to $f_i n_i$. - === CHARM === + == External links == + * [http://www.imtek.uni-freiburg.de/simulation/mathematica/imsReferencePointers/FVM_introDocu.html The Finite Volume Method (FVM) - An introduction] by Oliver Rübenkönig of Albert Ludwigs University of Freiburg, available under the GNU Free Document License\|GFDL. - === VONOS === - == Numerical examples == + Return to [[Numerical methods \| Numerical Methods]] - === Smith-Hutton test === + {{stub}} ## Latest revision as of 09:15, 3 January 2012 The Finite Volume Method (FVM) is one of the most versatile discretization techniques used in CFD. Based on the control volume formulation of analytical fluid dynamics, the first step in the FVM is to divide the domain into a number of control volumes (aka cells, elements) where the variable of interest is located at the centroid of the control volume. The next step is to integrate the differential form of the governing equations (very similar to the control volume approach) over each control volume. Interpolation profiles are then assumed in order to describe the variation of the concerned variable between cell centroids. The resulting equation is called the discretized or discretization equation. In this manner, the discretization equation expresses the conservation principle for the variable inside the control volume. The most compelling feature of the FVM is that the resulting solution satisfies the conservation of quantities such as mass, momentum, energy, and species. This is exactly satisfied for any control volume as well as for the whole computational domain and for any number of control volumes. Even a coarse grid solution exhibits exact integral balances. FVM is the ideal method for computing discontinuous solutions arising in compressible flows. Any discontinuity must satisfy the Rankine-Hugoniot jump condition which is a consequence of conservation. Since finite volume methods are conservative they automatically satisfy the jump conditions and hence give physically correct weak solutions. FVM is also preferred while solving partial differential equations containing discontinuous coefficients. ## Integral form of conservation law Consider a single partial differential equation in conservation form $\frac{\partial u}{\partial t} + \nabla \cdot f(u) = 0$ where $u$ is a conserved quantity and $f(u) = [ f_1(u), \ldots, f_d(u)]$ is the flux vector. Integrating this over any volume $V$ in $R^d$ we get the integral form of the conservation law $\frac{\partial}{\partial t} \int_V u dx + \oint_{\partial V} f_i n_i ds = 0$ where the divergence theorem has been used to convert the divergence term to a surface integral over the surface $\partial V$ and $(n_1, \ldots, n_d)$ is the unit outward normal to $\partial V$. The above equation is the starting point for the finite volume method. ## Finite volume formulation The computational domain $V_h$ is divided into non-overlapping cells or finite volumes $V_r, r=1,\ldots, N, \quad V_h = \cup_r V_r$ Usually these cells are polygons (triangles, quadrilaterals) in 2-D and polyhedra (tetrahedron, hexahedron, prims, etc) in 3-D. Introduce the cell average value over the cell $V_r$ $u_r = \frac{1}{\|V_r\|} \int_{V_r} u(x) dx$ which is the basic unknown quantity in the finite volume method. Let $N(r) = \{ \mbox{Set of cells which share a common face with } V_r \}$ Then the integral conservation law for cell $V_r$ is $\|V_r\| \frac{du_r}{dt} + \sum_{s \in N(r)} \int_{V_r \cap V_s} f_i n_i ds = 0$ It now remains to approximate the flux integral. This can be achieved using gaussian quadrature. Taking $p$ gaussian points we get $\int_{V_r \cap V_s} f_i n_i ds = \Delta s_{rs} \sum_{m=1}^p \omega_m F^m_{rs}$ where $\omega_m$ are the gaussian weights and $F$ is an approximation to $f_i n_i$.	4,821	16,957	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.0625	3	CC-MAIN-2017-26	latest	en	0.931242
https://electowiki.org/wiki/Smith,IRV	1,722,826,427,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00318.warc.gz	181,807,845	14,466	Smith//IRV (Redirected from Smith,IRV) The method 1. Eliminate all candidates not in the Smith set. 2. Perform an IRV tally among remaining candidates.[1] Example: ```49: A>B 3: B 48 C>B``` B pairwise beats A (51 to 49) and C (52 to 48), so B is the only candidate in the Smith set i.e. is the Condorcet winner. Therefore, A and C are eliminated, and B, being the only remaining candidate, wins. Notes When there are only 3 candidates in the Smith set, Smith//IRV elects the pairwise winner between the two candidates with the most 1st choices after eliminating everyone outside of the Smith set and redistributing support. Smith//IRV passes ISDA (Independence of Smith-dominated Alternatives) but fails mono-add-plump (adding in ballots that bullet vote the winner shouldn't make the winner lose), which is the opposite of several other Condorcet-IRV hybrids.[2] Smith//IRV's precinct-summability depends on how large the Smith Set is: while a non-summable method like IRV requires O(c!) space in the worst case, Smith//IRV only requires O(k!), where k is the number of candidates in the Smith set. However, it's not possible to decide beforehand how large the Smith set will be; any such restriction will cause the method to violate universal domain. Thus, Smith//IRV is non-summable in full generality. However, once the precincts report the pairwise table and the Smith Set is known, at most the precincts only have to do a second pass of the ballots (if the Smith Set is 3 candidates or larger). In general, if the Smith Set has n candidates in it, then each precinct only has to report the top n-2 preferences each voter has over the Smith Set candidates. When the Smith Set has 3 candidates (which is expected to be the case most of the time that it is larger than 2 candidates), only each voter's 1st choice in the Smith Set need be known, since IRV will eliminate the candidate with the fewest 1st choices, and then there will be only 2 candidates remaining; the winner in IRV when only 2 candidates remain is guaranteed to be the winner of the pairwise matchup between the two candidates, so only the pairwise table need be consulted then to determine the winner. If there are 4 candidates in the Smith Set, each voter's 1st and 2nd choice in the Smith Set must be known, since once the candidate with the fewest 1st choices is eliminated, 3 candidates will remain, and then the 2nd choices of the voters who ranked the just-eliminated candidate 1st must be known to determine who to eliminate next, then there will be only 2 candidates remaining, and their pairwise matchup in the table determines the winner, etc. If the Smith Set was rather large (say, 10 candidates or greater), only then would it likely be necessary to conduct centralized counting as is done with IRV. One hybrid of Smith//IRV and Benham's method would be "eliminate everyone outside the Smith Set, then do IRV, but before each elimination, elect the Condorcet winner (based solely on pairwise matchups between uneliminated candidates) if there is one." Some discussion on Smith//IRV and other Condorcet-IRV hybrids (names differ in the linked article):[3] It is also possible to do Condorcet//IRV: "elect the Condorcet winner if there is one, otherwise elect the IRV winner." Neither Smith//IRV nor Condorcet//IRV passes dominant mutual third burial resistance,[4] though both pass when the DMT set consists of a single candidate. Both methods elect from the resistant set, granting them greater resistance to burial strategy.	819	3,525	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2024-33	latest	en	0.915813
http://www.finderchem.com/what-effect-does-friction-have-on-motion.html	1,490,329,646,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218187690.11/warc/CC-MAIN-20170322212947-00396-ip-10-233-31-227.ec2.internal.warc.gz	507,891,122	6,989	# What effect does friction have on motion? ... motion must be sufficient to overcome static friction. To remain in motion, ... in effect, increases static friction and thus ... Does "Friction of ... - Read more ... What effect does minimizing friction have upon the ... the graphs should be labeled and communicate the effect of minimizing friction upon the subsequent motion ... - Read more ## What effect does friction have on motion? resources ### Effect of Friction on Objects in Motion - Science Buddies Okita, N.A., 2003. "Effect of Friction on Objects in Motion," California State Science Fair Abstract ... What effect does friction have on the speed of a rolling object? ### What effect does friction force have on motion? - qfak.com What effect does friction force have on motion? a. it counteracts the acceleration of an object or .b. it opposes the motion of all objects Neither of the above ### What effect does friction force have on motion?a. It ... What effect does friction force have on motion? a. It opposes the motion of all objects. b. It counteracts the acceleration of an object c. It prevents surfaces from ... ### How Does Friction Affect The Motion of a Bowling Ball? \| eHow How Does Friction Affect The Motion of a Bowling Ball?. The less friction a ... How Does Friction Affect The Motion of a ... How to Do Special Paint Effects called ... ### What effect does friction have on a moving object? All kinds of friction cause moving objects to slow down or stop. Friction can be quantified as a force resisting motion which is proportional to the weight of the ... ### What effect does friction have on co2 cars? \| Answerbag What effect does friction have on co2 cars? CO2 cars use jet propulsion to move, and do not require any amount of traction to transfer power to the ground. ### Forces and Motion - Newton's Laws of Motion Laws of motion are used to calculate precisely the effects of forces on the motion of ... With friction we have two ... Friction and Inertia. Why does a car need to ... ### Free Lesson Plan \| Rules of Forces and Motion Lesson Plan ... Understand that friction and other forces have an effect on speed and motion; ... Ask students provide examples of friction. How does rain affect the roads? ### what effect does gravity have on an object in motion? What effect does gravity have on motion? ... How does gravity in space effect motion? ... What effect does friction have on the speed of a rolling ... ### Does friction have a negative effect on the mechanical ... Does friction have a negative effect on the mechanical advantage in Biology, Chemistry & Other Homework is being discussed at Physics Forums ### Section 3 Friction: A Force That Opposes Motion motion. Friction is also a force that, ... The Effect of Force on Friction ... and the motion of the object does not change. ### Physics4Kids.com: Motion: Friction - Rader's PHYSICS 4 ... This doesn't involve sliding surfaces like friction does, ... When you have a high coefficient of friction, you have a lot of friction ... Mechanics & Motion \| Friction ### BBC - Schools Science Clips - Friction ... effect of friction ... have both Javascript enabled and Flash installed. Visit BBC Webwise for full instructions. Other KS2 Science activities: Try... KS2 ... ### Friction - Georgia State University ... the surface friction does no work against the motion of the wheel and no energy ... Figures of 0.02 to 0.06 have been reported as effective coefficients of ... ### GCSE PHYSICS - What is Friction? - Where does Friction ... Forces and Motion. Friction. ... Where does Friction Occur? ... but skates on ice have very low friction (they slide easily). ### How Does Friction Work? - Pennsylvania State University How Does Friction Work ... phenomena in the physical world Surface resistance to relative motion; ... to move the mouse and have it respond ... ### Force of Friction, Force and Laws of Motion \| Tutorvista.com ... is found to have ... still the block does not move as the force of friction increases ... as the limiting friction. When there is no relative motion ... ### Sliding Friction, Coefficient of Sliding Friction ... It does not effect on friction. ... to resistance in motion in ... along because of the fact that effect of rotational friction is far less than ... ### Does friction oppose force or motion? - Physics Forums Does friction oppose force or motion? in Classical Physics is being discussed at Physics Forums ### FearOfPhysics.com: Friction What is Friction? Friction is the "evil" of all motion. No matter which direction something moves in, friction pulls it the other way. Move something left, friction ... ### Friction \| Define Friction at Dictionary.com Friction definition, surface resistance to relative motion, ... the act, effect, or an instance of rubbing one object against another: ### What Are the Effects of Massage on the Muscular System? \| eHow Not only does a massage soothe ... and after your massage session range of motion in the affected area ... What Effect Does Muscular Exercise Have on Metabolic ... Related Questions Recent Questions	1,081	5,155	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2017-13	latest	en	0.914725
https://republicofsouthossetia.org/question/a-2-by-2-array-with-2-more-cubes-is-6-cubes-15161247-36/	1,632,818,065,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00366.warc.gz	512,739,595	13,060	## A 2-by-2 array with 2 more cubes is 6 cubes. Question A 2-by-2 array with 2 more cubes is 6 cubes. in progress 0 3 weeks 2021-09-09T10:51:27+00:00 1 Answer 0 ## Answers ( ) 1. this is correct. a 2-by-2 array is the same as 2 x 2. 2 x 2 = 4. 4 + 2 = 6.	112	259	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.421875	3	CC-MAIN-2021-39	latest	en	0.842564
https://homework.cpm.org/category/CON_FOUND/textbook/ac/chapter/11/lesson/11.1.2/problem/11-22	1,726,841,659,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00011.warc.gz	271,737,635	14,972	### Home > AC > Chapter 11 > Lesson 11.1.2 > Problem11-22 11-22. Find the equation of the line with slope $m=-\frac{4}{3}$ that passes through the point $\left(12,-4\right)$. Substitute the slope and the coordinates into the Slope-Intercept form of a line to solve for $b$. Then, rewrite the equation with only the values for $m$ and $b$ substituted in. $y=-\frac{4}{3}x+12$	119	378	{"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2024-38	latest	en	0.756798
https://www.runup.ca/calculating-time-to-critical-point/	1,726,342,583,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00387.warc.gz	926,081,457	37,477	# How to Calculate Distance and Time to the Point of Equal Time and How to Calculate Time and Distance to the Point of Safe Return ## Calculating distance and time to the point of equal time (PET) ### Distance to the point of equal time (DPET) To calculate DISTANCE to the point of equal time (DPET) you need: 1. the total distance from your point of origin to your destination (D) 2. your ground speed home (GSH) #### Example 1. 560 NM (D) 2. 180 KTS (GSH) 3. 220 KTS (GSO) DPET = (D x GSH) / (GSH + GSO) DPET = (560 x 180) / (180 + 220) DPET = 252 NM from your point of origin ### Time to the point of equal time (TPET) To calculate TIME to the point of equal time (TPET) you need: 1. distance to point of equal time (DPET) #### Example 1. 252 NM 2. 180 KTS TPET = DPET / GSO TPET = 252 / 220 TPET = 1 hour 24 minutes from your point of origin (1.4) ## Calculating Time and Distance to Point of Safe Return (PSR) ### Time to Point of Safe Return (TPSR) To calculate TIME to the point of safe return you need: 2. your ground speed home (GSH) #### Example 1. 5 hours 2. 180 KTS 3. 220 KTS TPSR = (E x GSH) / (GSH + GSO) TPSR = (5 x 180) / (180 + 220) TPSR = 2 hours 15 minutes from your point of origin (2.25) ### Distance to Point of Safe Return (DPSR) To calculate DISTANCE to the point of safe return you need: 1. time to point of safe return (TPSR)	427	1,375	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2024-38	latest	en	0.728291
https://hyperleap.com/topic/Calculus/Geometry/Mathematics	1,670,275,036,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00214.warc.gz	340,527,033	23,470	# A report onGeometry, Mathematics and Calculus Geometry is, with arithmetic, one of the oldest branches of mathematics. - Geometry Mathematics is an area of knowledge that includes such topics as numbers (arithmetic, number theory), formulas and related structures (algebra), shapes and the spaces in which they are contained (geometry), and quantities and their changes (calculus and analysis). - Mathematics Calculus, originally called infinitesimal calculus or "the calculus of infinitesimals", is the mathematical study of continuous change, in the same way that geometry is the study of shape, and algebra is the study of generalizations of arithmetic operations. - Calculus This was a necessary precursor to the development of calculus and a precise quantitative science of physics. - Geometry ## Area Quantity that expresses the extent of a region on the plane or on a curved surface. Quantity that expresses the extent of a region on the plane or on a curved surface. In mathematics, the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number. For shapes with curved boundary, calculus is usually required to compute the area. In addition to its obvious importance in geometry and calculus, area is related to the definition of determinants in linear algebra, and is a basic property of surfaces in differential geometry. ## Differential geometry Differential geometry is a mathematical discipline that studies the geometry of smooth shapes and smooth spaces, otherwise known as smooth manifolds. The first systematic or rigorous treatment of geometry using the theory of infinitesimals and notions from calculus began around the 1600s when calculus was first developed by Gottfried Leibniz and Isaac Newton. ## Integral In mathematics, an integral assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. Along with differentiation, integration is a fundamental, essential operation of calculus, and serves as a tool to solve problems in mathematics and physics involving the area of an arbitrary shape, the length of a curve, and the volume of a solid, among others. Area can sometimes be found via geometrical compass-and-straightedge constructions of an equivalent square. ## Mathematical analysis Analysis is the branch of mathematics dealing with continuous functions, limits, and related theories, such as differentiation, integration, measure, infinite sequences, series, and analytic functions. Analysis evolved from calculus, which involves the elementary concepts and techniques of analysis. Analysis may be distinguished from geometry; however, it can be applied to any space of mathematical objects that has a definition of nearness (a topological space) or specific distances between objects (a metric space).	547	2,904	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2022-49	latest	en	0.918341
https://www.coursehero.com/file/5906688/Exam-3/	1,495,724,343,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00040.warc.gz	854,594,062	24,788	# Exam 3 - Version 440 – Exam 3 – Laude –(53755 1 This... This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Version 440 – Exam 3 – Laude – (53755) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 6.0 points If an MP3 player does 200 kJ of work and releases 100 kJ of heat, what is the change in internal energy for the MP3 player? 1. 300 kJ 2. 100 kJ 3.- 300 kJ correct 4.- 100 kJ Explanation: Δ U = q + w =- 100 kJ +- 200 kJ =- 600 kJ 002 6.0 points You observe that carbon dioxide sublimes. Which of the following statements about the signs of this process is/are true? I) Work (w) is positive. II) Heat (q) is negative. III) Change in Gibbs free energy (Δ G ) is pos- itive. IV) Change in entropy (Δ S ) is positive. 1. IV only correct 2. I only 3. III and IV 4. I and II 5. II and III 6. I, II and III Explanation: Sublimation results in a significant increase in the volume of the system, allowing it to do work on its surroundings, i.e. the pressure-volume work function is negative. Sublimation is also an endothermic process, making heat positive. Since the process de- scribed ”happens” as a given in the problem, the change in free energy must be negative. Change in entropy must be positive since a solid is becoming a gas. 003 6.0 points Which of the following is not a definition of internal energy or change in internal energy? 1. A measure of the spontaneity of a reac- tion. correct 2. The total energy content of a system. 3. A measure of the change in heat of a system at constant volume. 4. The sum of heat (q) and work (w). Explanation: Gibbs free energy is a measure of the spon- taneity of a reaction. All of the other state- ments are mathematical identities describing internal energy or prose restatements of the same. 004 6.0 points Based on enthalpy of formation data, cal- culate Δ H for the reaction: 3H 2 O 2 (l) + 2Fe(s) ←→ 3H 2 O(l) + Fe 2 O 3 (s) 1.- 992 kJ · mol − 1 2.- 1 , 118 kJ · mol − 1 correct 3. 1 , 118 kJ · mol − 1 4. 992 kJ · mol − 1 Explanation: Δ H rxn = ΣΔ H f,products + ΣΔ H f,reactants = (3 · Δ H f,H 2 O ( l ) + Δ H f,Fe 2 O 3 ( s ) )- (3 · Δ H f,H 2 O 2 ( l ) + 2 · Δ H f,Fe ( s ) ) Version 440 – Exam 3 – Laude – (53755) 2 = (3 ·- 285 . 83 kJ · mol − 1 +- 824 . 2 kJ · mol − 1 )- (3 ·- 187 . 78 kJ · mol − 1 + 2 · 0) =- 1 , 118 kJ · mol − 1 005 6.0 points If a system at 273 ◦ C absorbs 1000 J of heat, what is its change in entropy? 1. 1 . 83 J · K − 1 correct 2.- 1 . 83 J · K − 1 3. 3 . 66 J · K − 1 4.- 3 . 66 J · K − 1 Explanation: Heat absorbed by a system will increase its entropy by an amount equal to q/T , in this case 1000J / 546K = 1 . 83 J · K − 1 . 006 6.0 points Which of the following would experience the smallest increase in temperature if 1 kJ of heat were added to it?... View Full Document ## This note was uploaded on 06/06/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas. ### Page1 / 9 Exam 3 - Version 440 – Exam 3 – Laude –(53755 1 This... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	1,108	3,449	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.359375	3	CC-MAIN-2017-22	longest	en	0.86145
https://datascience.stackexchange.com/questions/110147/plotting-the-confidence-interval-for-a-plot-in-python	1,718,551,529,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00559.warc.gz	176,775,779	38,388	# Plotting the confidence interval for a plot in python I have a curve and I want to create the confidence interval for the curve. Here, I provide a simple example: mean, lower, upper = [],[],[] ci = 0.2 for i in range (20): a = np.random.rand(100) MEAN = np.mean(a) mean.append(MEAN) std = np.std(a) Upper = MEAN+cistd Lower = MEAN-cistd lower.append(Lower) upper.append(Upper) plt.figure(figsize=(20,8)) plt.plot(mean,'-b', label='mean') plt.plot(upper,'-r', label='upper') plt.plot(lower,'-g', label='lower') plt.xlabel("Value", fontsize = 30) plt.ylabel("Loss", fontsize = 30) plt.xticks(fontsize= 30) plt.yticks(fontsize= 30) plt.legend(loc=4, prop={'size': 30}) In the above example, I drew %80 confidence interval. I have two questions: 1- Could you please tell me that this way of calculating and plotting the confidence interval is true? 2- I want to identify the area of the confidence interval. I have attached a figure, I want some thing like that. Could you please tell me if you have any solution? Thanks for your help.	286	1,047	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2024-26	latest	en	0.63575
https://kr.mathworks.com/matlabcentral/cody/problems/1678-count-consecutive-0-s-in-between-values-of-1/solutions/1876052	1,591,148,477,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00526.warc.gz	420,080,382	15,697	Cody # Problem 1678. Count consecutive 0's in between values of 1 Solution 1876052 Submitted on 16 Jul 2019 by Allen This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass x = 1; y_correct = 0; assert(isequal(zero_count(x),y_correct)) 2 Pass x = [0 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 0 0]; y_correct = [1 3 1 0 0 3 1 0 2]; assert(isequal(zero_count(x),y_correct)) 3 Pass x = [0 0 0 0 0 0]; y_correct = 6; assert(isequal(zero_count(x),y_correct)) 4 Pass x = [0 0 1 0 0 1]; y_correct = [2 2]; assert(isequal(zero_count(x),y_correct))	255	653	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2020-24	latest	en	0.684154
https://magedkamel.com/12-solved-problem-5-2-for-local-buckling	1,723,634,250,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00288.warc.gz	309,849,908	34,814	# 12- Solved problem 5-2 for local buckling of columns. Last Updated on March 21, 2024 by Maged kamel ## A Solved problem 5-2 for local buckling of columns. ### Detailed data for the Solved problem 5-2 for local buckling of columns. We are going to see, the first Solved problem for local buckling 5-2, in which, the arrangements of the solution. The Qs, Qa will be considered as = 1 as a start, solved problem 5-2 is From Prof. Jack McCormac’s handbook. For the column shown in fig 5.8, if 50 ksi steel is used. a) Using the column critical stress values in table 4-22 of the Manual, determine the LRFD design strength Φ Pn and the ASD allowable strength P/Ωc, for the column shown in fig 5.8. b) Repeat the problem by using table 4-1 of the Manual. c) Solve by using the AISC – equation E3. #### 1-Get the properties of section W12x72 from table 1-1 for the Solved problem 5-2 for local buckling Referring to the given section of W12x72, Height is 15 feet, fixed from the bottom and hinged from top, kx=ky=0.7. But, the recommended values for kx, Ky are taken equal to 0.80. Using table 1-1 for the properties of the section of Wt, the section as shown in the sketch, bf=12 inch, the thickness of the flange is 0.67, the web thickness is equal to 0.43, the overall height d=12.3 inch. #### 2-Check whether buckling is controlled by minor or major axis for the Solved problem 5-2 for local buckling We have to check, whether the column is short or long, so, we start by estimating the (Kl/ry) max (Kl/ry) at y, where ry is the minimum radius of gyration, we have k=0.8, L=15 x12=180 inch, ry=3.04 inch 4, kl/r at the y is 47.37, for Fy =50 ksi. Our equation 4.71sqrt(E/Qfy), is known to be =113.43 for Fy=50 ksi, the kl/r at the y-direction is less than 113.43. #### 3-Check the bf/2tf value for the W-section-unstiffened section. The controlling bf/2tf is =0.56sqrt(E/Fy). refer to equation E-7-4. For hot rolled when b/t<= 0.56 sqrt (E/fy), then the Qs is =1, 0.56 sqrt (E/fy),E=29000 ksi, Fy =50 ksi. The limiting Kl/r = 0.56 sqrt (E/fy)=0.56 sqrt (29000/50) = 13.49. Please refer to the next slide image for the given data for the solved problem 5-2 for local buckling of columns for a more detailed estimation. #### 4-Check the hw/2tw value for the W-section-stiffened section. The next step is checking the web from the previous check of the flange, Qs=1, for the Qa relating to the stiffened element, the web from the next slide, the overall height d is 12.3 inch, k-des is obtained from the table, kdes=1.27 inch multiplied by 2, and subtract from 12.3 inches,d-2ks=12.3-(21.27), then divide by the web thickness, 0.43, the result is 22.7. Refer to the coefficient λr for stiffened element case number 5. The detailed illustration when b/t is more than 1.49 sqrt (E/Fy). The ratio (1.49 sqrt(E/fy)=1.49sqrt(29000/50)=35.89, the W section stiffened part is case Number -5. Our web ratio hw/tw= 22.7 < 35.89. Then the section is non-slender, stiffened regarding the web, and non-slender unstiffened regarding the flange. let us review all the findings, first, the column is short, secondly, the flange, is non-slender, and also the web is non-slender, then Q=1. #### 5-Using table 4-22 for the available critical stress for compression members. Refer to the table for the kl/r and refer to the fy=50 ksi, our kl/r =47.37, this value is within the range of 47 and 48. The table will give Fcr corresponding to both the LRFD and ASD. For LRFD, the value is between 38.3 and 38.0, by interpolation. We will get the value of 38.189 ksi For the LRFD of the solved problem 5-2 for local buckling. While for ASD, the value is between 25.5 and 25.3. The final value is 25.426 for the ASD, to estimate Φc* Pn and Pn/Ω. For the ASD, Fcr/ Ω=25.426 ksi as stress will be multiplied by the area, and the value of Pn/Ω=536 kips approximately. Finally, the Φpn =806 kips=806 kips, while for the ASD Pn/Ω=536 kips, thus we have completed part a, as for table 4-22. ΦFcr =38.189 ksi as stress will be area which is 21.1, then ΦcPn =805.80 kips approximated to 806 kips. #### If Using Table 4-1 for the solved problem 5-2 for local buckling. Then for part b, for the solved problem 5-2, using table 4-1, as part b for the solved problem the main difference between the two tables is that table 4-1 is using the coefficient Kl, not Kl/r as used by table 4-22, previously our Kl/r=47.37, while, our kl= 0.8* 15=12 feet as a reminder, our column is fixed-fixed and k recommended =0.8 and the height is 15 feet log in by 12 ft from the left side, and from the top select the W12x72, as a vertical line. The intersection between the two lines, the value of 806 and 536 kips and 536 kips, the same values obtained from table 4-22 item a, these values matching each other. #### If Using the general provision for the available strength. In the third step, it is required to estimate the values by the general equation, which is part c for the solved problem 5-2, this is the general equation, for the short column we are dealing with the curve at the left side of the 113.43 line. The equation of the curve is given by Fcr= 0.658^(Qfy/fe)QFy, since Qs=Qa=1 then Q=1, First for Fe=Pi^2E/(kl/r)^2=(22/7)^2(29000)/(47.37)^2, fe=127.6557 ksi,. Fcr=0.658^(50/127.6557), all multiplied by(150), fcr=42.455 ksi, to be multiplied by phi=0.9 by area=21.1. We get ΦPn =806 kips for the LRFD. In the case of ASD, for Ω=1.67, then Pn/Ω=42.455*(21.1)/1.67 Pn/Ω=536.5 kips, which is matching with the values obtained from table 4-22 and table 4-1. This is the pdf file used in the illustration of this post. The next post: solved problem 5-3 for local buckling. This is a great external reference for limit state, here is the link: Limit State of Flexural Buckling for Slender Sections Scroll to Top	1,816	5,831	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2024-33	latest	en	0.831054
http://spotidoc.com/doc/273216/	1,539,926,979,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583512323.79/warc/CC-MAIN-20181019041222-20181019062722-00245.warc.gz	349,041,848	7,456	# Document 273216 ```Details on the Sample Size Calculator for the Cochran-Armitage Test The Cochran-Armitage trend test in proportions involves an ordered set of groups for which we test for a linear trend in the proportions responding as we go across the ordering. The linear trend in the probabilities of response is given by the expression: pi  a  bd i , where we test for a non-zero slope relating the probabilities to the numeric assigned to each of the ordered groups. For details on the test statistic see Agresti (2002) and see similar details as well as SAS implementation of the test at this link: http://www.lexjansen.com/pharmasug/2007/sp/sp05.pdf Nam (1987) notes that the test statistic is equivalent to using a linear logistic model instead and testing for the slope associated with ordered levels in this logistic model. Nam evaluates the sample size using this logistic model. Using the same notation as in Nam, we define pi as the probability of response, qi as the probability of non-response, di as the ordered numeric assigned to a group (typically the actual dose when looking at dose levels or ordinally ordered as equal to the subscript i) and ri as the multiple of the sample size in a group ni to that in the control n0 for i= 0 to (k-1) (k groups including control). Further d is the average of the di’s, z1-α is the (1-α)th quantile the standard normal distribution, z1-β is the (1-β)th quantile of the standard normal distribution, Δ is the difference between successive di’s and p   ri pi /  ri  . Nam (1987) provide the following expression for the continuity corrected sample size for the Cochran-Armitage Trend test in the control arm   2 n0  (n0* / 4) 1  1  2 /( An0* ) , where n0* is the sample size in the control arm without the continuity correction and is give by  n0*  z1   pq  ri (d i  d ) 2  z1  r p q (d i i i i  d )2  /A , 2 2 for A   ri pi (d i  d ) . References: 1) Nam Jun-mo (1987), A simple approximation for calculating sample sizes for detecting linear trend in proportions. Biometrics. Vol 43, No 3, pp701-705. 2) Agresti Alan (2002), Categorical Data Analysis. John Wiley and Sons, Hoboken, NJ. Pp181-182. ```	664	2,196	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2018-43	latest	en	0.885202
https://support.nag.com/numeric/nl/nagdoc_latest/flhtml/g01/g01fef.html	1,726,549,165,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00666.warc.gz	510,818,121	6,151	# NAG FL Interfaceg01fef (inv_cdf_beta) ## ▸▿ Contents Settings help FL Name Style: FL Specification Language: ## 1Purpose g01fef returns the deviate associated with the given lower tail probability of the beta distribution. ## 2Specification Fortran Interface Function g01fef ( p, a, b, tol, Real (Kind=nag_wp) :: g01fef Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: p, a, b, tol #include <nag.h> double g01fef_ (const double p, const double a, const double b, const double tol, Integer *ifail) The routine may be called by the names g01fef or nagf_stat_inv_cdf_beta. ## 3Description The deviate, ${\beta }_{p}$, associated with the lower tail probability, $p$, of the beta distribution with parameters $a$ and $b$ is defined as the solution to $P(B≤βp:a,b)=p=Γ(a+b) Γ(a)Γ(b) ∫0βpBa-1(1-B)b-1dB, 0≤βp≤1;a,b>0.$ The algorithm is a modified version of the Newton–Raphson method, following closely that of Cran et al. (1977). An initial approximation, ${\beta }_{0}$, to ${\beta }_{p}$ is found (see Cran et al. (1977)), and the Newton–Raphson iteration $βi=βi-1-f(βi-1) f′(βi-1) ,$ where $f\left(\beta \right)=P\left(B\le \beta :a,b\right)-p$ is used, with modifications to ensure that $\beta$ remains in the range $\left(0,1\right)$. ## 4References Cran G W, Martin K J and Thomas G E (1977) Algorithm AS 109. Inverse of the incomplete beta function ratio Appl. Statist. 26 111–114 Hastings N A J and Peacock J B (1975) Statistical Distributions Butterworth ## 5Arguments 1: $\mathbf{p}$Real (Kind=nag_wp) Input On entry: $p$, the lower tail probability from the required beta distribution. Constraint: $0.0\le {\mathbf{p}}\le 1.0$. 2: $\mathbf{a}$Real (Kind=nag_wp) Input On entry: $a$, the first parameter of the required beta distribution. Constraint: $0.0<{\mathbf{a}}\le {10}^{6}$. 3: $\mathbf{b}$Real (Kind=nag_wp) Input On entry: $b$, the second parameter of the required beta distribution. Constraint: $0.0<{\mathbf{b}}\le {10}^{6}$. 4: $\mathbf{tol}$Real (Kind=nag_wp) Input On entry: the relative accuracy required by you in the result. If g01fef is entered with tol greater than or equal to $1.0$ or less than (see x02ajf), the value of is used instead. 5: $\mathbf{ifail}$Integer Input/Output On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected. A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not. If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $-1$ is recommended since useful values can be provided in some output arguments even when ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit. On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6). ## 6Error Indicators and Warnings If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf). Errors or warnings detected by the routine: Note: in some cases g01fef may return useful information. If on exit ${\mathbf{ifail}}={\mathbf{1}}$ or ${\mathbf{2}}$, then g01fef returns $0.0$. ${\mathbf{ifail}}=1$ On entry, ${\mathbf{p}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{p}}\le 1.0$. On entry, ${\mathbf{p}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{p}}\ge 0.0$. ${\mathbf{ifail}}=2$ On entry, ${\mathbf{a}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{b}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{a}}>0.0$. On entry, ${\mathbf{a}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{b}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{a}}\le {10}^{6}$. On entry, ${\mathbf{a}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{b}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{b}}>0.0$. On entry, ${\mathbf{a}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{b}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{b}}\le {10}^{6}$. ${\mathbf{ifail}}=3$ The solution has failed to converge. However, the result should be a reasonable approximation. Requested accuracy not achieved when calculating beta probability. You should try setting tol larger. ${\mathbf{ifail}}=4$ The requested accuracy has not been achieved. Use a larger value of tol. There is doubt concerning the accuracy of the computed result. $100$ iterations of the Newton–Raphson method have been performed without satisfying the accuracy criterion (see Section 9). The result should be a reasonable approximation of the solution. ${\mathbf{ifail}}=-99$ See Section 7 in the Introduction to the NAG Library FL Interface for further information. ${\mathbf{ifail}}=-399$ Your licence key may have expired or may not have been installed correctly. See Section 8 in the Introduction to the NAG Library FL Interface for further information. ${\mathbf{ifail}}=-999$ Dynamic memory allocation failed. See Section 9 in the Introduction to the NAG Library FL Interface for further information. ## 7Accuracy The required precision, given by tol, should be achieved in most circumstances. ## 8Parallelism and Performance g01fef is not threaded in any implementation. The typical timing will be several times that of g01eef and will be very dependent on the input argument values. See g01eef for further comments on timings. ## 10Example This example reads lower tail probabilities for several beta distributions and calculates and prints the corresponding deviates until the end of data is reached. ### 10.1Program Text Program Text (g01fefe.f90) ### 10.2Program Data Program Data (g01fefe.d) ### 10.3Program Results Program Results (g01fefe.r)	1,819	5,926	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 66, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.984375	3	CC-MAIN-2024-38	latest	en	0.561905
https://www.transtutors.com/questions/1-find-a-grammar-in-chomsky-normal-form-equivalent-to-s-gt-aad-a-gt-ab-bab-b-gt-b-d--846478.htm	1,637,979,485,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00292.warc.gz	1,155,960,866	12,787	# 1. Find a grammar in Chomsky Normal form equivalent to S->aAD;A->aB/bAB; B->b, D->d. 2.. 1. Find a grammar in Chomsky Normal form equivalent to S->aAD;A->aB/bAB; B->b, D->d. 2. Convert to Greibach Normal Form the grammar G=({A1,A2, A3},{a,b},P,A1 ) where P consists of the following. A1 ->A2 A3, A2 ->A3 A1 /b,A3 ->A1 A2 /a. 3. Construct given CFG into GNF where v={S,A},T={0,1} and P is S Æ AA/0 , AÆSS/1 4. Convert the grammar S->AB, A->BS/b, B->SA/a into Greibach NormalForm. 5. Construct a equivalent grammar G in CNF for the grammar G1 where G1 =({S,A,B},{a,b},{S->bA/aB,A->bAA/aS/a, B->aBB/bS/b}, S) 6. Obtain the Chomsky Normal Form equivalent to the grammars S ÆaAbB , A ÆaA/a, BÆbB/b	280	699	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2021-49	latest	en	0.616866
https://en.grandgames.net/nonograms/picross/sovushka_2	1,561,434,633,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627999787.0/warc/CC-MAIN-20190625031825-20190625053825-00268.warc.gz	426,276,243	20,288	# Nonogram - Owl Size: 40x50 Rating: (4.89) Reviews: 568 Solved times: 1262 Added: 2015-02-04 Complexity: Possible solutions: 1 Author: belko61 Av. speed: 1h.35m.39s.Comments: 20 Topics: birds The nonogram has a unique solution and can be solved without guesswork? #птица #bird #сова #owl Help How to solve Nonograms? Printing Get Code Cross for the numbers: Cross for cells: Сoordinates Counter shaded cells Move the top panel Move Left Panel Automatic line crosses Autosize Autocheck Autofill Highlighting Numbers Transparency 4 3 1 2 5 3 3 1 2 1 3 2 2 1 1 5 2 3 2 3 3 2 1 2 2 3 2 2 1 3 2 1 2 3 6 3 3 5 1 3 4 3 6 2 2 7 1 4 3 1 2 1 2 1 3 3 2 2 5 1 4 3 2 2 2 3 2 2 2 2 10 2 3 1 9 2 1 3 4 3 2 2 1 1 6 3 6 1 4 2 1 1 10 3 3 2 2 3 2 2 1 4 2 4 2 3 2 4 2 3 2 5 4 5 4 5 3 2 3 2 15 2 2 4 1 4 4 3 4 2 4 7 2 2 8 7 2 5 8 8 2 1 1 1 1 1 4 3 1 3 4 1 7 4 3 5 5 6 3 4 5 3 1 2 5 2 1 2 5 4 1 2 3 2 2 4 4 3 1 6 5 8 4 2 3 3 2 3 11 1 2 5 2 7 3 5 7 1 5 4 5 1 5 5 4 1 1 4 1 2 2 2 2 1 4 2 2 2 13 4 3 11 4 1 2 6 1 3 6 5 3 4 8 2 9 5 4 3 3 4 3 1 1 2 2 3 1 1 2 2 5 5 1 3 2 1 2 9 3 5 1 2 4 1 2 1 2 4 5 5 3 9 2 2 2 2 1 1 1 4 3 2 4 2 2 5 4 1 3 4 2 1 2 2 2 3 2 2 2 9 5 6 6 2 1 2 2 2 1 6 6 6 2 3 2 2 2 3 2 1 4 5 3 4 2 2 3 3 1 2 6 10 1 1 2 2 3 3 3 3 3 2 1 1 4 2 1 5 3 2 4 2 5 6 4 3 3 3 3 3 2 6 1 1 6 3 2 11 6 1 2 2 1 2 7 3 2 2 1 2 3 4 3 3 2 6 5 2 5 3 1 3 2 3 1 3 3 1 1 1 1 2 9 3 5 3 1 2 4 3 2 2 3 3 2 2 3 1 2 2 4 2 1 2 1 2 2 1 2 3 2 3 2 2 1 4 1 1 6 2 2 3 3 1 3 3 6 4 3 7 3 6 1 5 1 2 1 1 4 5 4 1 2 1 1 4 3 3 4 2 1 2 1 7 2 4 2 2 3 2 1 2 4 3 3 2 2 2 2 1 3 3 4 2 1 2 2 2 1 4 4 3 1 1 3 3 3 2 1 4 1 3 2 1 1 2 3 3 1 2 2 4 2 4 3 1 3 3 2 1 2 3 4 2 7 3 2 1 2 1 1 1 9 2 12 2 5 3 1 4 1 3 3 6 3 4 2 1 1 1 2 3 2 3 4 2 1 2 1 1 2 1 2 3 3 2 3 3 1 6 2 1 3 3 3 3 3 6 1 6 1 9 3 2 4 15 1 3 4 2 3 2 3 4 2 2 5 3 2 1 1 1 2 3 4 2 2 2 2 6 2 2 2 4 2 2 3 2 6 3 3 1 2 1 1 3 5 3 5 7 5 4 6 4 6 11 4 2 1 3 1 7 9 3 2 3 5 1 3 3 3 12 2 8 6 12 10 2 4 1 4 4 3 2 1 1 5 11 2 2 3 2 2 2 3 9 3 2 10 2 2 3 3 4 3 2 4 4 5 1 5 3 2 1 8 9 2 2 5 10 2 2 Color grid: The main background: ninakiz Last login: 2017-09-12 09:28:37 118 201 #26882 2015-02-04 06:33 Хорошая совушка и кроссворд хороший. Спасибо. Like 2 0 Share hapa Last login: 2018-11-03 17:26:51 256 556 #26889 2015-02-04 11:01 спасибо, красивый хороший кроссворд. В принципе простой. но требует большой концентрации из-за пестроты. Like 2 0 Share justice Last login: 2017-04-26 12:58:33 6 7 #26893 2015-02-04 12:47 супер белка рисуй почаще Like 2 0 Share mumof Last login: 2019-06-24 22:39:50 1766 10167 #26908 2015-02-04 18:36 симпатично Like 1 0 Share margaRita Last login: 2019-06-24 23:34:53 1035 1725 #26928 2015-02-04 21:12 Шикарно! Спасибо,было интересно. Like 1 0 Share #27714 2015-02-19 16:18 очень интересный кроссворд, спасибо Like 1 0 Share v07 Last login: 2019-06-25 01:23:27 552 1337 #28001 2015-02-27 21:32 Замечательно! Спасибо! Like 1 0 Share Jurist1948 Last login: 2019-06-24 22:32:32 1002 2909 #36122 2015-08-25 16:41 Замечательно!! Like 0 0 Share olchichik Last login: 2017-11-17 17:52:06 261 488 #36938 2015-09-11 13:14 Прикольная... Like 1 0 Share VicNep Last login: 2019-03-03 08:55:06 102 513 #39843 2015-10-26 12:14 Вначале сложновато, затем почти как по маслу. Like 1 0 Share 8marta Last login: 2016-04-16 19:59:47 7 28 #55562 2016-02-09 01:22 Главное не потерять нить. Like 1 0 Share moroz Last login: 2016-12-11 20:29:50 11 35 #66892 2016-04-24 10:54 Не сложно, но интересно. Like 1 0 Share lesly Last login: 2019-06-13 13:22:56 82 1 206 #129006 2016-12-16 22:50 Отличный кросс, красивая картинка, 100% решаемость. Получила истинное удовольствие. Спасибо! Like 1 0 Share Irbischan Online 2472 141 1887 #160743 2017-04-29 16:42 Ай, да, я ! Собрала такую пеструшку и всего 1 раз ошиблась. Ну просто умница. Like 2 1 Share SDFGHJ Last login: 2019-06-25 00:02:50 2017-04-29 22:03 1 Молодцом,так держать!Сам себя не похвалишь,... а от других можно и не дождаться!Поэтому себя хвалить--получается хорошо и полезно! moroz73s Last login: 2017-09-27 20:01:34 1 18 #178452 2017-07-20 12:07 Красиво. Спасибо автору. Like 1 0 Share brickshooter3 Last login: 2019-06-24 22:24:15 1298 3487 #186879 2017-09-03 11:36 Классно! Начало сложное, а дальше просто очень внимательно. Картинка красивая. Классно! Like 1 1 Share belko61 Last login: 2019-06-24 22:49:15 2017-09-03 12:30 1 moroz73 Last login: 2018-01-16 12:16:25 1 10 #196014 2017-10-19 21:52 Все просто замечательно! Like 0 0 Share morozded Last login: 2019-05-27 19:37:48 1 14 #216171 2018-02-13 23:35 Like 0 0 Share kazan85 Last login: 2019-05-15 07:45:19 9 28 #217459 2018-02-23 18:56 Трудно было начинать, а уж потом - как кружева плести!!! Хорошо! Like 0 0 Share Nasturcia Last login: 2019-02-16 15:25:33 39 77 #219391 2018-03-08 02:01 Самое сложное сначала, а потом невозможно остановиться. Спасибо! Решать было интересно. Like 0 0 Share New on site Nonogram №30954 Nonogram №30953 Nonogram №30952 Nonogram №30950 Nonogram №30946 Nonogram №30944 Nonogram №30931 Nonogram №30928 Nonogram №30923 Nonogram №30922 Nonogram №30918 Nonogram №30917	3,018	5,139	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2019-26	latest	en	0.230287
http://users.ics.aalto.fi/harri/thesis/valpola_thesis/node38.html	1,708,617,257,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00764.warc.gz	38,248,199	2,746	Next: Algorithms Up: Linear Gaussian factor analysis Previous: Linear Gaussian factor analysis ### Neural network interpretation of the model It is somewhat artificial to call the linear factor analysis model a neural network, but it serves as a good starting point for the later development. The structure of neural networks is usually represented graphically by showing the computational elements, neurons, of the network. Each node corresponds to one neuron and the arrows usually denote weighted sums of the values from other neurons. It should be noted that although this representation bears resemblance to graphical models, a graphical model represents the conditional dependences while the neural network representation shows the computational structure. In general, these representations are therefore different. The linear factor analysis model can be represented as a neural network with two layers. On the first layer there are the factors and the second layer consists of linear neurons which compute a weighted sum of their inputs. A network interpretation of a model with two-dimensional factors and four-dimensional observations is depicted schematically in figure 5a. The weights Aij are shown as links between the nodes but the biases ai are not shown. Linear neurons as building blocks for larger networks are too simplistic because adding extra layers of linear neurons does not increase the representational power of the network. This is easily seen by considering the model x(t) = B (A s(t) + a) + b + n(t), (29) shown in figure 5b. By setting A' = BA and a' = Ba + b the model can be written as x(t) = A' s(t) + a' + n(t), (30) which, interpreted as a neural network, has only one layer of linear neurons. Next: Algorithms Up: Linear Gaussian factor analysis Previous: Linear Gaussian factor analysis Harri Valpola 2000-10-31	370	1,859	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2024-10	latest	en	0.925658
https://www.studypool.com/discuss/1054546/which-equation-represents-a-line-parallel-to-the-line-shown-on-the-graph?free	1,481,101,039,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00307-ip-10-31-129-80.ec2.internal.warc.gz	1,002,906,219	14,376	##### Which equation represents a line parallel to the line shown on the graph? Algebra Tutor: None Selected Time limit: 1 Day Jun 26th, 2015 (-8, 6) and (-6, 0) are 2 points on the line. Slope of the line, $\\ m=\frac{y_2-y_1}{x_2-x_1}\\ \\ m=\frac{0-6}{-6-(-8)}\\ \\ m=\frac{-6}{-6+8}\\ \\ m=\frac{-6}{2}\\ \\ m=-3$ Parallel lines have the same slope. So, a line parallel to the given line will have a slope of -3. So, out of the given options the equation of the parallel line is y = -3x + 3 ANSWER: y = -3x + 3 Jun 26th, 2015 ... Jun 26th, 2015 ... Jun 26th, 2015 Dec 7th, 2016 check_circle	228	608	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2016-50	longest	en	0.798982
https://physics.stackexchange.com/questions/710052/question-about-field-configurations-on-the-boundary-of-mathcali	1,713,221,077,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00536.warc.gz	419,268,875	39,635	# Question about field configurations on the boundary of $\mathcal{I}^+$ I am reading Strominger's lecture notes "The infrared structure of gravity and gauge theory" (https://arxiv.org/abs/1703.05448). In chapter two, while trying to derive an expression about the conserved charges $$Q_{\varepsilon}$$, the author assumes some that the behavior of the fields at $$\mathcal{I}^+$$. Specifically, he assumes that there are no long range magnetic fields at the future of $$\mathcal{I}^+$$, which he denotes as $$\mathcal{I}^+_+$$ and neither on its past, which he denotes as $$\mathcal{I}^+_-$$. Later, he claims that if we wish to have finite energy, the electromagnetic potential better be pure gauge. Why is that? I have also seen the stack exchange question (Electromagnetic radiation flux through null infinity) but I can not connect the arguments in the link with the gauge field being pure gauge. Also, I can not fully understand why the flux is given by the following formula in the link: $$\int_{\mathcal{I}^+}*(T_{\mu\nu}X^{\nu}dx^{\mu})$$ I have read the original paper, which I think Strominger bases its lecture notes regarding this part (https://arxiv.org/abs/1407.3789), and the author choose configurations which revert to the vacuum at $$\mathcal{I}^+_+$$ and hence $$F_{ur}\|_{\mathcal{I}^+_+}=0$$ and $$F_{uz}\|_{\mathcal{I}^+_+}=0$$. Is there a physical intuition related to this choice? Can someone elaborate on our motivation for choosing those boundary conditions? Any help will be appreciated. The energy flux through a cut $$[u_0,u_1]$$ of $${\cal I}^+$$ and charge flux through $$i^+$$ is given by $$E_{[u_0,u_1]} \sim \int_{u_0}^{u_1} du \int d^2 z \| F_{uz} \|^2 , \qquad Q \sim \int_{{\cal I}^+_+} d^2 z r^2 F_{ur}$$ Since the authors want the system to revert to pure vacuum in the far future, the energy flux must vanish as $$u_1,u_0\to\infty$$ and the charge flux must also vanish. This implies $$F_{uz} \|_{{\cal I}^+_+} = 0 , \qquad F_{ur} \|_{{\cal I}^+_+} = 0.$$ These conditions only make physical sense in the absence of massive charged particles. In later papers (should also be in the lecture notes), the effect of massive states has been studied and these conditions have been relaxed somewhat. • okay thanks Prahar. Can you elaborate a bit on why the absence of massive charged particles provides some physical sense to these conditions?? I mean, I know that charged massive particles can not travel to $i^+$, since they always follow timelike trajectories. But what is the exact relation between $i^+$ and $\mathcal{I}^+_+$?? From the arguments used so far, it seems to me that they are one and the same? Is this true? and why so? May 28, 2022 at 20:28 • $\partial i^+ = {\cal I}^+_+$ May 28, 2022 at 20:32 • $i^+$ is a single point in the conformal compactification. That is not a true representation of the asymptotic structure of a spacetime. $i^+$ and $i^-$ are 3-dimensional Euclidean hypersurfaces. ${\cal I}^+$ and ${\cal I}^-$ are 3-dimensional null hypersurfaces and $i^0$ is a 3-dimensional timelike hypersurface. Oct 12, 2023 at 12:32	881	3,084	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2024-18	latest	en	0.881932
https://www.people.vcu.edu/~rhammack/math105/archive/1A00sol.html	1,516,574,091,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084890893.58/warc/CC-MAIN-20180121214857-20180121234857-00721.warc.gz	1,032,968,720	4,237	Finite Math Test #1 Oct. 9, 2000 A Track R. Hammack Name: ________________________ Score: _________ (1) (10 points) Multiply the matrices. [ 1 3 ] [ 0 1 2 ] = 7 1 -5 2 5 [ 1(0)+3(-5) 1(1)+3(2) 1(2)+3(5) ] = [ -15 7 17 ] 7(0)+1(-5) 7(1)+1(2) 7(2)+1(5) -5 9 19 (2) (15 points) Sketch the solutions of the following system of inequalities. 2x1 + x2 ≤ 6 x1 + x2 ≤ 4 x1 ≥ 0 x2 ≥ 0 The solution is sketched on the right. Notice that the corner points are (0,0), (0,4), (3,0), and (2,2). (3) (15 points) Maximize subject to ... P = x1 + x2 2x1 + x2 ≤ 6 x1 + x2 ≤ 4 x1 ≥ 0 x2 ≥ 0 You may use any method. (However, notice that you sketched the feasible region in the previous problem. Feel free to use that information to solve this problem.) Using the work from the previous problem, we make a table: Corner point P = x1 + x2 (0, 0) 0 + 0 = 0 (0, 4) 0 + 4 = 4 (3, 0) 3+ 0 = 3 (2, 2) 2 + 2 = 4 From the table it's clear that the maximum value of P will be 4, and that happens for either (x1, x2) = (2, 2) or (x1, x2) = (0, 4) (4) (30 points) Use Gauss-Jordan Elimination to solve the following system of equations: 2x1 + 2x2 + x3 = 9 2x1 + x2 + 2x3 = 11 x1 + x2 + x3 = 6 [ 2 2 1 \| 9 ] 2 1 2 \| 11 1 1 1 \| 6 R1 <--> R3 [ 1 1 1 \| 6 ] 2 1 2 \| 11 2 2 1 \| 9 -2R1 + R2 --> R2 -2R1 + R3 --> R3 [ 1 1 1 \| 6 ] 0 -1 0 \| -1 0 0 -1 \| -3 -R2 --> R2 [ 1 1 1 \| 6 ] 0 1 0 \| 1 0 0 -1 \| -3 -R2 + R1 -->R1 [ 1 0 1 \| 5 ] 0 1 0 \| 1 0 0 -1 \| -3 -R3 --> R3 [ 1 0 1 \| 5 ] 0 1 0 \| 1 0 0 1 \| 3 -R3 + R1 --> R1 [ 1 0 0 \| 2 ] 0 1 0 \| 1 0 0 1 \| 3 (Reduced) Now you can read off the solution x1 = 2, x2 = 1, x3 = 3. (5) (30 points) Use the simplex method to solve the following problem. A hippie farmer wants to sell three crops -- apples, beans, and corn -- at the local Farmer's Market. He reckons it will take him 1 hour to harvest each bushel of apples, 2 hours to harvest each bushel of beans, and 1 hour to harvest each bushel of corn. Each bushel of apples weighs 20 pounds, each bushel of beans weighs 10 pounds, and each bushel of corn weighs 5 pounds. Apples sell for \$15 per bushel, beans sell for \$10 per bushel, and corn sells for \$4 per bushel. He has a maximum of 40 hours in which to harvest the crops. Given that his aging pick-up truck can haul at most 1000 pounds of produce, how many bushels of apples, beans, and corn should he take to the Farmer's Market to realize a maximum profit? (Assume he sells everything he brings to the market.) First, let's organize all the data into a table. Apples Beans Corn Maximum Weight 20 10 5 1000 pounds Time 1 2 1 40 hours Profit \$15 \$10 \$4 Let x = bushels of apples. Let y = bushels of beans. Let z = bushels of corn. ... so we want to maximize profit... P = 15x +10y + 4z Subject to.... Weight constraint ... 20x +10y +5z ≤ 1000 Time constraint... x + 2y + z ≤ 40 Nonnegative constraint... x, y, z ≥ 0 There are two problem constrains, so there will be two slack variables. The problem is transformed into the following system of equations. 20x + 10y + 5z + s1 = 1000 1x + 2y + z + s2 = 40 -15x -10y -4z + P = 0 At this point we can see how to set up the simplex tableau: x y z s1 s2 P s1 20 10 5 1 0 0 \| 1000 1000/20 = 50 s2 1 2 1 0 1 0 \| 40 40/1 = 40 P -15 -10 -4 0 0 1 \| 0 Now it's time to do a pivot operation. We select the pivot column (1st column) and the pivot row (second row). Thus, x will enter and s2 will exit. The pivot element is already 1 (see how nicely I made it work out?) so we just need to get 0's elsewhere in the pivot column. Thus we do -20R2 + R1 ---> R1 and 15R2 + R3 ---> R3 x y z s1 s2 P s1 0 -30 -15 1 -20 0 \| 200 x 1 2 1 0 1 0 \| 40 P 0 20 11 0 15 1 \| 600 There are no more negative indicators, so we can just read off the solution. The old guy will make a \$600 profit if he picks and sells x=40 bushels of apples, y=0 bushels of beans, and z=0 bushels of corn. That's the best he can do given the constraints.	1,524	3,921	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2018-05	latest	en	0.844051
https://astronomy.stackexchange.com/questions/44390/how-did-the-measured-diameter-of-ceres-evolve-over-time?noredirect=1	1,656,219,915,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00447.warc.gz	165,230,646	63,408	# How did the measured diameter of Ceres evolve over time? Based on the Wikipedia page of Ceres, Herschel measured a diameter of 260 kilometers for Ceres, and Schröter measured 2613 kilometers. I do remember seeing at least a half-dozen estimates of the diameter of Ceres that were made over time, but I can't seem to put my hands on the source again. Which astronomers attempted to measure the diameter of Ceres, and what values did they find? How did the measured diameter of Ceres evolve over time? Which astronomers attempted to measure the diameter of Ceres, and what values did they find? ### Background If astronomers can see an object as a disc instead of point of light, they can measure the apparent diameter of the disc. For example, they can have thin wires stretched across the telescope at regular intervals, and see how many of those intervals the object fills. Thus they can calculate how much of the diameter of their field of view the object occupied. And knowing the angular diameter of the field of view at he magnification used, they can calculate the angular diameter of the object. And knowing the distance to the object, and its angular diameter, they can calculate its physical diameter. So as instruments to measure angular diameters of objects improved, and more accurate measurements and calculations of the distances of those objects were made, calculations of the physical diameters of those objects improved over time. Using the heliocentric theory, the relative distances of the planets from the Sun were easy to calculate, and more and better observations of the apparent motions of the planets resulted in better calculations of their actual motions and better determinations of their relative distances from the Sun. But until the absolute distance between two planets could be discovered, the absolute distances between all of the planets were unknown. Astronomers measured the relative distances between planets in Astronomical Units (AU). One AU was the distance between Earth and the Sun. So a major quest for astronomers was to estimate, calculate, or measure the value of an AU, and to do it more and more accurately over time. Here: https://en.wikipedia.org/wiki/Astronomical_unit#History[1] is a link to a discussion of the history of the AU, with a table of various historical values for it. The first approximately correct values for the AU came in the 17th century, and today radar is used to measure the distance from Earth to other planets and thus find the value of an AU. Of course some object are too small and distant to appear as measurable discs in Earthbound telescopes. The solution to the problem of measuring their diameters is to create better earthbound telescopes to view those objects, or to send small telescopes much, much closer to those objects in space probes, which has been done many times in the last 60 years. Many astronomers had no belief that there would ever be a space age, or in some cases that it would arrive as soon as it did. So they looked for other ways to find the diameters of objects which were mere points of light in their telescopes. If the brightness of an object's image could be estimated or measured by instruments, and its its distance known with greater or lesser accuracy, the total brightness of all the light reflected from its surface could be calculated with greater or lesser accuracy. If an astronomer estimated what the albedo, or reflectivess of the object was, and knew the total amount of light it reflected, he could calculate the surface area and diameter of the object. So that was one method which astronomers used to calculate the diameters of various solar system objects. The diameter of a particular object might be calculated many times over decades and centuries, with a somewhat different value each time. So different astronomy books published before space probes passed close to an object would have different values for its diameter. I noticed, for example, that different books disagreed on whether Ganymede, Callisto, Titan, or Triton was the largest moon in the solar system. My copy of Exploration of the Universe: Brief Edition, George Abell, 1964, 1969, has an Appendix 11, Satellites of Planets, on page 463. It give the diameters of Jupiter's moons Ganymede and Callisto as 4,900 kilometers (3,044.7 miles) and 4,570 kilometers (2,839.666 miles) respectively, and Saturn's Titan as 4,950 kilometers (3,075.7 miles) and Neptune's Triton as 4,000 kilometers (2,485.4 miles), thus making Titan the most titanic moon. My copy of The Guinness Book of Astronomy Facts & Feats: Second Edition, Ptrick Moore, 1979, 1983 says that Ganymede has a diameter of 5,276 kilometers (3,278.3 miles), Callisto has a diameter of 4,820 kilometers (2,295 miles) on page 97, Titan has a diameter of 5,140 kilometers 3,193.8 (miles) on page 110, and Triton has a diameter of 6,000? kilometers (3,728.2 miles) on page 123, thus making Ganymede the largest unless Triton actually is about 6,000 kilometers in diameter. My copy of Men of Other Planets Kenneth Heuer, 1951, 1954, has a table on page 156 listing the surface areas in square miles of the planets and other large bodies in the solar system. It gives Ganymede a surface area of 33,576,000 square miles (86,961,440.781 square kilometers), Callisto 30,959,000 square miles (80,183,441.897 square kilometers), Titan 39,572,000 square miles (102,491,009.489 square kilometers), and Triton 105,630,000 square miles (273,580,444.059 square kilometers) - which is between the surface areas of Mars and Venus. So the diameters would be Ganymede 3,269 miles (5,261 kilometers), Callisto 3,139 miles (5,052 kilometers), Titan 3,549 miles (5,711 kilometers), and Triton 5,560 miles (8,947.9 kilometers). My guess is that this diameter of Triton was calculated from its brightnes and assuming it had an albedo similar to that of Earth's moon. The first attempt to measure the diameter of Triton was made by Gerard Kuiper in 1954. He obtained a value of 3,800 km. Subsequent measurement attempts arrived at values ranging from 2,500 to 6,000 km, or from slightly smaller than the Moon (3,474.2 km) to nearly half the diameter of Earth.[73] Data from the approach of Voyager 2 to Neptune on August 25, 1989, led to a more accurate estimate of Triton's diameter (2,706 km).[74] https://en.wikipedia.org/wiki/Triton_(moon)#Observation_and_exploration[2] So presumably all figures for Triton's diameter before 1954 were calculated from it's brightness and assumed albedo. As the Voyager 2 space probe approached Neptune in 1989, it observed Triton from closer and closer distances, and the measured diameter got smaller and smaller every time. So the project scientists joked that Triton would disappear before the probe reached it. In 2021, Triton is listed as having a diameter of 1,680 miles (2,700 kilometers). https://solarsystem.nasa.gov/moons/neptune-moons/triton/in-depth/[3] https://en.wikipedia.org/wiki/Triton_(moon)[4] So the diameter of Triton in Exploration of the Universe was about 1.47 times 2,716.8 kilometers, the diameter in Guinness Book of Astronomy Facts and feats was about 2.2 times that, and the diameter in Men of Other Planets was about 3.29 times that. ### Ceres Ceres is listed as having a mean radius of 469.73 kilometers,and thus a mean diameter of 939.46 kilometers. https://en.wikipedia.org/wiki/Ceres_(dwarf_planet)[5] So Herschel's diameter of 260 kilometers was 0.276 of the present value, and Schröter's diameter of 2,613 kilometers was 2.78 of the present value, as well as being 10.05 Herschel's. And I think that the examples of the differing estimated diameters of the largest moons of the giant planets, made 150 years after Herschel and Schröter's estimates of the diameter of Ceres, show that such large discrepancies are not that odd. • +1 for such a comprehensive answer! I added some headings that help readers find the part about Ceres; please feel free to edit further. – uhoh Jun 18, 2021 at 3:50 • This is an interesting answer, but it does not answer the question I asked. Jun 18, 2021 at 7:41	1,897	8,132	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.265625	3	CC-MAIN-2022-27	longest	en	0.956202
https://www.hindawi.com/journals/mpe/2012/345093/	1,653,689,766,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00034.warc.gz	939,637,793	144,523	Research Article \| Open Access Volume 2012 \|Article ID 345093 \| https://doi.org/10.1155/2012/345093 Qingsheng Wang, Aifan Ling, Tao Huang, Yong Jiang, Min Chen, "A Trend-Switching Financial Time Series Model with Level-Duration Dependence", Mathematical Problems in Engineering, vol. 2012, Article ID 345093, 20 pages, 2012. https://doi.org/10.1155/2012/345093 # A Trend-Switching Financial Time Series Model with Level-Duration Dependence Revised28 Nov 2012 Accepted28 Nov 2012 Published25 Dec 2012 #### Abstract The financial time series model that can capture the nonlinearity and asymmetry of stochastic process has been paid close attention for a long time. However, it is still open to completely overcome the difficult problem that motivates our researches in this paper. An asymmetric and nonlinear model with the change of local trend depending on local high-low turning point process is first proposed in this paper. As the point process can be decomposed into the two different processes, a high-low level process and an up-down duration process, we then establish the so-called trend-switching model which depends on both level and duration (Trend-LD). The proposed model can predict efficiently the direction and magnitude of the local trend of a time series by incorporating the local high-low turning point information. The numerical results on six indices in world stock markets show that the proposed Trend-LD model is suitable for fitting the market data and able to outperform the traditional random walk model. #### 1. Introduction The traditional Brownian motion [1] or random walk model [2] has rich applications in physics, biology, and economics, especially in many classical financial theories for capturing the uncertainty of financial data (e.g., efficient market hypothesis). However, the dynamic behavior of Brownian motion is symmetrical and the probability of up and down at any time is always 50%. This is not consistent with the dynamic of price of any risky asset. So Brownian motion is not efficient in capturing the uncertainty for financial time series. To overcome the shortcoming of Brownian motion and capture the uncertainty in financial market, some sophisticated mean-reversion models are proposed, such as, arithmetic Ornstein-Uhlenbeck model [3], geometric Ornstein-Uhlenbeck [4], and autoregressive moving average (ARMA) model [5]. In addition, Engle and Granger [6] developed an error correct model (ECM) in which the short-term trend can be corrected by the deviation from the equilibrium of the long-term cointegration. However, Granger and Engle’s model has a clear disadvantage: when the process (e.g., prices or returns of financial securities) deviates from a long-term mean level or equilibrium, the direction and magnitude of the local trend will immediately change and it will go back to the long-term mean level or equilibrium. This makes that the change of the trend is too frequent. However, in many time series, such as, sunspot numbers, Canada lynx data, business cycle series, the change of local trend does not occur frequently in a small amplitude or a short period due to the inertia of movement and these time series often present the nonlinearity and asymmetry. In the last two decades, in order to capture the properties of the real world time series, some regime-switching models, for instance, the threshold autoregressive (TAR) model [7, 8], the smoothing transition autoregressive (STAR) model [9, 10], and the Markov-switching autoregressive (MSAR) model [11, 12] are paid close attention. The switching between different regimes is often determined by the lagged variables, exogenous variables, or latent variables. For TAR model, the switching from one regime to another depends on the lagged variables or the others in different fixed intervals, such that the threshold values are fixed and the switch of the regimes is abrupt. STAR model is actually a weighted two-regime TAR model, where the weight is a function of the lagged variable or the others such that the transition between regimes is smooth. For MSAR model, The choice of the regime is based on a latent random variable which follows a Markov process. The regime-switching models are more flexible and asymmetric and able to overcome the shortcomings of the mean-reversion models and therefore are used widely in finance (see [1318]). TAR and STAR models are suitable for the fluctuation data with fixed thresholds or regular cycle, but not for the case that the thresholds and durations of different regimes are time-varying or longer-term dependent, since the change of the regime is still too frequent. While MSAR model characterizes that the choice of regimes is random and only dependent on current state. Hence, in these regime-switching models, the change of the regime (trend) is always abrupt and does not incorporate the longer-term historical information. Generally speaking, in financial practice, the measure of the volatility of the returns of risky assets is based on the range of daily price [1923]. On one hand, for the local high (low) level, technical analysts and investors often exploit the past local high-low levels or resistant-support levels of financial data for predicting the next one. For example, when investors forecast the direction of up-down regime, they often refer to the past local high (low) levels of the technical indicators, for example, BIAS (the relative deviation from the moving average), MACD (moving average convergence divergence), and RSI (relative strength index); on the other hand, for the duration (the arrival time between consecutive events), Engle and Russell [24] proposed an autoregressive conditional duration (ACD) model, and then Bauwens and Giot [25] introduce the logarithmic-ACD model. Zhang et al. [26] extended ACD model to a threshold autoregressive conditional duration (TACD) model. The first two models allow the expected duration to depend linearly on the past ones, but TACD model nonlinearly. Furthermore, Bauwens and Veredas [27] propose the stochastic conditional duration (SCD) model that is more flexible. The applications of the ACD models have been reported in numerous papers (see [2838]); moreover, for both level and duration, there are numerous empirical applications about cycle (e.g., business cycle, see [3942] and solar cycle, see [43]), and turning point [4446]. Durland and McCurdy [47] proposed a class of duration-dependent markov-switching model. The model can incorporate the new information (duration) into the transition probabilities of the choice of regimes; however, the new information is not longer term. In this paper, we introduce a two-regime model that depends on a local high and low turning-point process (local peaks and troughs). This model can be transformed into two processes, a high-low level process and an up-down duration process. The local high (low) level and the up (down) duration of the current regime, which is widely used in many fields, such as economics and finance, depend on those of the past ones. To exploit the past longer-term key information, a new class of trend-switching model that incorporates both a local high-low level process and an up-down duration process is discussed and we call it the trend-switching model which depends on both level and duration (Trend-LD). In fact, both the high-low level process and the up-down duration process are threshold processes. Therefore, Trend-LD model is also called a time-varying double-threshold regime-switching model. The rest of the paper is organized as follows. In Section 2, we introduce the model specifications. Section 3 gives the estimations of the parameters. In Section 4, we compare the forecasting performances of the proposed model with that of random walk using six stocks indices from different stock markets. #### 2. Model Specifications In this section, we first provide the definition of local high-low turning point and derive a local high-low level subseries and an up-down duration subseries by which the change in the local trend is determined. And then the equations satisfied by the two subseries are put forward. A nonlinear specifications of the Trend-LD model are proposed. Let denote stock index series at time , and where . Clearly, stands for the -order moving average of the logarithm of index from time to , and is the relative bias of the daily stock index at time . In the bias series , the long-term trend is removed from the time series of index. As a result, it is relative stable and fluctuate around zero, and it can better represent the short-term relative gains or losses, but not absolute gains or losses, than price . Figures 1 and 2 give an intuitional example of DJI (US Dow Jones Industrial Average index). In Figure 2, the solid line is the relative bias of the DJI (); the circles and squares are the local high and low turning-point series, respectively (window radius ). Definition 2.1. For the relative bias series , is a local high (low) level, if it is the maxima (minima) of all samples in for a given window radius , and we denote the high (low) level by . The pair is called a local high (low) turning point. The local high (low) turning-point process can be obtained by scanning all samples of . However, it may not be a strictly alternate high-low turning-point series because two neighbor points may be all high (low). In this case, we delete a high (low) point which is not greater (less) than the neighboring point. For simplicity, without loss of generality, suppose the first and last are two local high points, we obtain a strictly alternate local high-low turning-point series (see the circles and squares in Figure 2), where the subseries is a series consisting of time and is another series consisting of the levels of the local high and low points. Furthermore, we can obtain the difference series: let , , , , , . We call the strictly alternate duration of the up-down trend and an up-down duration subseries. Subseries pair is called Level-Duration (LD) series and will be considered as our main objective. ##### 2.1. Level-Duration Equations Clearly, LD series controls the direction and magnitude of the local trend of index series . Because the current local high (low) level and the up (down) duration depend on the past ones. Then, each can be expressed as the linear combination of the past high and low level; that is, we have Similarly, for no nonnegativity constraint on coefficients estimation, we use the logarithm of durations rather than durations to make sure that the range of the logarithm of duration is real . the duration subseries can be specified by the logarithmic-ACD model [25] where the error terms , , , and are independent normal distributions with mean 0, variance , , , and , respectively. and are the number of lag order. Let be the information sets. Assume that conditional expectations , , , and . Then, we have By straightforwardly computing, it follows from (2.3) that In the rest of this paper, we will denote the local high-low level (2.2) by , and the up-down duration (2.3) by . ##### 2.2. Trend-Switching Model with LD Dependence In this study, we derive the relative bias by removing the middle-to-long term trend (moving-average) from the price. is relative stable and fluctuate around 0, so is better at reflecting the short-term relative profit or loss of investors, but not absolute profit or loss, than the price . Furthermore, in finance, researchers and practitioners often pay closer attention to daily return than daily price itself. Similarly, We are interested in the differences of the relative bias rather than the relative bias itself. Let Then, the dynamic characteristics of the relative bias series can be indirectly represented by the difference sequences . And assume that the difference sequences have the following expression: where is an indicator function and satisfies This means that if is in the th upward phase, and if is in the th downward phase. And where and represent respectively the upward and downward slope in the th phase; , and are the parameters that need be estimated. The innovation and are mutual independent normal error terms with mean 0, variance and , respectively. We call (2.8) . It is not hard to find that the local trend () consists of three components, that is, the constant term (), the short-term (, ), and the phase term (, ). The direction and magnitude of the local trend incorporate the key medium- and long-term information, for example, the local high-low levels and the up-down durations. Therefore, the dynamic behavior of the equation depends on both the level equation and the duration equation . It is integrally called model, that is, Trend-LD model. In fact, the alternate local high-low turning points (the local high-low levels, the up-down durations) are the two-dimensional thresholds of the Trend-LD and therefore it is also called the time-varying regime-switching model. Interestingly, if the parameters of the model satisfy some constrain conditions, such as, , , , . Then model is a classical pure autoregressive (AR) model. In addition, if the local high-low levels and up-down durations are substituted by some constants, model is a periodically changing AR or a particular form of TAR. #### 3. Estimation of Parameters In this section, we will estimate the parameters of the Trend-LD model using a two-stage method. We first estimate the parameters of the local high-low level equation and up-down duration equation by the method of ordinary least squares (OLS) in the first stage. And then, in the second stage, we estimate the parameters of the equation by maximum likelihood (ML). In fact, the ML is equivalent to the OLS under the assumption that error terms are normally distributed. We use the ML in the second stage because the statistical inferences of parameter estimations are similar to that in [48]. The estimation in the first stage is straightforward. Let be the high-low levels and up-down durations series samples data. Then the parameters in equation and equation can be estimated directly using OLS method. After the parameters in equation are estimated, we can estimate using (2.5) and forecast the next local high-low levels. Similarly, in terms of the estimated parameters in equation , can be obtained using (2.6), and the next local up-down durations can be forecasted. Then, the next local high (low) turning point (threshold point) can be obtained, which is helpful for predicting the change of trend. In the second stage, we can obtain the estimation of the slope of the up-down phase, in the equation using For the purpose of our analysis, denote and , . Next we estimate the parameters in equation . To this end, we denote the parameter set by with For data set , , note that and are normal series. Then for the given indicator , the conditional logarithm likelihood function can be written as where To obtain the maximum likelihood estimation (MLE) of the conditional logarithm likelihood function on parameter set , we only need to maximize and on parameter set and , respectively. If the process is stationary, the estimates fall within the boundaries of the allowable parameter space, then the estimators are consistent [48]. #### 4. Hypotheses of Asymmetry and Nonlinearity In this section, we focus on the asymmetry and nonlinearity of Trend-LD model and consider the following three hypotheses. Hypothesis 1. The dynamic behavior of the high-low level process does not have asymmetric effects on space (level), namely Hypothesis 2. The dynamic behavior of the up-down duration process does not have asymmetric effect on time (duration), namely Hypothesis 3. The dynamic of the local trend is linear and does not depend on local high-low turning point (level and duration), namely If Hypothesis 1 is true, the local high-low level equation can be further simplified to the equation . The truth of Hypothesis 2 will show that the up-down duration equation can be further simplified to the equation . Hypothesis 3 means that equation can be further simplified to a linear autoregressive model AR. The three Hypotheses are tested by the likelihood ratio statistic which is asymptotic Chi-square distribution and its degree of freedom is the number of the constrained parameters. #### 5. Empirical Applications Six stock indices with different degree of market development are used to test the performance of our models in this section. These indices are US Dow Jones Industrial Average (DJI), US Standard Poor 500 Index (SP500), British Financial Times Stock Exchange 100 Index (FTSE), Singapore Straits Times Index (STI), Indian Bombay Stock Exchange Sensex Index (BSE), and Chinese Shenzhen Component Index (SZCI). ##### 5.1. Data We collect the daily close prices of the above six indices from ‘‘http://www.finance.yahoo.com.’’ Each complete data sample is divided into two subsamples, estimating subsample (ES) and forecasting subsample (PS). The details about the data can be found in Table 1. It is worth mentioning here that we preliminarily use of training set for each index and then select a local high point as the separating point which is nearest from the point that divided the total daily index by since the end of the local high-low turning-point series , is a local high point. Index ES Sizes PS Sizes DJI 1928.10.01–1993.08.26 16286 1993.08.27–2010.08.18 4276 SP500 1950.01.03–1997.12.04 12062 1997.12.05–2010.08.18 3193 FTSE 1984.04.02–2005.07.07 5372 2005.07.08–2010.08.18 1293 STI 1987.12.28–2005.10.12 4445 2005.10.13–2010.08.18 1215 BSE 1990.01.01–2006.10.17 3932 2006.10.18–2010.08.18 943 SZCI 1991.04.03–2006.07.04 3757 2006.07.05–2010.08.18 1008 According to the definition of local high-low turning point, we computed its level data and , duration data and for each index and provide their descriptive statistics (mean and standard deviation) in Table 2, where moving average order , window radius (The choice of these two parameters is related to the commonly used parameters of Dow's trend theory and Elliott's wave theory in technical analysis. This study is to model the short-term trend and fluctuation of the stock index and to compare with the random walk hypothesis. Therefore, we choose some short-term parameters). In terms of the high-low level, the vibration amplitudes of the later three indices (STI, BSE, SZCI) are greater than those of the first three indices (DJI, SP500, FTSE). Among 6 indices, the fluctuation of the SP500 is the smallest and that of the SZCI is the largest. However, it seems that the characteristics of the up-down durations are not significantly different. Therefore, the local high-low level maybe have more informative than the up-down duration. statistic DJI SP500 FTSE STI BSE SZCI high mean 0.041 0.036 0.041 0.049 0.078 0.097 level std 0.027 0.017 0.02 0.035 0.05 0.091 low mean −0.046 −0.037 −0.042 −0.055 −0.074 −0.091 level std 0.046 0.035 0.04 0.055 0.056 0.069 level range 0.087 0.073 0.083 0.104 0.152 0.188 up mean 24.489 23.427 23.349 23.464 25.653 23.685 duration std 13.438 12.452 12.934 12.225 13.637 12.466 up mean 27.817 27.073 28.287 27.071 25.221 27.88 duration std 14.511 15.079 15.762 16.309 13.017 13.871 duration cycle 52.306 50.5 51.636 50.535 50.874 51.565 Note: level range = mean of local high level − mean of local low level, duration cycle = mean of up duration + mean of down duration. ##### 5.2. Estimates and Tests For the convenience of comparisons, we preliminarily consider the simple case of the proposed models, -, which are enough to capture the dynamics of the data (in financial application, first- or second-order AR model is normally used, especially in modeling stock returns. For our Trend-LD, we use second-order lag for the level equation, first-order lag for the duration equation, and the autoregressive term of the local trend equation. We did not provide concise explanation how we choose the orders. In general, the choice of the order is based on AIC or BIC criteria. We believe that these criteria are limited as the penalty on the number of parameters is somewhat arbitrary. We estimated the results on different combinations of parameters in unreported tests. For the local trend equation, we found the coefficients were not significantly different from 0 by -test when the order of autoregressive term was 2 or higher, and the conditional expectation of the difference is mainly determined by the second term related to turning points. Therefore, we only use the first order of the autoregressive term as well as the choice of lag order of other parameters). Then we estimate the parameters and conduct the asymmetry and nonlinearity tests for the proposed models. Ultimately, the structure of the model is chosen as - after the tests. For brevity, we consider two representative indexes, the developed market index DJI and developing market index STI and focus our discussion on results pertaining to the two indexes. The other four stock indexes have some similar properties. We first consider the OLS estimations (OLE) and asymmetric tests on high-low level equation and up-down duration equation . The results are presented in Tables 3 and 4, respectively. We find for equation under the null hypothesis (in Panel C) and alternative hypothesis (in Panel A and B), the statistics are very large for , and its values of -statistics are very small. From the statistics of the coefficients, the current high-low level is significantly dependent on the past (lagged first-order and second-order) high-low levels. (see Table 3). However, for duration equation under the null hypothesis (in Panel C) and alternative hypothesis (in Panel A and B), The statistics are very small for duration equation, and its values of -statistcs are comparatively large (see Table 4). From the statistics of the coefficients, the current up-down duration is weakly dependent on the past (lagged first-order) up-down durations. The constant estimations are all significantly greater than zero. That means there is inertia in the movement of the stock indices after each high (low) turning point. The findings show that the dependency of the current up-down duration on the past ones is much weaker than that of the current local high-low level on the past ones. The reason may be that the profit of investors is more directly related to local high-low level of the relative bias of stock price than to the up-down duration. The nonsymmetry test in Panel D shows as follows: at the significance level, the null hypothesis should be rejected, namely, the level equation is not symmetric , but asymmetric , it means that there is asymmetric effect on high-low level process (space); however, the null hypothesis fails to be rejected, namely, the duration equation is not asymmetric , but symmetric , it means that there is symmetric effect on up-down duration process (time). Panel A: high level equation ( ) Index -stat. value DJI 0.017* −0.171* −0.084 −0.236* 0.117 0.022 0.230 22.868 0.001 (6.44) (−4.838) (−1.582) (−6.346) (2.307) STI 0.010 −0.137* −0.032 −0.291* 0.327* 0.029 0.316 9.387 0.001 (1.550) (−1.765) (−0.321) (−3.594) (3.501) Panel B: low level equation ( ) Index -stat. value DJI −0.02* 0.039 0.175* −0.056 0.224* 0.035 0.099 8.400 0.001 (−4.816) (0.436) (3.026) (−0.668) (3.586) STI −0.029* 0.350 0.442* −0.164 0.030 0.041 0.191 4.728 0.002 (−3.193) (2.342) (3.905) (−1.139) (0.251) Panel C: level equation Index -stat. value DJI −0.001 −0.209* 0.170* −0.256* 0.244* 0.030 0.560 195.946 0.001 (−0.710) (−5.376) (4.409) (−6.658) (6.272) STI −0.0004 −0.100 0.301* −0.279* 0.186 0.039 0.538 48.248 0.001 (−0.093) (−1.310) (4.116) (−3.824) (2.467) Panel D: asymmetry test -stat. d.f. value DJI 121.701 6 0.001 STI 46.604 6 0.001 Note: Asterisks , , indicate significance at the 1%, 5%, and 10% level, respectively. -statistics are given in parentheses beneath the parameter estimates. Panel A: . Panel B: . Panel C: . Panel A: Up duration equation ( ) Index -stat. value DJI 3.029* 0.011 0.002 0.511 0.0001 0.021 0.979 (12.058) (0.203) (0.034) STI 3.080* −0.021 0.0007 0.514 0.0001 0.025 0.975 (6.466) (−0.220) (0.006) Panel B: Down duration equation ( ) Index -stat. value DJI 2.824* −0.023 0.085 0.536 0.023 3.620 0.028 (16.194) (−0.562) (2.107) STI 3.988* −0.158 −0.130 0.593 0.035 1.485 0.233 (7.739) (−1.241) (−1.201) Panel C: Duration equation Index -stat. value DJI 2.924* −0.023 0.085 0.116 0.008 2.413 0.090 (16.194) (−0.562) (2.107) STI 3.555* −0.090 −0.071 0.553 0.012 1.041 0.355 (10.240) (−1.181) (−0.924) Panel D: Asymmetry test -stat. d.f. value DJI 8.967 4 0.062 STI 4.340 4 0.362 Note: Asterisks *, , * indicate significance at the 1%, 5%, 10% level, respectively. -statistics are given in parentheses beneath the parameter estimates. Panel A: ln . Panel B: ln . Panel C: ; ; ; . Next, we consider the maximum likelihood estimate (MLE) and nonlinearity test on equation . The results are presented in Table 5. From the statistics and the value of the -statistics in Panels A and B, the coefficient estimations of the up-down phase slopes are significant greater than zero under 1% significant level and are the greatest among of . This means that the local up-down trend is mainly affected by the up-down phase trend . By the non-symmetry test in Panel D, at the significance level, the null hypothesis is rejected. It shows the dynamic of the local trend nonlinearly depends on local high-low turning point (level and duration). Thus, the structure of trend part of the model is considered as , not first-order autoregressive model AR. Panel A: up trend equation ( ) Index -stat. value DJI 0.0007* −0.0080 0.671* 0.011 0.018 67.954 0.001 (2.956) (−0.674) (11.606) STI 0.0004 0.013 0.893* 0.012 0.040 42.387 0.001 (0.097) (0.574) (8.840) Panel B: down trend equation ( ) Index -stat. value DJI −0.001* −0.113* 0.538* 0.012 0.0150 64.694 0.001 (−3.926) (−9.761) (6.605) STI −0.002*** 0.008 0.227* 0.013 0.0017 1.965 0.140 (−3.759) (0.389) (1.925) Panel C: trend equation Index -stat. value DJI 0.0265* 0.012 0.001 11.334 0.001 (−0.019) (3.367) STI 0.124 * 0.013 0.015 67.537 0.001 (−0.067) (8.218) Panel D: nonlinearity Test -stat. d.f. value DJI 1035.713 5 0.001 STI 300.714 5 0.001 Note: Asterisks *, , * indicate significance at the 1%, 5%, 10% level, respectively. -statistics are given in parentheses beneath the parameter estimates. Panel A: . Panel B: . Panel C: . In summary, the structure of the Trend-switching model for the indices is selected as . ##### 5.3. Comparisons with Random Walk In this section, we will consider the forecast performance of Trend-LD model and compare our model with the traditional random walk model. When the nonlinear model is used in predicting next day local trend (), it is necessary to know whether the direction of next day local trend is up or down, is 1 or 0. If the most recent turning point is high (low), next day local trend is down (up), . In the definition, the high (low) turning point is the highest (lowest) points within the window of 20 days in radius. However, the most recent local maximum or minimum point is not necessarily a local high or low turning point, since number of the observations on the right of the most recent local maximum or minimum point is less than 20. So we decide whether to accept the most recent high (low) point based on the current available observation in two different cases. Denote the current observation by . Note that the data on the right side of are unknown. The following gives two cases to decide whether a most recent local maximum or minimum point is a local turning point.(1)If the most recent observation is a local maxima (minima) in a given window . From Definition 2.1, is not seen as the local high (low) point in window , since the observations between are yet unknown. In this case, we assume that the direction of the next observation will continue to go up (down).(2)If the current observation is neither a local maxima nor minima in a given window . Denote the window distance between the current observation and the most recent local maxima or minima point (referred to as ) by . The choice of a reference distance is important for our comparison. For the relative bias and the window radius (20 days), we take days (one-fifth of the window radius ) as the referent distance. If , then is regarded as a high or low turning point, and if , we use the estimated local high-low level and up-down duration equation to predict the next local high turning point or low turning point , and then the direction of local trend is determined by comparing the most recent observed local maxima (minima) point with the predicted one. If the most recent observed local high (low) level is greater (less) than the predicted one and the most recent up (down) duration greater the predicted one, then is viewed as a high (low) turning point, and vice versa. Using the above rules, we estimate the direction of the local trend. Denote the return of stock index by and the change of relative bias by . Then, from , we have Rearrange the above formula, we derive the relationship between the return of stock index and the change of the relative bias as follows: Then, the one-step forecast of the return can be written as where is a predictor of , and is a predictor of in (5.2). The forecast performance can be measured by comparing the real return with forecasted return on the direction and magnitude. It is well known that the daily return of stock index is very uncertain and difficult to predict. However, in practice, due to transaction costs, most investors do not trade frequently, but focus on the direction and magnitude of local trend, with which our Trend-LD model is capable of dealing. However, the direction and magnitude of daily returns do not represent those of the local trend. In order to reduce the impact of random shocks on the local trend, we use the three-day moving average of the change of the bias (the real return can be seen as two parts: the unknown, deterministic expected local trend part and random shock part. As the stock prices can be affected by investors’ complicated behavior biases, their behavior is difficult to be predicted, and the daily return does not reflect real trend. It is normal that the -squared is very small in regression on short-term stock return, because the variation generated by random shock is responsible for most of the variation in stock return. When the time interval is smaller (daily), the effects of random shock on real return is greater, the accuracy of the prediction is clearly weaker, and the deterministic local trend part is relatively smaller or even submerged by the random shock. To reduce the effects of random shock, we use 3-day moving average of the real return as the proxy of the real local trend. Even the moving average is different from the local trend of real return, it is a better local trend proxy than real return, as the random shock is mostly averaged out. The purpose of our model is to predict the local trend of real return, not the real return that is easily affected by the random shock. The reason to use 3-day is that if the longer the moving average is, the more it may lag behind the real local trend; if the shorter the moving average is, the weaker the random shock effects are removed), ), as a proxy of the local trend of the relative bias, that is, substitute into in formula (5.2), we have a proxy of the local trend of daily return, denoted by , namely The following two subsections are the forecasting performance comparisons of the direction and magnitude between Trend-LD model and random walk model. Firstly, we consider the performance comparison of forecasting direction for our Trend-LD model with random walk model. A common directional measure [49] is to use a contingency table that summarizes the numbers of ‘‘rights’’ and ‘‘errors’’ of the model in predicting ups and downs of local trend of the return in the forecasting subsample, see Table 6. and indicate correct forecasts of the up and down direction, respectively. and indicate incorrect forecasts of the up and down direction, respectively. The Chi-squared statistic derived from the above contingent table is computed by which can be used to evaluate the performance of the model. A large signifies that the model outperforms the chance of random choice. Under some mild conditions, has an asymptotic chi-squared distribution with 1 degree of freedom. Forecast Up Down Total Up Trend Down Total Table 7 shows results of the numbers of ‘‘rights’’ and ‘‘errors’’ of the model in 1-step predicting ups and downs of local trend of the return in the six forecasting subsamples. Table 8 summarizes the right and error rates of directional prediction and its tests. the DJI has the smallest right rate, and the SP500 has the largest, . The values of statistic are almost zero. This shows that the dynamic behavior of Trend-LD model outperforms significantly the random walk model for all the six stock indices. In addition, the right rates of up direction () are greater than those of down direction () except the BSE. Indeed, this can be observed intuitively from Figure 1. In this figure, we find that, when the price rises, the local trend of index is more regular and easily predicted than the case when the price falls. This may result from the complicated, asymmetric psychology factors of most of investors, such as diminishing marginal utility, greed, and fear. Moreover, interestingly, the best and worst rate of forecast of up direction is of the BSE and of the SZCI, respectively. For the down direction, conversely, the right rate of the SZCI is and the BSE is , respectively. That is, the best and worst rate of forecast of the up-down direction always appears in the market of underdeveloped countries. DJI SP500 FTSE forecast forecast forecast Up Down Total Up Down Total Up Down Total Up 1594 782 2376 1216 505 1721 473 225 698 Trend Down 797 1103 1900 591 885 1472 245 350 595 Total 2391 1885 4276 1807 1386 3193 718 575 1293 STI BSE SZCI forecast forecast forecast Up Down Total Up Down Total Up Down Total Up 468 203 671 319 202 521 423 143 566 Trend Down 237 307 544 130 292 422 223 219 442 Total 705 510 1215 449 494 943 646 362 1008 DJI SP500 FTSE STI BSE SZCI 0.671 0.707 0.678 0.698 0.612 0.747 0.581 0.599 0.588 0.564 0.692 0.496 0.631 0.657 0.637 0.638 0.648 0.637 270.697 300.586 91.954 84.546 86.514 63.584 value 0.001 0.001 0.001 0.001 0.001 0.001 Secondly, we consider the comparison of forecasting magnitude for the two models above. Assume that the predicting error of the Trend-LD model is the difference between forecast return and the proxy of local trend , and the predicting error of random walk hypothesis is the difference between and . Denote , the forecast performance of the magnitude of local trend is tested by a DM statistic [50] where is the sample mean of the loss differential series is a consistent estimate of , is the spectral density of series . Under the null hypothesis of equal predictive accuracy, the DM statistic is asymptotically standard normal distribution . In Table 9, the values of the DM statistic show that the magnitude forecasts of local trend model all significantly outperform the random walk model. DJI SP500 FTSE STI BSE SZCI DM −4.524 −4.612 −2.207 −2.793 −2.659 −3.060 value 0.001 0.001 0.014 0.003 0.004 0.001 In Figure 3 (as there are 4276 forecasted daily returns, the figure will be too congested to convey any details if we plot the total time series. As a result, we only show part of the forecasted results), the dash-dot line is the proxy of the real local trend , the solid line is the forecast of the real local trend . Even if the forecast of the direction and magnitude of local trend have some errors, the overall forecast is good. Furthermore, the direction and magnitude of the local trend and its prediction show inertia (duration), which is different from the dynamic of the mean-reversion model and the error correction model that the direction and magnitude of the local trend will immediately change and it will go back to the long-term mean level or equilibrium when the process deviates from a long-term mean level or equilibrium. To summarize, the proposed Trend-LD model can efficiently reflect the fluctuation characteristics of financial time series, such as the inertia or the duration of the local up-down trend, the asymmetric dependency of the alternate local high-low level, the level and duration clustering and has many properties that beyond the traditional random walk model. #### 6. Conclusions Trend-LD model, in essence, is a conditional regime-switching (or time-varying two-dimensional thresholds) model, its threshold (local high-low levels and up-down durations or the high-low turning points) is not fixed, but conditionally dependent on the past ones. This means that the proposed model can incorporate longer-term, salient history information into trend-switching. Therefore, it is more flexible than the traditional random walk, mean-reversion, TAR, STAR and MSAR models. Our model may introduce to some more interesting further studies: The choices of the order of moving average and the window radius , and the determination of the local trend change is a challenge and interesting problem for further exploring. Comparisons of our model with mean reversion, TAR, STAR, and MSAR models for other time series data are also meaningful. #### Acknowledgments This work is supported by National Natural Science Foundations of China (no. 71001045, 10971162), Natural Science Foundation of Jiangxi Province of China (no. 20114BAB211008), and Jiangxi University of Finance and Economics Support Program Funds for Outstanding Youths. The authors thank the Editor and referees for their helpful comments. #### References 1. B. 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View at: Google Scholar Copyright © 2012 Qingsheng Wang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.	11,851	46,876	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2022-21	longest	en	0.879634
https://www.education.com/science-fair/article/mass-effect-projectile-motion/	1,571,273,371,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00216.warc.gz	865,661,596	48,201	Learning Library # Projectile Distance as Related to Mass of an Arrow and other Factors (35 ratings ) ### Research Questions: • Why does air resistance determine whether the mass of an arrow has any effect on distance? • Does the mass of an arrow have any effect on the distance an arrow travels? Taking air resistance into account, do you think a high-mass arrowhead would penetrate a target deeper than a low-mass arrowhead? • Does the angle from which an arrow is shot affect the distance an arrow travels? Is there a point of diminishing returns in which a greater angle minimizes distance? In an ideal universe, such as the ones in which most physics experiments take place, mass does not matter in studies of projectile motion because where is no air resistance. However, in real life, mass exerts a large effect. The angle from which an arrow is shot also determines the arc traversed by the arrow and how far it will go. By systematic experimentation, students can take various measures of this phenomenon and graph their results. ### Materials: If students use a computer-simulated bow and arrow, the only needed equipment is a computer, Internet access and an online computer-simulator, like the one listed in the Bibliography. Students using genuine equipment will require arrows of various weights, a regular bow and arrow, and measuring tapes. They should use a recognized archery range and have distances marked on the ground to determine how far the arrow was shot. While this equipment is available online and in specialty sport shops, students are encouraged not to spend money on this experiment if they do not have the necessary equipment. ### Experimental Procedure: Simulated 1. Keeping velocity at 50 meters/sec, air resistance “off” and a 70-degree angle (these are your constant variables), vary the mass of the projectile by taking it to first the minimum mass and then to the maximum. When you fire the simulator, note whether there is a difference. Can you explain this result? 2. Turn the “air resistance” feature on. Repeat step #1 a second time, varying the mass in 5 kg increments until mass no longer has an effect. Note how far the arrow went each time. Graph your results with mass being the y-axis and distance being the x-axis. 3. Keeping velocity at 50 mg/sec, air resistance “off” and mass at 19.05 kg, experiment by shooting the arrow from a 90 degree angle. However far did the arrow go? 4. Repeat step 3, varying the angle by 5 degree increments. At what point does lowering the angle become a point of diminishing returns? Repeat the experiment with air resistance on. Is there a difference? Real Bow and Arrow Protocol-Experiment#1 1. Identify a safe place such as an archery range to conduct this experiment. Measure the distances from the shooting range so that you can determine how far an arrow has flown. 2. Weigh your arrows so that you have at least using three or four arrows of different weights. The weight of an arrow varies depending upon the materials from which the shaft is made and the weight of the arrowhead (traditionally measured in grains). 3. Fire each three times, keeping the angle you hold the bow and distance you pull the bow string as constant as possible. Measure the distances and average your results together. 4. Repeat Step #3 with all of your arrows. 5. Graph your results. The y-axis can represent mass of the arrow and the x-axis can represent distance. Real Bow and Arrow Protocol-Experiment#2 1. Use the same place you used in Experiment #1. Select a weight of arrow for this experiment. Both the weight of the arrow and the distance you pull the bow-string are the constants of this experiment. 2. Fire a single arrow three times, keeping the arrow parallel to the ground. Measure how far the arrow traveled. Average your results together. 3. Repeat Step #2, varying the angle and shooting the arrow at least three times for each angle you select. 4. Graph your results. The y-axis can represent the angle and the x-axis can represent distance. Terms/Concepts: Projectile motion; Air resistance; Mass; Aangle and Distance Disclaimer and Safety Precautions Education.com provides the Science Fair Project Ideas for informational purposes only. 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For further information, consult your state's handbook of Science Safety. Create new collection 0 ### New Collection> 0 items What could we do to improve Education.com?	1,073	5,350	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.3125	4	CC-MAIN-2019-43	longest	en	0.934903
http://techsciencenews.com/referencetopics/inventor_invention/inventors/Quadrant_(instrument).html	1,579,605,666,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00556.warc.gz	160,114,029	8,278	### HomeABCDEFGHIJKLMNOPQRSTUVWXYZ #### Art \| Business Studies \| Citizenship \| Countries \| Design and Technology \| Everyday life \| Geography \| History \| Information Technology \| Language and Literature \| Mathematics \| Music \| People \| Portals \| Religion \| Science \| African Inventors \| Invention Timeline \| Space (Astronomy) \| Main Menu A quadrant is an instrument that is used to measure angles up to 90°. First proposed by Ptolemy as a better kind of astrolabe.[1] ## Contents There are several types of quadrants: • Mural quadrants used for measuring the altitudes of astronomical objects. • Large frame-based instruments used for measuring angular distances between astronomical objects. • Geometric quadrant used by surveyors and navigators. • Davis quadrant a compact, framed instrument used by navigators for measuring the altitude of an astronomical object. They can also be classified as[2] • Altitude - The plain quadrant with plumb line, used to take the altitude of an object. • Gunner's - The quadrant used by an artillery officer to set the angle of a gun barrel. • Gunter's - A quadrant used for time determination. Invented by Edmund Gunter in 1623. • Islamic - King identified four types of quadrants that were produced by Muslim astronomers.[3] 1. The sine quadrant - also known as the "Sinecal Quadrant", (the Arabic term for it is "Rubul Mujayyab") – was used for solving trigonometric problems and taking astronomical observations. It was developed by al-Khwarizmi in 9th century Baghdad and prevalent until the nineteenth century. Its defining feature is a graph-paper like grid on one side that is divided into sixty equal intervals on each axis and is also bounded by a 90 degree graduated arc. A cord was attached to the apex of the quadrant with a bead at the end of it to act as a plumb bob. They were also sometimes drawn on the back of astrolabes. 2. The universal (shakkāzīya) quadrant – used for solving astronomical problems for any latitude: These quadrants had either one or two sets of shakkāzīya grids and were developed in the fourteenth century in Syria. Some astrolabes are also printed on the back with the universal quadrant like an astrolabe created by Ibn al-Sarrāj. 3. The horary quadrant – used for finding the time with the sun: The horary quadrant could be used to find the time either in equal or unequal (length of the day divided by twelve) hours. Different sets of markings were created for either equal or unequal hours. For measuring the time in equal hours, the horary quadrant could only be used for one specific latitude while a quadrant for unequal hours could be used anywhere based on an approximate formula. One edge of the quadrant had to be aligned with the sun, and once aligned, a bead on the end of a plumbline attached to the centre of the quadrant showed the time of the day. 4. The astrolabe/almucantar quadrant – a quadrant developed from the astrolabe: This quadrant was marked with one half of a typical astrolabe plate as astrolabe plates are symmetrical. A cord attached from the centre of the quadrant with a bead at the other end was moved to represent the position of a celestial body (sun or a star). The ecliptic and star positions were marked on the quadrant for the above. It is not known where and when the astrolabe quadrant was invented, existent astrolabe quadrants are either of Ottoman or Mamluk origin, while there have been discovered twelfth century Egyptian and fourteenth century Syrian treatises on the astrolabe quadrant. These quadrants proved to be very popular alternatives to astrolabes. The geometric quadrant is a quarter-circle panel usually of wood or brass. Markings on the surface might be printed on paper and pasted to the wood or painted directly on the surface. Brass instruments had their markings scribed directly into the brass. For marine navigation, the earliest examples were found around 1460. They were not graduated in degrees but rather had the latitudes of the most common destinations directly scribed on the limb. When in use, the navigator would sail north or south until the quadrant indicated he was at the destination's latitude, turn in the direction of the destination and sail to the destination maintaining a course of constant latitude. After 1480, more of the instruments were made with limbs graduated in degrees.[4] Along one edge there were two sights forming an alidade. A plumb bob was suspended by a line from the centre of the arc at the top. In order to measure the altitude of a star, the observer would view the star through the sights and hold the quadrant so that the plane of the instrument was vertical. The plumb bob was allowed to hang vertical and the line indicated the reading on the arc's graduations. It was not uncommon for a second person to take the reading while the first concentrated on observing and holding the instrument in proper position. The accuracy of the instrument was limited by its size and by the effect the wind or observer's motion would have on the plumb bob. For navigators on the deck of a moving ship, these limitations could be difficult to overcome. ### Solar observations Drawing of a back observation quadrant. This instrument was used in the manner of a backstaff to measure the altitude of the sun by observing the position of a shadow on the instrument. In order to avoid staring into the sun to measure its altitude, navigators could hold the instrument in front of them with the sun to their side. By having the sunward sighting vane cast its shadow on the lower sighting vane, it was possible to align the instrument to the sun. Care would have to be taken to ensure that the altitude of the centre of the sun was determined. This could be done by averaging the elevations of the upper and lower umbra in the shadow. In order to perform measurements of the altitude of the sun, a back observation quadrant was developed.[4] With such a quadrant, the observer viewed the horizon from a sight vane (C in the figure on the right) through a slit in the horizon vane (B). This ensured the instrument was level. The observer moved the shadow vane (A) to a position on the graduated scale so as to cause its shadow to appear coincident with the level of the horizon on the horizon vane. This angle was the elevation of the sun. A large frame quadrant at the ancient Beijing Observatory. It was constructed in 1673. Large frame quadrants were used for astronomical measurements, notably determining the altitude of celestial objects. They could be permanent installations, such as mural quadrants. Smaller quadrants could be moved. Like the similar astronomical sextants, they could be used in a vertical plane or made adjustable for any plane. When set on a pedestal or other mount, they could be used to measure the angular distance between any two celestial objects. The details on their construction and use are essentially the same as those of the astronomical sextants; refer to that article for details. ## References 1. ^ The history of the telescope Henry C. King, Harold Spencer Jones Editor Harold Spencer Jones Courier Dover Publications, 2003 ISBN 0486432653, 9780486432656 2. ^ Gerard L'E. Turner, Antique Scientific Instruments, Blandford Press Ltd. 1980 ISBN 0-7137-1068-3 3. ^ King, D. (1987), ‘Islamic Astronomical Instruments’, Variorum, London, repr. Aldershot: Variorum, 1995. 4. ^ a b May, William Edward, A History of Marine Navigation, G. T. Foulis & Co. Ltd., Henley-on-Thames, Oxfordshire, 1973, ISBN 0 85429 143 1 • Maurice Daumas, Scientific Instruments of the Seventeenth and Eighteenth Centuries and Their Makers, Portman Books, London 1989 ISBN 978-0713407273	1,673	7,691	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2020-05	latest	en	0.943498
https://ndltd.ncl.edu.tw/cgi-bin/gs32/gsweb.cgi/login?o=dnclcdr&s=id=%22107NCKU5028011%22.&searchmode=basic	1,627,513,462,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00217.warc.gz	430,469,297	18,996	臺灣博碩士論文加值系統 (35.172.136.29) 您好！臺灣時間：2021/07/29 07:04 ::: 詳目顯示 : • 被引用:0 • 點閱:19 • 評分: • 下載:0 • 書目收藏:0 本文研究目的為探討一塊壓電三明治Mindlin Plate的動態響應，此三明治板結構的上下層為鋁合金板，中間為壓電材料(PZT-5H)。　　本文經由Mindlin Plate理論，計算出結構的應力、應變，再藉由應力、應變推算出動能及應變能方程式。通過邊界條件以及形狀函數計算出運動方程式，再以Hamilton’s Principle 理論計算出壓電三明治之統馭方程式，應用解析法於求出壓電複合板之模態頻率。　　施加一個集中型移動負載於結構上，獲得其位移與電壓，並探討改變移動負載的速度對於位移與電壓的影響。 The purpose of this paper is to investigate the dynamic response of a piezoelectric sandwich Mindlin Plate. The upper and lower layers of the sandwich plate structure are aluminum alloy plates with piezoelectric material (PZT-5H) in the middle. The stresses and strains of the structure in this paper are calculated via the Mindlin plate theory. The kinetic energy and strain energy are derived by stress and strain. The governing equations of the piezoelectric sandwich plate are derived by Performing Hamilton's Principle. The modal frequencies of the piezoelectric composite plate are obtained by analytic method. Applying a concentrated moving load on the structure obtain the displacement of the plate and the voltage on the piezoelectric layer. The velocity effect on the displacement and voltage is investigated in the thesis. 目　錄摘要……………………………………………………………………………...I英文摘要………………………………………………………………………..II誌謝…………………………………………………………………………...VII目錄………………………………………………………………………….VIII圖目錄……………………………………………………………………...….XI表目錄……………………………………………………………………….XIII第一章緒論……………………………………………………………………11-1 前言……………………………………………………………………11-2 文獻回顧………………………………………………………………31-3 本文大綱………………………………………………………………6第二章研究架構………………………………………………………………72-1 研究流程………………………………………………………………72-2 基本假設………………………………………………………………7第三章壓電複合板之運動方程式……………………………………………83-1 研究模型設定…………………………………………………………83-2 鋁板位移函數…………………………………………………………93-3壓電板位移函數……………………………………………………...103-4 鋁板之動能與應變能………………………………………………..143-5壓電板之動能與電焓………………………………………………...163-6 壓電複合板之運動方程式…………………………………………..17第四章壓電複合板之振動分析……………………………………………..194-1 邊界條件……………………………………………………………..194-2 自由震動……………………………………………………………..204-3 強迫振動……………………………………………………………..254-4 移動負載作用………………………………………………………..28第五章研究數據分析與討論…………………………………………..……295-1 材料設定……………………………………………………………..295-2 振動分析……………………………………………………………..30 5-2-1 自由震動……………………………………………………....30 5-2-2 強迫振動………………………………………………………315-3 位移與時間之關係………………………………………………......31 5-3-1 不同速度下位移與時間之關係……………………………....31 5-3-2 速度與位移之比較……………………………………………385-4 電壓與時間之關係…………………………………………………..42 5-4-1 不同速度下電壓與時間之關係………...…………………….42 5-4-2 速度與電壓之比較……………………………………………47第六章總結…………………………………………………………….…….516-1 結論…………………………………………………………………..516-2 未來展望……………………………………………………………..52參考文獻………………………………………………………………………53附錄……………………………………………………………………………57 1. E. Reissner, “The effect of transverse shear deformation on the bending of elastic plates, Journal of Applied Mechanics, Vol. 12, pp. 69-77, 1945.2. R. D. Mindlin, “Influence of rotary inertia and shear on flexural motions of isotropic elastic plates, Journal of Applied Mechanics, Vol. 18, pp. 31-38, 1951.3. R. D. Mindlin, A. Schacknow and H. Deresiewicz, “Flexural vibrations of rectangular plates, Journal of Applied Mechanics, Vol. 23, pp. 431-436, 1955.4. E. Reissner and Y. Stavsky, “Bending and stretching of certain types of heterogeneous aeolotropic elastic plates, Journal of Applied Mechanics, Vol. 28, pp. 402-408, 1961.5. J. M. Whitney and N. J. Pagano, “Shear deformation in heterogeneous anisotropic plates, Journal of Applied Mechanics, Vol. 37, pp. 1031-1036, 1970.6. A. W. Leissa, “The free vibration of rectangular plates, Journal of Sound and Vibration, Vol. 31, pp. 257-293, 1973.7. N. D. Phan and J. N. Reddy, “Analysis of laminated composite plates using a higher‐order shear deformation theory, International Journal for Numerical Methods in Engineering, Vol. 21, pp. 2201-2219, 1985.8. N. S. Putcha and J. N. Reddy, “Stability and natural vibration analysis of laminated plates by using a mixed element based on a refined plate theory, Journal of Sound and Vibration, Vol. 104, pp. 285-300, 1986.9. J. N. Reddy, Mechanics of Laminated Composite Plates: Theory and Analysis, CRC Press, 1997.10. L. X. Luccioni and S. B. Dong, “Levy-type finite element analyses of vibration and stability of thin and thick laminated composite rectangular plates, Composites Part B: Engineering, Vol. 29, pp. 459-475, 1998.11. J. A. Gbadeyan and S. T. Oni, “Dynamic behaviour of beams and rectangular plates under moving loads, Journal of Sound and Vibration, Vol. 182, pp. 677-695, 1995.12. J. S. Wu, M. L. Lee and T. S. Lai, “The dynamic analysis of a flat plate under a moving load by the finite element method, International Journal for Numerical Methods in Engineering, Vol. 24, pp. 743-762, 1987.13. J. J. Wu, A. R. Whittaker and M. P. Cartmell, “The use of finite element techniques for calculating the dynamic response of structures to moving loads, Computers ＆ Structures, Vol. 78, pp. 789-799, 2000.14. C. C. Chao and Y. C. Chern, “Comparison of Natural Frequencies of Laminated by 3-D Theory, Journal of Sound and Vibration, Vol. 230, No. 5, pp. 985-1007, 2000.15. M. Aydogdu and T. Timarci, “Vibration Analysis of Cross-Ply Laminated Square Plates with General Boundary Condition, Composites Science and Technology, Vol. 63, pp.1061-1070, 2003.16. G. Bradfiled, “Ultrasonic transducers 1. Introduction transducers. Part A. Ultrasonic, pp.112-113, 1970.17. R. D. Mindlin, “Forced thickness –shear and flexural vibrations of piezoelectric crystal plates, Journal of Applied Physics, Vol. 22, pp. 83- 88, 1952.18. H. F. Tiersten, “Thickness vibrations of piezoelectric plates, The Journal of the Acoustical Society of America, Vol. 35, pp. 53-58, 1963.19. P. Lloyd and M. Redwood, “Finite‐difference method for the Investigation of the vibrations of solids and the evaluation of the equivalent‐circuit characteristics of piezoelectric resonators, The Journal of the Acoustical Society of America, Vol. 39, pp. 346-354, 1966.20. C. K. Lee and F. C. Moon, “Laminated piezopolymer plates for torsion and bending sensors and actuators, The Journal of the Acoustical Society of America, Vol. 85, pp. 2432-2439, 1989.21. E. F. Crawley and J. De Luis, “Use of piezoelectric actuators as elements of intelligent structures, AIAA Journal, Vol. 25, pp. 1373-1385, 1987.22. S. Brooks and P. Heyliger, “Static behavior of piezoelectric laminates with distributed and patched actuators, Journal of Intelligent Material Systems and Structures, Vol. 5, pp. 635-646, 1994.23. Y. K. Kang, H. C. Park, W. Hwang and K. S. Han, “Optimum placement of piezoelectric sensor/actuator for vibration control of laminatedbeams, AIAA Journal, Vol. 34, pp. 1921-1926, 1996.24. M. W. Lin, A. O. Abatan and C. A. Rogers, “Application of commercial finite element codes for the analysis of induced strain-actuated structures, Journal of Intelligent Material Systems and Structures, Vol. 5, pp. 869- 875, 1994.25. J. H. Huang and H. I. Yu, “Dynamic electromechanical response of piezoelectric plates as sensors or actuators, Materials Letters, Vol. 46, pp. 70-80, 2000. 國圖紙本論文推文當script無法執行時可按︰推文網路書籤當script無法執行時可按︰網路書籤推薦當script無法執行時可按︰推薦評分當script無法執行時可按︰評分引用網址當script無法執行時可按︰引用網址轉寄當script無法執行時可按︰轉寄無相關論文無相關期刊無相關點閱論文簡易查詢 \| 進階查詢 \| 熱門排行 \| 我的研究室	2,434	7,169	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2021-31	latest	en	0.32219
http://raiffeisen-cards.ru/calculate-growth-rate-dividends-6652552.html	1,519,047,380,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00542.warc.gz	278,729,040	17,425	# How to Calculate Growth Rate in Dividends by Carter McBride ; Updated July 27, 2017 A corporation may pay dividends out of its earnings to investors during the year, although not at a set rate. Investors will then calculate the dividend growth rate to see how much the dividends are growing or shrinking over a period of time. Usually, if dividends are growing, the company is doing well. To calculate the dividend growth rate, the investor needs dividend information that all corporations must disclose. Step 1 Determine the dividends per share from the beginning of the period examined and the dividends per share from the end of the period. For example, an investor wants to know a firm's dividend growth rate from Year 1 to Year 3. In Year 1, the firm paid dividends of \$1.25 per share. In Year 3, the firm paid dividends of \$1.68 per share. Step 2 Subtract the latest dividends per share from the older dividends per share. In our example, \$1.68 minus \$1.25 equals \$0.43. This is the change in dividends. Step 3 Divide the change in dividends by the older dividends per share to calculate the dividend growth rate. In our example, \$0.43 divided by \$1.25 equals 34.4 percent. #### References Carter McBride started writing in 2007 with CMBA's IP section. He has written for Bureau of National Affairs, Inc and various websites. He received a CALI Award for The Actual Impact of MasterCard's Initial Public Offering in 2008. McBride is an attorney with a Juris Doctor from Case Western Reserve University and a Master of Science in accounting from the University of Connecticut. #### Photo Credits • calculating image by timur1970 from	374	1,659	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.4375	3	CC-MAIN-2018-09	longest	en	0.908233
https://www.evilmadscientist.com/2007/0-999-equals-one/	1,720,966,890,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00381.warc.gz	666,782,518	28,318	# 0.999… equals one. But you knew that, right? Check out the Wikipedia article about different proofs that 0.999… equals 1. It’s really quite interesting to see the wildly different methods that can be used to arrive at the same conclusion. While I’m partial to the Cauchy sequence version, the first of the proofs is my favorite, beautiful in its simplicity: 0.333…. = 1/3 3* 0.333… = 3 * (1/3) 0.999… = (31)/3 = 3/3 = 1 Q.E.D. =) ## 70 thoughts on “0.999… equals one.” 1. Anonymous says: But 0.333…. itself does not equal one-third; it is an approximation, just as saying 0.999…. equals one. Technically, 0.3 = 3/10, 0.33 = 33/100, 0.333 = 333/1000, etc. etc. 1/3 = 0.33333…… This is an infinitely repeating decimal. You can’t use this arguement to "prove" 0.999… equals one because saying 0.333…. equals exactly 1/3 is already using an assumption. I don’t know if this makes sense, but it does in my head! :-D 1. Anonymous says: If you’ve ever taken Calculus, as the number of decimal places increases, 0.999… converges to 1, therefore it equals one. It is a limit. 2. Anonymous says: No. .3 is an approximation, .33 or .333 or .333333333333333 are approximations, but .333… (which indicates an infinite number of 3’s) is an exact representation of 1/3. I think the problem is the way we read the number from left to right one digit at a time, we imagine the number getting closer and closer to 1/3 as we read each digit. But in fact, the number is there all at once, and with an infinite number of 3’s it exactly represents 1/3. 1. Anonymous says: Actually, you can prove .9999…. = 1 matheimatically. (and you can prove .3333…. = 1/3 the same way. Try this. Let x = .99999 ….. Then 10x = 9.9999…… Now subtract. 10x – x = 9.9999…. – .9999… 9x = 9 x = 1 Since x = .9999…. and x = 1, .9999…. = 1. Similarly, let x = .3333….; 10x then = 3.3333….. 10x – x = 3.3333…. – .3333….. 9x = 3 x = 3/9 = 1/3 Therefore, .3333…. = 1/3 It’s not an approximation, it’s not a limit – it’s an equality. It’s 7th grade algebra (if I can recall all the way back to 7th grade ;-) 3. Anonymous says: well it does make sense in my head too! :D 4. Anonymous says: you’re right. although 0,333 tends to get to infinite (which could stop anywhere if needed), it cannot be 1/3; ever!!! 1. No– "0,333" is a definite number. It does not "tend to" anything. Reminds me of a joke, though" 2 = 3, for a sufficiently large value of 2. Windell H. Oskay 2. Anonymous says: Your proof is wrong! 1/3 is actually not 0.333 but 0.3(3 occuring infinite times!). Simply calculate 1/3 in your scientific calculator and try change it into a fraction.. You will see what I mean. 3. Anonymous says: The "…" in this case means infinitely continuing. Same as writing .9 with a line over the 9. It’s just impossible to type it that way. 1. Anonymous says: the problem with the equation is that 3 0.333… does NOT = 0.999… 3 * 0.333… = 1 3 * 0.333 (not recurring) DOES = 0.999 (not recurring) and that’s why it looks like it works. If that made any sense at all! 4. Anonymous says: it does make sense .333 equaling 1/3 is just a close estimation, or theory, and not a actual proven fact Im not saying its right or wrong just that it is possible ‘theory’ and ‘fact’ do not necessarily stand in opposition of each other and the .999… theory equaling 1 is just a theory 1. Anonymous says: you may have also proven that 3/3 is not equal to 1 because 0.333… times 3 is 0.999… and 0.999 is not equal to 1 5. Anonymous says: The proof is wrong – at least in my opinion, here’s why : 0,333333333…. no matter how many 3’s you add to the number it is not equal to 1/3 it is just nearly there(forgive the language, but i’m no mathematician) :D (0,333…)3 /= 3(1/3) 0,333…=333…/1000… right? .Yes so (333…../1000…..)3 /= 1 – thus does not equal 3(1/3) than : 0,(3)3 = 0,(9) thank you and good night :)))) Tankowiec signing off 1. Anonymous says: lolz. i agree 2. Anonymous says: Unfortunately, .33333… does equal 1/3 .3333… is the summation of repeating geometric series (3)(.1)^(n-1). By doing some simple calculus, we find that the sum of this series is equal to: lim n=> infinity, the series will add up to equal: .3/(1-1/10)=1/3 likewise, we prove that .9999… is equal to one the same way. the geometric series for this is 9(.1)^(n-1) as n=> infinity, the series will add up to equal .9/(1-1/10)= 1 6. Anonymous says: This is mathematically proven to be true. Another proof is: Let x = 0.9999…. Multiplying both sides by 10, we get: 10x = 9.999…. Subtracting the first equation from the second: 9x = 9 => x = 1 Therefore, 0.9999…. = 1 <Trumpet call> End of proof! ;) 1. Anonymous says: No. Because your second equation contains one fewer digit after the decimal, no matter how many digits of precision you choose. Let’s look at a simple case: ```Let x = 0.99 10x = 9.9 10x-x = 9.9-0.99 9x/9 = 8.91/9 x = 0.99``` Busted! You cannot prove this without trickery, without fooling the reader somewhere along the line. All of these proofs rely on tricking us somewhere, and the more successful proofs hide the falsehood deep in the math where most people won’t catch it. 0.999… does not equal 1 and never will. The best you can say is that the approximation 0.999… approaches one as the decimal precision increases. 1. When we write "0.999…", we mean an infinitely repeated decimal, sometimes written as 0.9 (with a bar over the nine). By definition, this mathematical object has a non-finite number of decimal places so it is quite meaningless for you to talk about what happens "as the decimal precision increases." The only falsehood I see here is your supposed equivalence between an object with a fixed number of decimal places and an infinitely repeating decimal. Windell H. Oskay 1. Anonymous says: No. What you’re suggesting requires "infinity + 1" because for your math to work, you must have LESS precision on the "let x" line and MORE precision on the "10x" line in order for the math to work. If the precision is equal, i.e. infinite, as you say, then the math is broken. This is a trick, plain and simple. It’s not mathematically useful to discuss decimals with "infinite precision" because any number that cannot be represented with a finite number of decimal places cannot be accurately written in decimal. This only means that decimal is a lousy format for writing certain numbers. It doesn’t mean the numbers are weird. The value 1/3, for example, cannot be accurately written in decimal. Only an approximation can be written, but the true value is finite and well-understood. In base-3, one can write 1/3 accurately. 1. >No. What you’re suggesting requires “infinity + 1” because for your math to work, you must have LESS precision on the “let x” line and MORE precision on the “10x” line in order for the math to work. If the precision is equal, i.e. infinite, as you say, then the math is broken. This is a trick, plain and simple. No, there is no trick here. If you think that there is, then you simply aren’t fully grasping what it means to say “0.999…” When we write “0.999…”, we mean an infinite sum: Summation, from i = 1 to infinity, of 9/(10^i) This series does not terminate. There is* no last digit. If there were, then there could be a finite, non-zero difference between 0.999…. and 1. The term “precision,” loosely speaking, means the number of digits– if you think that the term precision applies, (or in other words, that there is a last digit) then you still aren’t getting it. Your argument is exactly equivalent to stating that infinity is some large but unspecified number– can you see the problem with that argument yet? You might want to consider reading the original article that I linked to, which discusses a number of other common misconceptions as well. There is also a special FAQ section to deal with basic questions like these. Windell H. Oskay 1. Anonymous says: Windell, you seem like a bright guy. I cannot believe that you are STILL arguing that 0.999… = 1. Seriously? What’s next? 2=3? I understand infinity, and I understand asymptotic functions that APPROACH a value but NEVER reach it. That’s what we’re talking about here. They are NOT equal. Period. I rest my case and will stop arguing with you. If you still believe that 0.999… = 1 then I have a bridge in Brookly that I’d like to sell you. 1. >Windell, you seem like a bright guy. I cannot believe that you are STILL arguing that 0.999… = 1. Seriously? What’s next? 2=3? You can drop the implied slur on my intelligence. If you don’t understand a point in mathematics, turning to ad hominem attacks doesn’t really help your argument much. > I understand infinity, and I understand asymptotic functions that APPROACH a value but NEVER reach it. You understand part of what’s going on here, which is a start. You aren’t understanding what it means for something to approach an asymptote. What it means, is that the value infinitely far away IS equal to that asymptotic value– otherwise it would not be approaching that asymptote, but instead diverging or converging to a different number. From wikipedia (and your math book): "A function f(x) is said to be asymptotic to a function g(x) as x -> infinity" if f(x) − g(x) -> 0. (Side note: We have been talking about a number, not a function. If you want to represent the question as a function– you brought up functions here– we can do so, but it gets a little more challenging yet since it’s a function of discrete values. So, let f(m) = sum(1:m)(1/(10^m)). The domain of the function is the set of natural numbers, and the range is the set of natural numbers. Now, for any finite value m, f(m) = 0.999…9. There is a final, terminating 9 in the sequence. In the limit as m goes to infinity, you need to use the limit, which brings us right back to the sum that I had mentioned earlier. ) Windell H. Oskay 1. Anonymous says: I concede. I read more at the Wikipedia article, and I see many many people are arguing this but the mathematicians all seem to agree with you. It intuitively SEEMS as though 0.999… must be infinitesimally smaller than 1, and it appears as though some of these "proofs" simply try to obfuscate the infinitesimal difference. But there is math far too complicated for me that supposedly proves it. I still don’t understand it, but I’ll take the mathematicians’ words for it, including yours. Please accept my apology for the Brooklyn Bridge comment. :-( 2. Anonymous says: Um, actually, your argument is wrong, because the entire proof is based on the fact that even if you multiply 0.999… by 10, getting 9.999…, you STILL have infinitely many 9’s after the decimal point. So my proof is still valid. :) 1. Wow– that was remarkably ambiguous– no way to tell whether you are arguing for a difference between 0.999…. and 1 or not. If you at least logged in as a user, we might be able to tell which way you were arguing from your name or something. Windell H. Oskay 2. Anonymous says: 0.999 does not equal 1. When you represent 1/3 as 0.333 you are using a format that approximates the value (less precision). The reason that your calculator outputs 0.333 when you input 1/3 is because 1/3 can not be represented exactly due to the way numbers are stored in a computer. Therefore it uses an approximation. In the actual world however, there are no need for approximations as we are not limited by having to store numbers using a binary format so instead we use fractions. The Calculator provided in Microsoft Windows XP also does this. From the help section: "Understanding Extended Precision – Extended Precision, a feature of Calculator, means that all operations are accurate to at least 32 digits. Calculator also stores rational numbers as fractions to retain accuracy. For example, 1/3 is stored as 1/3, rather than .333." 0.9999 can never equal one. It will always be less due to it being an approximation. 1. We are talking about "0.999…", not "0.999" There is a very big difference between the two, and it sounds like you’ve missed the point entirely. Windell H. Oskay 3. Anonymous says: The idea of any decimal repeating i.e. 0.999….. is simply put is like a 35 year old nerd who lives in his mom’s basement… he’ll just never make it all the way. 0.9999…. itself is an active number, approaching one, but never actually getting there. Just like an asymptote on a graph, the line approaches a number infinitely, but can never and will never equal said number. nuff said. 1. > 0.9999…. itself is an active number, approaching one, but never actually getting there. There is not any mathematical object that I am aware of called an "active number." 0.999… has an absolute and real value that is completely fixed– it does not vary, and does not approach any other number– it’s just a fixed number like 2, pi, sqrt(2), or so forth. (It just happens to be another way of writing "1".) > Just like an asymptote on a graph, the line approaches a number infinitely, but can never and will never equal said number. No. You have a fundamental misunderstanding about what it means to be as asymptote. If a function asymptotically approaches a given value, that means that the limit, infinitely far away, is that value. If that were not the case, then the function would be limiting to some other value– and not asymptotically approaching it after all. For any finite value of your independent variable, the function can only approach the final value; that analysis certainly does not apply in the limit that you look at an infinite number of digits. Windell H. Oskay 7. Anonymous says: 0.999 does not equal 1. When you represent 1/3 as 0.333 you are using a format that approximates the value (less precision). The reason that your calculator outputs 0.333 when you input 1/3 is because 1/3 can not be represented exactly due to the way numbers are stored in a computer. Therefore it uses an approximation. In the actual world however, there are no need for approximations as we are not limited by having to store numbers using a binary format so instead we use fractions. The Calculator provided in Microsoft Windows XP also does this. From the help section: "Understanding Extended Precision – Extended Precision, a feature of Calculator, means that all operations are accurate to at least 32 digits. Calculator also stores rational numbers as fractions to retain accuracy. For example, 1/3 is stored as 1/3, rather than .333." 0.9999 can never equal one. It will always be less due to it being an approximation. 1. Anonymous says: Very true and you explained it like pro. If i understand corectly though isnt it actually imposible to work this sort of thing exactly? By definition the reocurring 3 in 1/3 as a decimal figure (3.333…) is never ending so if a computer tried to work it out without limiting itself to a number of decimal places, it could never complete the calculation as it would be repeatedly adding 3’s to the equation literally forever. 2. >0.999 does not equal 1. Correct you are. You will never get an argument from me, or anyone else that knows a little math, about that! Windell H. Oskay 3. Anonymous says: the number .9999… or .333… is not an approximation. you missed the fact that … means repeats forever. Also its not a good idea to look at the workings of a computer to understand mathematical concepts. In the case of .999…=1 it is not designed to recognize the abstract logic it takes to come to that conclusion which is in fact correct. The only way to get a computer to tell you .999…=1 would be for a person to program it to do so, it will never come to this conclusion on its own. If you know some calculus, there are more complex proofs of .999…=1 that are more satisfying if you think the other proofs are lacking. 8. Anonymous says: A way to think about this in words and logic instead of equations is that given the fact that .9999… has an infinite number of 9s at the end of it there is no number you can add to .999… to equal 1. Try it if you don’t believe. .999…+.001= 1.000999… etc. Since there is no number between .999… and 1 you can conclude it is the same number. It sounds weird when you first hear it, but then again, no one ever guaranteed that common sense is always right. 1. Anonymous says: Yup, that’s how I thought of it too when I first heard this proof in a math lec! :) 0.9 + 0.1 = 1 0.99 + 0.01 = 1 0.999 + 0.001 = 1 and so, 0.999…. + 0.000… = 1 where the 0.000… essentially = 0 because the ‘1’ comes after infinite digits, and therefore doesn’t come at all. I don’t think this is a valid proof though, but I think it kinda helps in ‘understanding’ so to speak. 1. Thanks– that’s an interesting and intuitive approach; might be a good way to approach the topic for youngsters, even. [With a little limit calculus (or real analysis, if you’re hardcore!) the proof that you are outlining can actually be put on a rigorous, mathematically sound basis as well.] Windell H. Oskay 9. Anonymous says: Uhh… I’m just a code monkey, not some math genius, but… 1=0.999… 1-0.999…=0 but actually executing that operation gives 1-0.999…=0.000…1 so 0.000…1=0 for f(x)=1/n; f(0.1)=10, f(0.01)=100, f(0.001)=1000, etc. to f(0.000…1)=1000…0.0 1/0.000…1= +inf 1/0=fail as +inf!=fail, 0.000…1!=0, so 1!=0.999… Uhh… WTFBBQ??? I’m wondering if mixing open-form ideas like 0.999… and 0.000…1 with closed form ones like equals is fully valid? Or maybe I’ll just grab another banana and go back to my nice, simple, easy chars, ints and longs… 1. If you’re a code monkey, you know about bugs. (And you’ve just added one.) > 1=0.999… > 1-0.999…=0 > but actually executing that operation gives > 1-0.999…=0.000…1 That last line is not valid unless the "sequence of 9’s" terminates at some point. (Our "…" here means that it repeats forever; you can’t have something repeat forever and THEN have a final number after that!) Windell H. Oskay 1. Anonymous says: Bugs. Yes. They have a bad habit of getting into the bananas… yuck. 8-P What’s the difference between 0.999…999 and 0.999… ? Only 9s trail the decimal point. They both have an infinite number of nines trailing the decimal point. So each 9 in the first should have a corresponding 9 in the second, no? If each digit in one number has a corresponding digit in another, and they are all of the same value at the same place (in this case, 9) should not the two numbers be equal? So, if 0.999…999 = 0.999… and 0.999…999 does terminate, then 0.999… does also. And if 0.999…999 = 0.999… and 1-0.999…999=0.000…1 then 1-0.999…=0.000…1. Already established is that 1=0.999… so 1-0.999…=0, so 0.000…1 also must equal 0. Except that by definition, 0.999… does not terminate (so, does 0.999… even exist?). And what is the smallest infinitely small number greater than zero? Would it not be 0.000…1? And if it’s greater than zero, it can’t be equal to zero. Um… are you sure that using = with … is valid? (In other words, neither statement is true: 1=0.999… and 1!=0.999… ???) 1. Oh boy. >What’s the difference between 0.999…999 and 0.999… ? Only 9s trail the decimal point. They both have an infinite number of nines trailing the decimal point. […] No. There is no mathematical object "0.999…9"; it just doesn’t make sense. Something either goes on forever or it does not. Either there is no end and it’s a true repeating decimal (i.e., it goes on forever), or there is a last digit and it does not go on forever. Let’s rephrase your proposition. Suppose you come across a row of trees and a sign that tells you that it’s an infinitely long row of trees. What that means is that you could keep driving as long as you want– even forever– along this row of trees, and there would never be and end to it. Now suppose you drive along for a few days and come across a point where there’s a last tree in the row. Does that mean that you are now infinitely far away, and that there are infinitely many trees between your present position and where you started? No: It means that the sign was wrong! The row DID NOT go on forever, because it had an end. >so, does 0.999… even exist?. Yes, it’s one of the infinitely many ways to write the number "1." > And what is the smallest infinitely small number greater than zero? Would it not be 0.000…1? The only infinitely small number is zero. There is no "smallest number greater than zero." There can’t be. Suppose that my aunt thought she had found the smallest number greater than zero. I would take that number, divide it by two, and produce a smaller yet number that was still greater than zero, proving that her number was not the smallest. Since that can be repeated for any positive number, it shows that there is no such thing. >Um… are you sure that using = with … is valid? These symbols can be used in valid ways. I wish you would. ;) Windell H. Oskay 1. Anonymous says: Still can’t believe there are human beings trying to devolve our race by claiming things to be true via mathematical manipulation. 1 is not 0.999… I have plenty of arguments as to why it isn’t. (r–>) = recurring digits in the direction of the arrow 1.000(r–>) – 0.999(r–>) = 0.(<–r)0001 Are you going to argue that 0 = 1? That is exactly what you’re doing when you say 1 = 0.999… There is always going to be that 1 at the end when you subtract 0.999… from 1. Some may argue that if the 9’s don’t end, the 0’s don’t. Fine. They don’t. But how are you going to dictate which direction the recurring digits go? They would go left forever into the decimal point, making the value smaller and smaller but the 1 is still ALWAYS there. By arguing that 1 = 0.999…, you are saying something infinitely small does not exist. Then we don’t exist. The universe has infinite space, so compared to the universe, we are infinitely small. 1. >1 is not 0.999… I have plenty of arguments as to why it isn’t. The only trouble is that you don’t have any arguments that are correct. I am truly impressed by the density of mathematical and conceptual errors that you have managed to achieve in such a short comment! You’ve even managed to drag physics errors into this– that’s a new record, even for this discussion. What’s funny to me is that when I posted the original article, I just thought that all of the different proofs were neat– I didn’t appreciate whatsoever that there would be an significant number of people who found this non-obvious. I remember learning about repeating decimals in about sixth grade. Have so many people really forgotten? Windell H. Oskay 1. Anonymous says: Instead of just saying, "You’re wrong," how about you prove me wrong? 1. This whole article was about a list of different proofs that 0.999… = 1. Each and every one of those proofs is a proof that you are wrong. If that’s not enough, Wikipedia even has a whole FAQ section dedicated to the particular error that you are making. (Well, the error that you repeat the most, anyway.) Now, we’ve been over that particular error (it seems like) twenty times in other comments, and so has Wikipedia, (and your sixth grade teacher) but it comes down to this: If a digit repeats infinitely, it does not have a point at which it stops. That’s what it means for it to repeat infinitely. If it stops at some point, it’s not infinite. So long as you keep arguing “this goes on forever and then stops at some point,” you aren’t discussing an infinite situation: you’re discussing some make-believe hybrid between infinite and non-infinite. Mathematical objects have well-defined definitions– your mysterious hybrid number does not. Windell H. Oskay 2. Anonymous says: 1 is greater than 0. Want a "proof"? 1 + x > 0 + x Period. You are trying to manipulate math by using infinity in an equation. INFINITY IS NOT A NUMBER. How could you possibly think that you could plug it in and expect it to prove anything? That’s like ignoring the properties of 0. Infinity minus one is still infinity. Meaning: X – 1 = X Is that physically possible for any other "number"? No, therefore infinity should have its own set of properties. 3. >1 is greater than 0. I never argued otherwise. > INFINITY IS NOT A NUMBER. I never said it was. >Infinity minus one is still infinity. I agree. > infinity should have its own set of properties. It does. Out of curiosity, since we seem to agree on every single point, why are you arguing? ;) Windell H. Oskay 4. Anonymous says: I am saying it can’t just be plugged into equations to be considered "proof" because it is considered manipulation – not truth. In calculators, 0.999… might "be" 1 aso to pi "being" 22/7. Also: 1.000000000000… 0.999999999999… See the first digits in each line of numbers? That is the most important one. Period. 1.000… > 0.999… Arguing that an infitely small number is equal to zero is saying that we do not exist and seconds do not exist. You can always divide a second into something smaller, but will it ever be equal to no time passing at all? Also, our bodies are infinitely small when compared to the infinite universe. Does that mean we are 0 and that we don’t exist? Saying 0.999… is 1 is also saying 1/x = 0 where X = infinity. No. That 1 is still a value and no matter how large you make the denominator, the 1 will ALWAYS be there. 5. > See the first digits in each line of numbers? That is the most important one. Period. 1.000… > 0.999… You can’t just ignore what’s after the decimal place and pretend. Math doesn’t work that way. >Arguing that an infitely small number is equal to zero is saying that we do not exist and seconds do not exist. Where the heck do you get that idea? It’s simply not true. > You can always divide a second into something smaller, but will it ever be equal to no time passing at all? Wow– bringing Zeno’s paradox into this. Very impressive. I’m sure the ancient greeks would be impressed. >Also, our bodies are infinitely small when compared to the infinite universe. Does that mean we are 0 and that we don’t exist? Simply put, not true. It genuinely reflects that you DO NOT understand what it means for something to be infinite. Please stop abusing mathematics– learn the definition before you keep using the terms. Windell H. Oskay 6. Anonymous says: (Different Anonymous…) Windell, I want to congratulate you for your persistence in trying to persuade other Anonymous that you are correct, but I fear you are fighting a losing battle. I use this proof quite frequently with my (high school) students, and most of them seem to get the idea without too much effort, even though they want to resist the idea at first. Once they get it, you can see the light in their eyes, seeing that maths can actually be interesting. Makes a change from trying to correct errors like 7/0=0. 10. Anonymous says: When one constructs the real numbers, either from cauchy sequences or using supremums of sets, you actually create equivalence classes. Just as when one constructs the rational numbers. That is why 1/2 = 2/4, it is because they belong to the same equivalence class. Likewise you see when constructing the reals that 1 = .99999… P.S. 1/3 = .333… It is not an approximation. 1. Anonymous says: Think of it as a geometric sequence. From [n = 1 to infinity] (n = integers) the sum of 3 * (1/10)^n For a geometric sequence, the first term divided by (1-[rate of change]) = the sum. So, the first term = 3 / 10 Rate of change = 1/10 3/10/(1-1/10) = 1/3 2. Anonymous says: Thank you, sir. Dedekind cuts are indeed the only answer to this question. 11. alphachapmtl says: A simple proof: x = 0.999… 10x = 9.999… 10x – x = 9.999… – 0.999… 9x = 9 x=1 1. Anonymous says: .9999…. can be represented as an infinite geometric series (9/10 + 9/100 +9/1000…) of the form sum(c*X^n) where c=9 X=(1/10) and n goes from 1 to infinity. the sum of a power series is given by (first term in series)/(1-X) if AbsoluteValue(X) is less than one. in this case 1/10 <1. so we can use the equation to solve for the sum. if we plug the above values into this equation, we get (9/10)/(1-(1/10))=(9/10)/(9/10)=1. so .99999…(going on forever) does equal one. the same steps can be repeated to show that .33333…(going on forever) does equal 1/3 (c=3, X=1/10) 12. Anonymous says: Jesus.. "Arguing" about this is so silly. People who don’t "believe" it don’t grasp that the … means infinite repeating decimals. That simple. They think they’re somehow clever by noticing that 0.333 isn’t 1/3. Well, golly, really? 0.333… isn’t the same thing as 0.333, and there is no number such as "0.000…1". EOD 1. Anonymous says: YES!!!! There is a concept called Density of Real Numbers. It states between 2 real numbers there is an infinite number of reals between them. There are no numbers between 0.999… and 1, so they’re the same number. Same with the hypothetical 0.000…01 and 0. 1. Anonymous says: Actually its Dutch. InfoNu does not accept English articles. And I can’t write any good English anyway. I suggest you try an online-translation programme. They are not perfect, but you should get an impression of what it’s about. If you have any serious comments on the math, I would be glad to hear it. Regards, Bart van Donselaar 1. >Actually its Dutch. So it is! >If you have any serious comments on the math, I would be glad to hear it. I couldn’t translate the whole thing, but it looks like you’re possibly cooking up a non-standard analysis that doesn’t apply to real numbers (and therefore is off-topic to this discussion), or are just blatantly wrong. I’m not sure which. In either case, it certainly is not a valid refutation of the point of our article. Windell H. Oskay 1. Anonymous says: It’s clear to everybody who knows the subject that the real numbers 1 and 0.999… are equal. It’s no use discussing the point. The only reason this discussion about 1 and 0.999… is stil going on, is because many people would rather see 1 and 0.999… to be different. They are in fact looking for alternative interpretations of ‘1’ and ‘0.999…’. Now why not construct a theory or theories in which the signs ‘1’ and ‘0.999…’ have (besides there ordinary meaning) an alternative interpretation which makes them unequal? In that case people who want to delve into the problem can decide for themselves which theory they want to use, and on which occasion. This would probably – at least – result in interesting recreational mathematics. The theory I have used in the article is a precursor of Robinsons nonstandard analysis. It is simpler in that it is based on the commutative ring of infinite sequences of real numbers. No free ultrafilters or the like are used. Two translation-services I have found are: Regards, Bart van Donselaar 1. Unfortunately, it isn’t as clear as you and I would both hope– if you read through the comments above, you’ll find a few folks that argue for a difference in the real number values. I’ll agree that there is a time and a place for advanced and nonstandard analysis, but it’s important that we’re clear to separate that from the more basic issue, which is all that we’re addressing here. Windell H. Oskay 1. Anonymous says: I don’t think the Wikipedia-article ‘0.999…’ as a relatively simple explanation can be essentially improved. If people deny the equality of the real numbers 1 and 0.999… they usually have no idea of what real numbers mathematically are. Real numbers are seen as quantities given by infinite decimal expansions. So they think as long as you are talking in terms of infinite decimal expansions you are in the realm of real numbers. You consequently see a lot of nonsens and sometimes ideas that can be treated rigorously in terms of alternative number systems. I don’t think the battle wil end as long as the denial of the equality of 1 and 0.999… isn’t acknowledged as a theoretical possibility. Regards, Bart van Donselaar 13. Anonymous says: Completely correct 1. Well then, it’s a good thing that mathematics relies on logic and proof, rather than just beliefs! Windell H. Oskay	8,381	32,054	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2024-30	latest	en	0.904953
http://mathhelpforum.com/differential-equations/179769-laplaces-equation-2-dimension-print.html	1,524,229,208,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125937780.9/warc/CC-MAIN-20180420120351-20180420140351-00612.warc.gz	206,571,110	2,996	# Laplace's equation in 2 dimension • May 7th 2011, 05:20 AM leshields Laplace's equation in 2 dimension Hi there, I'm doing some revision for an upcoming test and have all the past papers we have answers for ask us to find separable solutions when dealing with the one dimensional heat equation which I think I can do now but then in other papers we don't have answers for it talks about Laplaces equation in two dimensions which I don't think we even toucjed on in class so other than part a) which I think I can get (elliptic) I'm pretty lost. Any help appreciated!! http://www.eussc.com/images/q3%202009%20cvd.jpg • May 7th 2011, 02:12 PM TheEmptySet Since you are looking for separable solutions assume it has the form. $\displaystyle u(x,y)=X(x)Y(y)$ This gives $\displaystyle u_{yy}=XY'' \quad u_{xx}=X''Y$ This gives $\displaystyle X''Y+XY''=0 \iff \frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$ With $\displaystyle \lambda^2 > 0$ (why?) This gives the system $\displaystyle X''+\lambda^2X=0 \quad Y''-\lambda^2 Y=0$ The boundary conditions gives $\displaystyle X(0)=X(\pi)=0$ The solutions to the X equation are $\displaystyle X =c_1\cos(\lambda x)+c_2\sin(\lambda x)$ This gives $\displaystyle X(0)=c_1=0$ $\displaystyle X =c_2\sin(\lambda x)$ $\displaystyle X(\pi)=0 =c_2\sin(\lambda \pi) \iff n\pi =\lambda \pi \implies \lambda = n$ with $\displaystyle n \in \mathbb{Z}^+$ $\displaystyle X(x)=\sin(n x)$ Now if you solve the Y equation you get $\displaystyle Y(y)=d_1\cosh(ny)+d_2\cosh(ny)$ This gives $\displaystyle u(x,y)=X(x)Y(y)=\sin(n x)( d_1\cosh(ny)+d_2\cosh(ny))$ For C just use this formula For D just mimic the process done in b)	553	1,673	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2018-17	latest	en	0.80577
https://www.hackmath.net/en/math-problem/5679	1,610,999,339,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00738.warc.gz	804,240,169	12,265	How many 1/4 cup servings are in 2 and 1/3 cups of lemonade? Correct result: n = 9 #### Solution: We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. Do you want to round the number? ## Next similar math problems: • Cups We have three cups. In the cups we had fluid and boredom we started to shed. 1 We shed one-third of the fluid from the second glass into the first and third. 2 Then we shed one quarter cup of liquid from the first to the second and to the third. 3 Then we • Caleb Caleb is making lemonade for a party. He has 5 gallons of lemonade. He is putting 1/3 cubic inch of lemonade in a cup for each guest. How many guests are going to be at the party? • Berry Smoothie Rory has 5/8 cup of milk. How much milk does she have left after she doubles the recipe of the smoothie? Berry Smoothie: 2 cups strawberries 1 cup blueberries 1/4 cup milk 1 tbsp (tablespoon) sugar 1/2 tsp (teaspoon) lemon juice 1/8 tsp (teaspoon) vanilla • Almonds Rudi has 4 cups of almonds. His trail mix recipe calls for 2/3 cup of almonds. How many batches of trail mix can he make? • Cups of punch Cyka made 6 19/20 cups of punch punch at two different types of juice in it. If the punch had 4 1/5 cups of one type of juice how many cups of the other type of juice did it have? • Numbers Determine the number of all positive integers less than 4183444 if each is divisible by 29, 7, 17. What is its sum? • Jill's Jill's mom stored some of the honey in a container that held 3/4 of a gallon. She used half of this amount to sweeten tea. How much honey, in cups was used in the tea. Express your answers in cups. • How many How many integers are greater than 547/3 and less than 931/4? • Quotient Determine the quotient (q) and the remainder (r) from division numbers 100 and 8. Take the test of correctness. • What is 11 What is the quotient of Three-fifths and 1 Over 10? • Remainder What is the remainder of the division of natural numbers 293 and 7? • Numbers division With what number should be divided mixed number 2 3/4 to get 11/12? • Barrels Peter has 42 barrels. One-sixth of them are filled with lemonade, one third is filled with wine and half is empty. Calculate how many barrels are with each filling. • Rounding Double round number 727, first to tens, then to hundreds. (double rounding) • Fractions 4 How many 2/3s are in 6? • Coffee Coffee from the machine in the cup cost 28 cents. Coffee is 20 cents more expensive than the cup. How much is the cup? • Brownies Mrs. Merritt made brownies, the recipe had to be cut in 1/2. The recipe called for 2 5/8 cups of sugar. How much sugar did she use?	767	2,801	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2021-04	longest	en	0.949696
http://sigmapedia.com/includes/term.cfm?word_id=1994&lang=ENG	1,571,521,677,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00489.warc.gz	142,824,466	6,308	Hypothesis Test Go Back Definition A statistical technique used to help disprove or reject a particular conjecture about a process or population based on evidence from the sample data. This conjecture or hypothesis usually assumes a stand of status quo or “no difference”, called the null hypothesis. The statement held to be true if the null is false is called the alternative hypothesis. Rejection of the null hypothesis results in concluding in favor of the alternative hypothesis. Application All hypothesis tests share these basic steps: 1. Select a significance level which will be used to decide the test. This must be done up front so as not to bias the test results. The significance level is the probability of committing a Type I error, i.e., the probability of rejecting the null hypothesis when it is in fact true. It is considered the more serious type of error, the other being Type II error, or failing to reject the null when it is fact false. 2. State the null and alternative hypotheses in terms of the population parameters. E.g. The means of the two populations under study are equal, or H0: µ1 = µ2 against the alternative that the means differ significantly, H1: µ1 ? µ2 3. Calculate the test statistic, chosen based on the hypotheses and test under consideration. This test statistic will form the decision criterion for rejecting the null hypothesis. 4. Obtain the p-value corresponding to the test statistic under the null hypothesis. This is the probability of observing a value as or more extreme than the test statistic assuming the null hypothesis is true. 5. Decide the test by either comparing the test statistic to a cut-off value prescribed by the significance level chosen in step 1, or by comparing the p-value to the significance level. Reject the null hypothesis if the test statistic falls in the rejection region defined by the cut-off value(s) or if the p-value is smaller than the significance level.	389	1,952	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.28125	3	CC-MAIN-2019-43	latest	en	0.883734
https://questioncove.com/updates/54357ae3e4b05d51093ea502	1,653,286,994,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00738.warc.gz	542,063,343	5,213	Mathematics OpenStudy (anonymous): negative 10 equals x minus 21 OpenStudy (studygurl14): -10 = x -21 Add 21 to both sides 11 = x OpenStudy (anonymous): wouldnt it be negative 11? OpenStudy (studygurl14): No it wouldn't because -10 + 21 = 11 OpenStudy (anonymous): ohh, no its positive i see it, thanks OpenStudy (studygurl14): you're welcome :)	107	356	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.28125	3	CC-MAIN-2022-21	latest	en	0.868814
https://byjusexamprep.com/engineering-mechanics--design-of-steel-structures-tension-members-i	1,720,885,282,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00144.warc.gz	119,668,517	70,267	# Engineering Mechanics & Design of Steel Structures: Tension Members By Ashutosh Yadav\|Updated : August 15th, 2021 BYJU'S Exam Prep Brings you 60 Days Study Plan for the preparation of DFCCIL Civil Engineering. This Study Plan will be free and will be very beneficial for the students preparing and targeting the DFCCIL Exam. Save this article as it will get updated on a daily basis as scheduled. ## Tension Member Tension members are linear members in which axial forces act so as to elongate (stretch) the member. A rope, for example, is a tension member. Tension members carry loads most efficiently since the entire cross section is subjected to uniform stress. Unlike compression members, they do not fail by buckling. Ties of trusses, suspenders of cable stayed and suspension bridges, suspenders of buildings systems hung from a central core (such buildings are used in earthquake prone zones as a way of minimizing inertia forces on the structure), and sag rods of roof purlins are other examples of tension members. #### Tesnsion Members in Structures Cross Sections of Tension Members ### Introduction 1. Tension member has no stability problem. 2. In tension, member net section will be effective whereas in compression member gross section is effective. #### Net Sectional Area (i) For plate Net area = (b x t) – nd't where, s1 = Distance between two consecutive rivets in the direction of load, also called pitch. g1 = Distance between two consecutive rivets perpendicular to the direction of load also called gauge. b = Width of the plate n = Number rivets at the section T = Thickness of the plate d' = Gross diameter of the rivet (ii) Single angle connected by one leg only. (a) where, A1 = Net cross-section of area of the connected leg. A2 = Gross cross-sectional area of unconnected leg. (out stand) (b) (c) (d) (e) (iii) For pair of angle placed back to back (or a signal tee) connected by only one leg of each angle (or by the flange of a tee) to the same side of a gusset plate: or it the two angles are tagged along a-a. where, A1 = Area of connected leg A2 = Area of outstand (unconnected leg) (c) The area of a web of tee = Thickness of web x (depth – thickness of flange) (d) The outstand legs of the pair of angles should be tacked by rivets of a pitch not exceeding 1 m. (iv) If two angles are places back to back and connected to both sides of the gusset plate. Then when tack riveted. If not tack riveted then both will be considered separately and case (ii) will be followed #### Permissible Stress in Design • The direct stress in axial tension on the effective net area should not exceed σat where σat = 0.6fy and fy = minimum yield stress of steel in MPa #### Lug Angle The lug angle is a short length of an angle section used at a joint to connect the outstanding leg of a member, thereby reducing the length of the joint. When lug angle is used k = 1 You can avail of Online Classroom Program for all AE & JE Exams: ## Online Classroom Program for AE & JE Exams (12+ Structured LIVE Courses and 160+ Mock Tests) You can avail of BYJU'S Exam Prep Test Series specially designed for all AE & JE Exams: ## BYJU'S Exam Prep Test Series AE & JE (160+ Mock Tests) Thanks, Team BYJU'S Exam Prep Sahi Prep Hai to Life Set Hai !!! Download BYJU'S Exam Prep APP , for best Exam Preparation , Free Mock Test, Live Classes write a comment ### AE & JE Exams AE & JEAAINBCCUP PoliceRRB JESSC JEAPPSCMPPSCBPSC AEUKPSC JECGPSCUPPSCRVUNLUPSSSCSDEPSPCLPPSCGPSCTNPSCDFCCILUPRVUNLPSPCLRSMSSB JEOthersPracticeMock TestCourse ### Featured Articles Follow us for latest updates GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 bepstudentsupport@byjus.com	941	3,780	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2024-30	latest	en	0.935059
https://socratic.org/questions/how-do-you-solve-t-3r-m-n-for-r#173425	1,721,878,858,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00785.warc.gz	446,305,478	6,234	# How do you solve T=(3R)/(M-N) for R? Oct 1, 2015 Remember $\frac{a}{b} = \frac{c}{d} \to a d = b c$ (cross-product) #### Explanation: $\frac{T}{1} = \frac{3 R}{M - N} \to T \cdot \left(M - N\right) = 1 \cdot 3 R \to$ $3 R = T \left(M - N\right) \to R = \frac{1}{3} T \left(M - N\right) = \frac{T \left(M - N\right)}{3}$ Oct 1, 2015 $R = \frac{T M - T N}{3}$ #### Explanation: First of you need to multiply both sides by $M - N$; $T \left(M - N\right) = \frac{3 R}{M - N} \left(M - N\right) = T M - T N = 3 R$ Then you need to divide both sides by three; $\frac{T M - T N}{3} = \frac{3 R}{3} = \frac{T M - T N}{3} = R$ Hope that helps :)	274	648	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.4375	4	CC-MAIN-2024-30	latest	en	0.639267
https://en.wikipedia.org/wiki/Trans_Lunar_Injection	1,558,650,240,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00239.warc.gz	459,195,572	14,802	# Trans-lunar injection (Redirected from Trans Lunar Injection) Fig. 1: Lunar transfer, perspective view. TLI occurs at the red dot near Earth. A trans-lunar injection (TLI) is a propulsive maneuver used to set a spacecraft on a trajectory that will cause it to arrive at the Moon. Typical lunar transfer trajectories approximate Hohmann transfers, although low-energy transfers have also been used in some cases, as with the Hiten probe.[1] For short duration missions without significant perturbations from sources outside the Earth-Moon system, a fast Hohmann transfer is typically more practical. A spacecraft performs TLI to begin a lunar transfer from a low circular parking orbit around Earth. The large TLI burn, usually performed by a chemical rocket engine, increases the spacecraft's velocity, changing its orbit from a circular low Earth orbit to a highly eccentric orbit. As the spacecraft begins coasting on the lunar transfer arc, its trajectory approximates an elliptical orbit about the Earth with an apogee near to the radius of the Moon's orbit. The TLI burn is sized and timed to precisely target the Moon as it revolves around the Earth. The burn is timed so that the spacecraft nears apogee as the Moon approaches. Finally, the spacecraft enters the Moon's sphere of influence, making a hyperbolic lunar swingby. ## Modeling Artist's concept of NASA's Constellation stack performing the trans-lunar injection burn ### Patched conics TLI targeting and lunar transfers are a specific application of the n body problem, which may be approximated in various ways. The simplest way to explore lunar transfer trajectories is by the method of patched conics. The spacecraft is assumed to accelerate only under classical 2 body dynamics, being dominated by the Earth until it reaches the Moon's sphere of influence. Motion in a patched-conic system is deterministic and simple to calculate, lending itself for rough mission design and "back of the envelope" studies. ### Restricted circular three body (RC3B) More realistically, however, the spacecraft is subject to gravitational forces from many bodies. Gravitation from Earth and Moon dominate the spacecraft's acceleration, and since the spacecraft's own mass is negligible in comparison, the spacecraft's trajectory may be better approximated as a restricted three-body problem. This model is a closer approximation but lacks an analytic solution,[2] requiring numerical calculation via methods such as Runge-Kutta.[3] ### Further accuracy More detailed simulation involves modeling the Moon's true orbital motion; gravitation from other astronomical bodies; the non-uniformity of the Earth's and Moon's gravity; including solar radiation pressure; and so on. Propagating spacecraft motion in such a model is numerically intensive, but necessary for true mission accuracy. ### Free return In some cases it is possible to design a TLI to target a free return trajectory, so that the spacecraft will loop around behind the Moon and return to Earth without need for further propulsive maneuvers.[4] Such free return trajectories add a margin of safety to human spaceflight missions, since the spacecraft will return to Earth "for free" after the initial TLI burn. ## History The first space probe to successfully perform TLI was the Soviet Union's Luna 1 on January 2, 1959. The first human-crewed mission to successfully perform this procedure, and thus becoming the first humans to leave the Earth's influence, was Apollo 8 on December 21, 1968. For the Apollo lunar missions, the restartable J-2 engine in the third (S-IVB) stage of the Saturn V rocket performed TLI. This particular TLI burn lasted approximately 350 seconds, providing 3.05 to 3.25 km/s (10,000 to 10,600 ft/s) of delta-v, at which point the spacecraft was traveling at approximately 10.4 km/s (34150 ft/s) relative to the Earth.[5] The Apollo 8 TLI was spectacularly observed from the Hawaiian Islands in the pre-dawn sky south of Waikiki, photographed and reported in the papers the next day.[6] In 1969, the Apollo 10 pre-dawn TLI was visible from Cloncurry, Australia.[7] It was described as resembling car headlights coming over a hill in fog, with the spacecraft appearing as a bright comet with a greenish tinge.[7]	913	4,276	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2019-22	latest	en	0.915341
https://juliaobserver.com/packages/LatticeSites	1,600,553,756,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00155.warc.gz	494,831,983	8,173	0 2 1 2 # LatticeSites.jl Type for different kind of sites on different lattices. ## Installation ``````pkg> add https://github.com/Roger-luo/LatticeSites.jl.git `````` ## Intro This package provides types for sites, which defines the configuration of a lattice. Binary configuration label is provided as: • `Bit`, refers to `0`/`1` • `Spin`, refers to `-1`/`+1` • `Half`, refers to `-0.5`/`+0.5` • `Clock`, refers to the 2D q-state clock model with `q` discrete spin values (`1:q`) • `Potts`, refers to the standard Potts model with values `-q, ..., q` • `Continuous`, is in development still and not ready. `Array`, `StaticArray` and etc. (e.g `SparseArray`) is supported for store configurations. It is simple, and you can use it like a `Number` (but it is not a `Number`) ``````julia> rand(Bit{Float64}) 0.0 julia> rand(Bit{Float64}, 2, 2) 2×2 Array{Bit{Float64},2}: 0.0 0.0 0.0 0.0 julia> using StaticArrays julia> rand(SMatrix{2, 2, Bit{Int}}) 2×2 SArray{Tuple{2,2},Bit{Int64},2,4}: 0 1 0 1 `````` and to support indexing, you can convert any `AbstractArray` contains sites to an integer (as long as this integer type does not overflow). ``````julia> convert(Int, rand(SMatrix{2, 2, Bit{Int}})) 12 `````` There is also a `HilbertSpace` iterator that help you iterate through the space. ``````julia> space = HilbertSpace{Bit{Int}}(2, 2) HilbertSpace{Bit{Int64},Tuple{2,2},2,4}(Bit{Int64}[0 0; 0 0]) julia> collect(space) 16-element Array{Array{Bit{Int64},2},1}: [0 0; 0 0] [1 0; 0 0] [0 0; 1 0] [1 0; 1 0] [0 1; 0 0] [1 1; 0 0] [0 1; 1 0] [1 1; 1 0] [0 0; 0 1] [1 0; 0 1] [0 0; 1 1] [1 0; 1 1] [0 1; 0 1] [1 1; 0 1] [0 1; 1 1] [1 1; 1 1] `````` We use the convention that the first index `a[1]` take the first digit position during the convention, which is opposite to natural notation `0b0101`, where the last digit in bit string take the first position. In short `0b011` is equivalent to `Bit[1, 1, 0]` 10/03/2018 7 months ago 44 commits	754	1,976	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2020-40	longest	en	0.727261
https://www.nagwa.com/en/videos/308158405078/	1,618,407,004,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00456.warc.gz	994,142,455	8,734	# Question Video: Discussing the Monotonicity of a Graphed Function Mathematics Which of the following statements correctly describe the monotony of the function represented in the figure below? [A] The function is increasing on (5, 8), constant on (−1, 5), and decreasing on (−2, −1). [B] The function is increasing on (−2, −1), constant on (−1, 5), and decreasing on (5, 8). [C] The function is increasing on (5, 8) and decreasing on (−2, 5). [D] The function is increasing on (−2, 5) and decreasing on (5, 8). 03:05 ### Video Transcript Which of the following statements correctly describe the monotony of the function represented in the figure below? Is it (A) the function is increasing on the open interval five to eight, constant on the open interval negative one to five, and decreasing on the open interval negative two to negative one? Is it (B) the function is increasing on the open interval negative two to negative one, constant on the open interval negative one to five, and decreasing on the open interval five to eight? Is it (C) the function is increasing on the open interval five to eight and decreasing on the open interval negative two to five? Or (D) the function is increasing on the open interval negative two to five and decreasing on the open interval five to eight. So by reading the question, we’ve probably inferred what we mean by the monotony of a function. The monotony of a function simply tells us if the function is increasing or decreasing. And of course, we recall that if a function is increasing over some interval, it has a positive slope. If it’s decreasing, it has a negative slope. And if it’s constant, well, that’s a horizontal line. So let’s look at the graph of our function. We see it has three main sections. The first section is between negative two and negative one. Then the next section is between negative one and five, whilst the third section is between five and eight. So let’s consider each section in turn. We can see that the slope of the first part of our function must be positive. It’s sloping upwards. We then have a horizontal line between 𝑥 equals negative one and five. And the third part of our graph has a negative slope. It’s sloping downwards. Our function is therefore increasing for sometime, it’s constant, and then finally it’s decreasing. We need to decide the intervals over which each of these occur. It has a positive slope between 𝑥 equals negative two and negative one. And so we define this using the open interval negative two to negative one. We are not going to use a closed interval. We don’t really know what’s happening at the endpoints of this interval. For instance, when 𝑥 is equal to negative one, the graph of our function has this sort of sharp corner. And so we’re going to leave 𝑥 equals negative two and 𝑥 equals negative one out of our interval. In a similar way, the function is constant over the open interval negative one to five. And it’s decreasing over the open interval five to eight. Once again, we don’t know what’s really happening at those endpoints, but we do have sharp corners. And so we can’t say whether it’s increasing, decreasing, or constant. And so the correct answer is (B): the function is increasing on the open interval negative two to negative one, constant on the open interval from negative one to five, and decreasing on the open interval five to eight.	762	3,387	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.65625	5	CC-MAIN-2021-17	latest	en	0.905003
https://www.solvedlib.com/divide-5-12-divied-by-4-35-answer-in-simplest-form,784	1,695,872,590,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00117.warc.gz	1,079,073,519	16,390	# Divide 5 1/2 divied by 4 3/5 answer in simplest form ###### Question: divide 5 1/2 divied by 4 3/5 answer in simplest form. does simplest form mean a mixed number or decimal? I got an answer of 1 9/46 or 1.195652... #### Similar Solved Questions ##### Evaluate the given integral by changing t0 polar coordinates Jf (x? +y' )dA , where D is the region bounded by the circle x? +y = 4,y=0,and lines y=X , y=0 (5 points) Set up the integral in polar coordinates. points) Evaluate the integral Evaluate the given integral by changing t0 polar coordinates Jf (x? +y' )dA , where D is the region bounded by the circle x? +y = 4,y=0,and lines y=X , y=0 (5 points) Set up the integral in polar coordinates. points) Evaluate the integral... ##### Review Topics] Draw the structure(s) of the major organic product(s) of the following reaction. 1. Dioxane... Review Topics] Draw the structure(s) of the major organic product(s) of the following reaction. 1. Dioxane / reflux 2. aqueous HCI . You do not have to consider stereochemistry . You do not have to explicitly draw H atoms. . Do not include lone pairs in your answer. They will not be considered in th... ##### A particle moves along the x-axis so that its velocity at time t is given by Hdtll 8r3 + 2 At time t = 0, the initial position of the particle is x = 5.Find (he acceleration of the particle at time t = 5b. Is the particle speeding up or slowing down at time / = 5.1? Justify your answer. Find all times on the interval 0 < ( < 5, that the particle is at rest. Write an expression involving an integral that gives the position x(t) Find the position of the particle at ( = 2.Calculate the tota A particle moves along the x-axis so that its velocity at time t is given by Hdtll 8r3 + 2 At time t = 0, the initial position of the particle is x = 5. Find (he acceleration of the particle at time t = 5 b. Is the particle speeding up or slowing down at time / = 5.1? Justify your answer. Find all t... ##### 9. Show that Mgo has the sodium chloride crystal structure and calculate the density of Mgo.... 9. Show that Mgo has the sodium chloride crystal structure and calculate the density of Mgo. Mg+2 = 0.066 nm and ro-2 = 0.132 nm, The atomic masses are 24.312 and 16 g/mol for magnesium and oxygen, respectively. (10%)... ##### BALANCE SHEET ANALYSIS Complete the balance sheet and sales information using the following financial data: Total... BALANCE SHEET ANALYSIS Complete the balance sheet and sales information using the following financial data: Total assets turnover: 1.5x Days sales outstanding: 33.5 daysa Inventory turnover ratio: 5x Fixed assets turnover: 3x Current ratio: 2.5x Gross profit margin on sales: (Sales - Cost of goods ... ##### A carbon ion with only one electron is in the n=3 state. (a) What are all... A carbon ion with only one electron is in the n=3 state. (a) What are all the possible values of the quantum numbers of the electron in the n=3 state? (b) If the electron decays to the ground state, what will be the energy and wavelength of the emitted photon? (c) What is the binding energy of the e... ##### Suppose you just received a shipment of 6 televisions to of the televisions are defective if... Suppose you just received a shipment of 6 televisions to of the televisions are defective if to to hella visions are randomly selected compute the probability that both televisions work what is a probability at least one of the televisions does not work... ##### 4iz51,0 min, 34.0% of & compound has decomposed What is the half-life of this reaction assuming first-ordcr kinctics?t1n 4iz51,0 min, 34.0% of & compound has decomposed What is the half-life of this reaction assuming first-ordcr kinctics? t1n... ##### Add the following: 3 + (12) + ( ~ 48) + + ( = 12288) equals Add the following: 3 + (12) + ( ~ 48) + + ( = 12288) equals... ##### Let Xi,Xz'_ XJL be i.i.d from some distribution with E[X ] = 0 e R unknown and VIX ] eR+ assumed known. Argue a)% lower bound large-sample confidence interva for can be given by(exp(x-2.4) Where Za represents the 1 - percentile of the N(0,1) distribution: Let Xi,Xz'_ XJL be i.i.d from some distribution with E[X ] = 0 e R unknown and VIX ] eR+ assumed known. Argue a)% lower bound large-sample confidence interva for can be given by (exp(x-2.4) Where Za represents the 1 - percentile of the N(0,1) distribution:... ##### Find b (real) to make the system ð‘¯(ð’”) = ðŸ/(ð’”ðŸ + ð’ƒð’” + ðŸ)stable. Find b (real) to make the system ð‘¯(ð’”) = ðŸ/(ð’”ðŸ + ð’ƒð’” + ðŸ) stable.... ##### For 0 <= x <= 360^@, how would you find the coordinates of the points of intersection with the coordinate axes of y = sin(x - 45^@)? For 0 <= x <= 360^@, how would you find the coordinates of the points of intersection with the coordinate axes of y = sin(x - 45^@)?... ##### For all integers and if albc then alb or alc For all integers and b, ifakb then a\|b. 30. For all integers and n, ifa/n? and then For all integers and if albc then alb or alc For all integers and b, ifakb then a\|b. 30. For all integers and n, ifa/n? and then... Id â‚¬ â‚¬8 (p Id zzz (4 4d L 99 ( Id â‚¬ ES] (1 ST J142p SIY} JO 2 voueuoedeo J41 $J1n31[ 341 LI LALOYS SB W 70 â‚¬ SQpBI JO q1a1ds Suuonpuoj I?IIBLS & YILM JIILUDJUO? UI â‚¬" 0 q snipex JO [[a4S Sunonpuov [BQuayds B Jo SISISuO? JouoBdeo [Bouayds V... 5 answers ##### Compare the thee data sets m the nght:8888: J0 1121314151610 M1213141516412IAAISGWiich data set has the least samp e standard deviation?Data set (), because ts more entnes that are close Ihe mean 0 B Dala sel (Ii). because has more entnes that are fanther away from the mean 0c Data set (), because has less entres that are tatther Dway fror Ihe mean() How are [le data sets thc same? How do they differ? The Jhee dala sets have the same Mean median and mode but have diflerent standard deviations 0 Compare the thee data sets m the nght: 8888: J0 11213141516 10 M1213141516 412IAAISG Wiich data set has the least samp e standard deviation? Data set (), because ts more entnes that are close Ihe mean 0 B Dala sel (Ii). because has more entnes that are fanther away from the mean 0c Data set (), beca... 1 answer ##### Question 6 Perform the operation and leave the result in trigonometric form, cos n+isin 7 cos... Question 6 Perform the operation and leave the result in trigonometric form, cos n+isin 7 cos n/3)+isina 7/3) 12... 1 answer ##### Inlet = 81,8 of Ou Het=76.805 79,50 Bulb Temp, a Whats the percentage of efficiency? Inlet = 81,8 of Ou Het=76.805 79,50 Bulb Temp, a Whats the percentage of efficiency?... 1 answer ##### The - a magnetic field intensity in the - region xxo, assumed to be free space,... The - a magnetic field intensity in the - region xxo, assumed to be free space, is given Ai = 4 årt sây+ 328 Alm. In the region xco, assumed to be occupied by material with relative permeability M, the magnetic field intensity is Ar= 2ax +3ây +4 az Alm. Determine My and the surface... 5 answers ##### UW I I0 8 8 Jil 1 1 1 1 { 2 1 1 3 3 8 1 5 Li F 1 3 L W 1 1 02 1 8 3 1 1 1 8 2 1 1 6 3 8 V I 3 3 3 1 8 U 01 1 UW I I0 8 8 Jil 1 1 1 1 { 2 1 1 3 3 8 1 5 Li F 1 3 L W 1 1 02 1 8 3 1 1 1 8 2 1 1 6 3 8 V I 3 3 3 1 8 U 0 1 1... 5 answers ##### [1 point-\| Prote that the stquence {Ia(1)}$ -1 defined by fa(1) =r conTerges unitornly Zero [unC [OM [1 point-\| Prote that the stquence {Ia(1)}` -1 defined by fa(1) =r conTerges unitornly Zero [unC [OM...	2,322	7,417	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2023-40	longest	en	0.852781
https://www.physics-world.com/ap-physics-1/4-circular-motion/	1,555,856,089,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578531984.10/warc/CC-MAIN-20190421140100-20190421162100-00143.warc.gz	795,569,848	12,747	# Uniform Circular Motion In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body. Examples of circular motion include: an artificial satellite orbiting the Earth at the constant height, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism. Since the object’s velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according to Newton’s laws of motion. Topical Notes, Problems, Presentations, Quiz, Test, Investigations and Videos Uniform Circular Motion Centripetal Acceleration and Centripetal Force Banked Curves Satellites in Circular Orbits Test your Understanding: Chapter 3 MCQ Quiz 1 Here Take Chapter 3 ReQuiz MCQ Quiz 2 Here In physics, uniform circular motion describes the motion of a body traversing a circular path at constant speed. Since the body describes circular motion, its distance from the axis of rotation remains constant at all times. Though the body’s speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body’s speed and its direction of travel. This changing velocity indicates the presence of an acceleration; this centripetal acceleration is of constant magnitude and directed at all times towards the axis of rotation. This acceleration is, in turn, produced by a centripetal force which is also constant in magnitude and directed towards the axis of rotation. In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis. error: Content is protected !!	470	2,467	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4	4	CC-MAIN-2019-18	latest	en	0.88071
https://msdn.microsoft.com/en-us/library/vstudio/wyk4d9cy(v=vs.100)	1,436,165,538,000,000,000	text/html	crawl-data/CC-MAIN-2015-27/segments/1435375098059.60/warc/CC-MAIN-20150627031818-00008-ip-10-179-60-89.ec2.internal.warc.gz	1,034,126,002	11,978	# Math.Round Method (Double) .NET Framework 4 Rounds a double-precision floating-point value to the nearest integral value. Namespace: System Assembly: mscorlib (in mscorlib.dll) ## Syntax ```public static double Round( double a ) ``` #### Parameters a Type: System.Double A double-precision floating-point number to be rounded. #### Return Value Type: System.Double The integer nearest a. If the fractional component of a is halfway between two integers, one of which is even and the other odd, then the even number is returned. Note that this method returns a Double instead of an integral type. ## Remarks The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding to nearest, or banker's rounding. It minimizes rounding errors that result from consistently rounding a midpoint value in a single direction. To control the type of rounding used by the Round method, call the Math.Round(Double, MidpointRounding) overload. If the value of a is Double.NaN, the method returns Double.NaN. If the value of a is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively. Notes to Callers Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double) method may not appear to round midpoint values to the nearest even integer. In the following example, because the floating-point value .1 has no finite binary representation, the first call to the Round(Double) method with a value of 11.5 returns 11 instead of 12. ``` using System; public class Example { public static void Main() { double value = 11.1; for (int ctr = 0; ctr <= 5; ctr++) Console.WriteLine(); value = 11.5; } { Console.WriteLine("{0} --> {1}", value, Math.Round(value)); return value + .1; } } // The example displays the following output: // 11.1 --> 11 // 11.2 --> 11 // 11.3 --> 11 // 11.4 --> 11 // 11.5 --> 11 // 11.6 --> 12 // // 11.5 --> 12 ``` ## Examples The following example demonstrates rounding to the nearest integer value. ``` using System; class Program { static void Main() { Console.WriteLine("Classic Math.Round in CSharp"); Console.WriteLine(Math.Round(4.4)); // 4 Console.WriteLine(Math.Round(4.5)); // 4 Console.WriteLine(Math.Round(4.6)); // 5 Console.WriteLine(Math.Round(5.5)); // 6 } } ``` The following example uses Round to assist in the computation of the inner angles of a given trapezoid. ``` /// <summary> /// The following class represents simple functionality of the trapezoid. /// </summary> class MathTrapezoidSample { private double m_longBase; private double m_shortBase; private double m_leftLeg; private double m_rightLeg; public MathTrapezoidSample(double longbase, double shortbase, double leftLeg, double rightLeg) { m_longBase = Math.Abs(longbase); m_shortBase = Math.Abs(shortbase); m_leftLeg = Math.Abs(leftLeg); m_rightLeg = Math.Abs(rightLeg); } private double GetRightSmallBase() { return (Math.Pow(m_rightLeg,2.0) - Math.Pow(m_leftLeg,2.0) + Math.Pow(m_longBase,2.0) + Math.Pow(m_shortBase,2.0) - 2* m_shortBase * m_longBase)/ (2(m_longBase - m_shortBase)); } public double GetHeight() { double x = GetRightSmallBase(); return Math.Sqrt(Math.Pow(m_rightLeg,2.0) - Math.Pow(x,2.0)); } public double GetSquare() { return GetHeight() m_longBase / 2.0; } { double sinX = GetHeight()/m_leftLeg; return Math.Round(Math.Asin(sinX),2); } { double x = GetRightSmallBase(); double cosX = (Math.Pow(m_rightLeg,2.0) + Math.Pow(x,2.0) - Math.Pow(GetHeight(),2.0))/(2xm_rightLeg); return Math.Round(Math.Acos(cosX),2); } public double GetLeftBaseDegreeAngle() { double x = GetLeftBaseRadianAngle() * 180/ Math.PI; return Math.Round(x,2); } public double GetRightBaseDegreeAngle() { double x = GetRightBaseRadianAngle() * 180/ Math.PI; return Math.Round(x,2); } static void Main(string[] args) { MathTrapezoidSample trpz = new MathTrapezoidSample(20.0, 10.0, 8.0, 6.0); Console.WriteLine("The trapezoid's bases are 20.0 and 10.0, the trapezoid's legs are 8.0 and 6.0"); double h = trpz.GetHeight(); Console.WriteLine("Trapezoid height is: " + h.ToString()); Console.WriteLine("Trapezoid left base angle is: " + dxR.ToString() + " Radians"); Console.WriteLine("Trapezoid right base angle is: " + dyR.ToString() + " Radians"); double dxD = trpz.GetLeftBaseDegreeAngle(); Console.WriteLine("Trapezoid left base angle is: " + dxD.ToString() + " Degrees"); double dyD = trpz.GetRightBaseDegreeAngle(); Console.WriteLine("Trapezoid left base angle is: " + dyD.ToString() + " Degrees"); } } ``` ## Version Information #### .NET Framework Supported in: 4, 3.5, 3.0, 2.0, 1.1, 1.0 #### .NET Framework Client Profile Supported in: 4, 3.5 SP1 #### Portable Class Library Supported in: Portable Class Library ## Platforms Windows 7, Windows Vista SP1 or later, Windows XP SP3, Windows XP SP2 x64 Edition, Windows Server 2008 (Server Core not supported), Windows Server 2008 R2 (Server Core supported with SP1 or later), Windows Server 2003 SP2 The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.	1,420	5,350	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.90625	3	CC-MAIN-2015-27	latest	en	0.756257
https://completesuccess.in/index.php/2017/03/24/ratio-proportion-5/	1,653,482,021,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00616.warc.gz	232,959,739	22,324	# Ratio & Proportion (5) 1. Sobha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents? 1. 6 years 2. 5 years 3. 4 years 4. 3 years Explanation :- Lets suppose Sobha’s age= x her brother’s age = x – 4 Sobha’s father’s age = x + 38 Sobha’s mother’s age = (x – 4) + 36 = x + 32 Sobha’s father’s age – Sobha’s mother’s age = (x + 38) – (x + 32) = 6 years 2. The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. What is the ratio of their present ages? 1. 7 : 3 2. 3 : 7 3. 9 : 4 5. 4 : 9 Explanation :- Let the age of son 10 years ago = x years Therefore Age of father 10 years ago = 3 x years According to question , (3x + 10) + 10 = 2 [(x + 10) + 10] 3x + 20 = 2x + 40 x = 20. Ratio of Present age of father and son = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3. 3. The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person? 1. 10 2. 20 3. 30 4. 40 Explanation :- Lets suppose the age of younger person = x years, Therefore, Elder person’s age = (x +16) Before , 6 years 3(x-6) = (x+16-6) 3x-18 = x+10 x = 14. So, Elder person’s age = x + 16 = 30 4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father’s age will be 10 years more than twice the age of the son. What is father’s present age? 1. 30 years 2. 31 years 3. 32 yeas 4. 33 years Explanation :- Let the Son’s peresent Age = x. Therefore, Father’s Present age = 3x + 3 After three years, Son’s Age = (x+3) & Father’s Present age = (3x + 3 + 3) = (3x + 6) Therefore , According to question, (3x + 6) = 10 + 2(x + 3) 3x + 6 = 10 + 2x + 6 x = 10 Father’s Present age = 3x + 3 = 310 + 3 = 33 5. Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal. 1. 40 years 2. 38 years 3. 42 years 4. 36 years Explanation :- Let Son’s age 8 years ago = x years Therefore, Kamal’s age 8 years ago = 4x After 8 years, Son’s age = (x + 8 + 8 ) = (x +16) years Kamal’s age = 4x + 8 + 8 = 4x +16 According to Question 416 (16) x = 8 Kamal’s Present Age = 4x + 8 = (4 8) + 8 = 40 years	901	2,383	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5	4	CC-MAIN-2022-21	latest	en	0.950709
https://doctormyessay.com/2022/04/24/1b-statistics-homework-help/	1,669,651,772,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00856.warc.gz	249,271,227	20,358	# 1b \| Statistics homework help Directions: For each of the following, identify the method used to collect the data (census, systematic, convenience, voluntary response, cluster, stratified, or simple random). Explain why you chose your answer and if the method will represent the population of interest or not? 1. The admissions department at a college wants to see how many of their students would be in favor of using a new program to register for classes. They put a link on their website so that any students that want to try out the program can. The students can then take a survey and say how well they like the new system. 2. Rick works for a sports equipment manufacturing company. He wants to compare the opinion of his older employees to the new employees. To do this, he separates all the employees into two groups, employees that have been with company five or more years and those that have been with the company less than five years. He then chooses 12 of his most trusted older employees and 16 new employees that have proven themselves and ask what they think about changing the medical insurance coverage. 3. Michelle, a teacher at a local high school, wants to see how many students at her high school will be attending community college. She gives the students in her one section of advanced placement U.S. History a questionnaire to fill out that asks where they will be attending college. 4. Jamie is working at the Republican recruiting committee in her city. She is curious how many people that live in her city will vote for the Republican candidate in the next election. She uses a computer to randomly select phone numbers in her city. She then calls those phone numbers to ask people about their voting preferences. 5. Rachael works at the Democrat recruiting center in her hometown. To determine what percent of people will vote for the Democratic candidate, she obtains a list of all residents in her town and decides to ask every 40th person on the list. 6. Laya is passionate about bringing an NFL football team to her city. She needs to take an opinion poll about how people in her city would feel about raising taxes in order to build a stadium for a professional football team. She randomly selects 75 streets in her city and asks every person living on those streets. 7. Micah is the CEO of large software development company. He wants to see if his employees have any ideas about areas of software development that the company should pursue. He has every single employee in his company fill out a questionnaire outlining his or her ideas. He gives the employees a stipend on their paycheck to pay them for their time it took to fill out the questionnaire. 8. Tara wants to collect data on people living in Portland Oregon. She wants to know how many cups a coffee they drink per day. She went to a few supermarkets close to her house and asked people as they were leaving the store. 9. Julius works for a company in Toronto, Canada that manufactures eyeglasses. He wants to know what styles of glasses people in Toronto prefer. He randomly selects phone numbers in Toronto and calls them to ask about glasses preference. 10. Hugo works at a public library and wants to collect data on all of the people that come to the library. He looks up every single person in the library database and notes the number of books that he or she has checked out in the last six months. 11. A company is designing a new type of smart phone. They want to know how much memory people prefer in their smart phones. The put a question up on several search engines and allow anyone to answer. 12. A college wants to collect data on their students to see how often they use the various student services offered by the college. They randomly select 60 classes and collect data from all of the students taking those classes. 13. A clothing store is designing a new line of athletic wear. They want to compare the percentage of teenagers that prefer the new line of athletic wear to the percentage of adults that prefer the new line of athletic wear. They take a random sample of teenagers and ask them about the new athletic wear. Then they take a random sample of adults and ask them about the new athletic wear. 14. Brian is collecting data for his statistics class project on the amount of time people spend on social media per day. He asks people in his college classes and at his church how many minutes they spend on social media per day. 15. A store that sells BBQ’s in North Carolina wants to know what percentage of people own a “smoker BBQ”. They ask every third person that enters the store if they own a smoker BBQ or not. ## Calculate the price of your order 550 words We'll send you the first draft for approval by September 11, 2018 at 10:52 AM Total price: \$26 The price is based on these factors: Number of pages Urgency Basic features • Free title page and bibliography • Unlimited revisions • Plagiarism-free guarantee • Money-back guarantee On-demand options • Writer’s samples • Part-by-part delivery • Overnight delivery • Copies of used sources Paper format • 275 words per page • 12 pt Arial/Times New Roman • Double line spacing • Any citation style (APA, MLA, Chicago/Turabian, Harvard) # Our guarantees Delivering a high-quality product at a reasonable price is not enough anymore. That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe. ### Money-back guarantee You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent. ### Zero-plagiarism guarantee Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in. ### Free-revision policy Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems. ### Fair-cooperation guarantee By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language. Order your essay today and save 10% with the coupon code: best10	1,334	6,468	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2022-49	latest	en	0.969999
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Conditional_probability	1,601,252,439,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00469.warc.gz	1,142,175,419	20,375	# Conditional probability In probability theory, conditional probability is a measure of the probability of an event occurring given that another event has (by assumption, presumption, assertion or evidence) occurred.[1] If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A \| B), or sometimes PB(A) or P(A / B). For example, the probability that any given person has a cough on any given day may be only 5%. But if we know or assume that the person has a cold, then they are much more likely to be coughing. The conditional probability that someone coughing is unwell might be 75%, then: P(Cough) = 5%; P(Sick \| Cough) = 75% The concept of conditional probability is one of the most fundamental and one of the most important in probability theory.[2] But conditional probabilities can be quite slippery and require careful interpretation.[3] For example, there need not be a causal relationship between A and B, and they don't have to occur simultaneously. P(A \| B) may or may not be equal to P(A) (the unconditional probability of A). If P(A \| B) = P(A), then events A and B are said to be "independent": in such a case, knowledge about either event does not give information on the other. P(A \| B) (the conditional probability of A given B) typically differs from P(B \| A). For example, if a person has dengue, they might have a 90% chance of testing positive for dengue. In this case what is being measured is that if event B ("having dengue") has occurred, the probability of A (test is positive) given that B (having dengue) occurred is 90%: that is, P(A \| B) = 90%. Alternatively, if a person tests positive for dengue they may have only a 15% chance of actually having this rare disease because the false positive rate for the test may be high. In this case what is being measured is the probability of the event B (having dengue) given that the event A (test is positive) has occurred: P(B \| A) = 15%. Falsely equating the two probabilities causes various errors of reasoning such as the base rate fallacy. Conditional probabilities can be reversed using Bayes' theorem. Conditional probabilities can be displayed in a conditional probability table. ## Definition ### Conditioning on an event #### Kolmogorov definition Given two events A and B, from the sigma-field of a probability space, with the unconditional probability of B (that is, of the event B occurring) being greater than zero, P(B) > 0, the conditional probability of A given B is defined as the quotient of the probability of the joint of events A and B, and the probability of B:[4] ${\displaystyle P(A\mid B)={\frac {P(A\cap B)}{P(B)}},}$ where ${\displaystyle P(A\cap B)}$ is the probability that both events A and B occur. This may be visualized as restricting the sample space to situations in which B occurs. The logic behind this equation is that if the possible outcomes for A and B are restricted to those in which B occurs, this set serves as the new sample space. Note that this is a definition but not a theoretical result. We just denote the quantity ${\displaystyle {\frac {P(A\cap B)}{P(B)}}}$ as ${\displaystyle P(A\mid B)}$ and call it the conditional probability of A given B. #### As an axiom of probability Some authors, such as de Finetti, prefer to introduce conditional probability as an axiom of probability: ${\displaystyle P(A\cap B)=P(A\mid B)P(B)}$ Although mathematically equivalent, this may be preferred philosophically; under major probability interpretations such as the subjective theory, conditional probability is considered a primitive entity. Further, this "multiplication axiom" introduces a symmetry with the summation axiom for mutually exclusive events:[5] ${\displaystyle P(A\cup B)=P(A)+P(B)-{\cancelto {0}{P(A\cap B)}}}$ #### As the probability of a conditional event Conditional probability can be defined as the probability of a conditional event ${\displaystyle A_{B}}$.[6] Assuming that the experiment underlying the events ${\displaystyle A}$ and ${\displaystyle B}$ is repeated, the Goodman–Nguyen–van Fraassen conditional event can be defined as ${\displaystyle A_{B}=\bigcup _{i\geq 1}\left(\bigcap _{j It can be shown that ${\displaystyle P(A_{B})={\frac {P(A\cap B)}{P(B)}}}$ which meets the Kolmogorov definition of conditional probability. Note that the equation ${\displaystyle P(A_{B})=P(A\cap B)/P(B)}$ is a theoretical result and not a definition. The definition via conditional events can be understood directly in terms of the Kolmogorov axioms and is particularly close to the Kolmogorov interpretation of probability in terms of experimental data. For example, conditional events can be repeated themselves leading to a generalized notion of conditional event ${\displaystyle A_{B(n)}}$. It can be shown[6] that the sequence ${\displaystyle (A_{B(n)})_{n\geq 1}}$ is i.i.d., which yields a strong law of large numbers for conditional probability: ${\displaystyle P\left(\lim _{n\to \infty }{\overline {A}}_{B}^{n}=P(A\mid B)\right)=100\%}$ ### Measure-theoretic definition If P(B) = 0, then according to the simple definition, P(A\|B) is undefined. However, it is possible to define a conditional probability with respect to a σ-algebra of such events (such as those arising from a continuous random variable). For example, if X and Y are non-degenerate and jointly continuous random variables with density ƒX,Y(x, y) then, if B has positive measure, ${\displaystyle P(X\in A\mid Y\in B)={\frac {\int _{y\in B}\int _{x\in A}f_{X,Y}(x,y)\,dx\,dy}{\int _{y\in B}\int _{x\in \mathbb {R} }f_{X,Y}(x,y)\,dx\,dy}}.}$ The case where B has zero measure is problematic. For the case that B = {y0}, representing a single point, the conditional probability could be defined as ${\displaystyle P(X\in A\mid Y=y_{0})={\frac {\int _{x\in A}f_{X,Y}(x,y_{0})\,dx}{\int _{x\in \mathbb {R} }f_{X,Y}(x,y_{0})\,dx}},}$ however this approach leads to the Borel–Kolmogorov paradox. The more general case of zero measure is even more problematic, as can be seen by noting that the limit, as all δyi approach zero, of ${\displaystyle P(X\in A\mid Y\in \bigcup _{i}[y_{i},y_{i}+\delta y_{i}])\approxeq {\frac {\sum _{i}\int _{x\in A}f_{X,Y}(x,y_{i})\,dx\,\delta y_{i}}{\sum _{i}\int _{x\in \mathbb {R} }f_{X,Y}(x,y_{i})\,dx\,\delta y_{i}}},}$ depends on their relationship as they approach zero. See conditional expectation for more information. ### Conditioning on a random variable Let X be a random variable; we assume for the sake of presentation that X is discrete, that is, X takes on only finitely many values x. Let A be an event. The conditional probability of A given X is defined as the random variable, written P(A\|X), that takes on the value ${\displaystyle P(A\mid X=x)}$ whenever ${\displaystyle X=x.}$ More formally, ${\displaystyle P(A\mid X)(\omega )=P(A\mid X=X(\omega )).}$ The conditional probability P(A\|X) is a function of X: e.g., if the function g is defined as ${\displaystyle g(x)=P(A\mid X=x),}$ then ${\displaystyle P(A\mid X)=g\circ X.}$ Note that P(A\|X) and X are now both random variables. From the law of total probability, the expected value of P(A\|X) is equal to the unconditional probability of A. ### Partial conditional probability The partial conditional probability ${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})}$ is about the probability of event ${\displaystyle A}$ given that each of the condition events ${\displaystyle B_{i}}$ has occurred to a degree ${\displaystyle b_{i}}$ (degree of belief, degree of experience) that might be different from 100%. Frequentistically, partial conditional probability makes sense, if the conditions are tested in experiment repetitions of appropriate length ${\displaystyle n}$ .[7] Such ${\displaystyle n}$-bounded partial conditional probability can be defined as the conditionally expected average occurrence of event ${\displaystyle A}$ in testbeds of length ${\displaystyle n}$ that adhere to all of the probability specifications ${\displaystyle B_{i}\equiv b_{i}}$, i.e.: ${\displaystyle P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\operatorname {E} ({\overline {A}}^{n}\mid {\overline {B}}_{1}^{n}=b_{1},\ldots ,{\overline {B}}_{m}^{n}=b_{m})}$[7] Based on that, partial conditional probability can be defined as ${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\lim _{n\to \infty }P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m}),}$ where ${\displaystyle b_{i}n\in \mathbb {N} }$ [7] Jeffrey conditionalization [8] [9] is a special case of partial conditional probability in which the condition events must form a partition: ${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\sum _{i=1}^{m}b_{i}P(A\mid B_{i})}$ ## Example Suppose that somebody secretly rolls two fair six-sided dice, and we wish to compute the probability that the face-up value of the first one is 2, given the information that their sum is no greater than 5. • Let D1 be the value rolled on die 1. • Let D2 be the value rolled on die 2. Probability that D1 = 2 Table 1 shows the sample space of 36 combinations of rolled values of the two dice, each of which occurs with probability 1/36, with the numbers displayed in the red and dark gray cells being D1 + D2. D1 = 2 in exactly 6 of the 36 outcomes; thus P(D1 = 2) = 636 = 16: Table 1 + D2 1 2 3 4 5 6 D1 1 234567 2 345678 3 456789 4 5678910 5 67891011 6 789101112 Probability that D1 + D2 5 Table 2 shows that D1 + D2 5 for exactly 10 of the 36 outcomes, thus P(D1 + D2 5) = 1036: Table 2 + D2 1 2 3 4 5 6 D1 1 234567 2 345678 3 456789 4 5678910 5 67891011 6 789101112 Probability that D1 = 2 given that D1 + D2 5 Table 3 shows that for 3 of these 10 outcomes, D1 = 2. Thus, the conditional probability P(D1 = 2 \| D1+D2 5) = 310 = 0.3: Table 3 + D2 1 2 3 4 5 6 D1 1 234567 2 345678 3 456789 4 5678910 5 67891011 6 789101112 Here, in the earlier notation for the definition of conditional probability, the conditioning event B is that D1 + D2 5, and the event A is D1 = 2. We have ${\displaystyle P(A\mid B)={\tfrac {P(A\cap B)}{P(B)}}={\tfrac {3/36}{10/36}}={\tfrac {3}{10}},}$ as seen in the table. ## Use in inference In statistical inference, the conditional probability is an update of the probability of an event based on new information.[3] Incorporating the new information can be done as follows:[1] • Let A, the event of interest, be in the sample space, say (X,P). • The occurrence of the event A knowing that event B has or will have occurred, means the occurrence of A as it is restricted to B, i.e. ${\displaystyle A\cap B}$. • Without the knowledge of the occurrence of B, the information about the occurrence of A would simply be P(A) • The probability of A knowing that event B has or will have occurred, will be the probability of ${\displaystyle A\cap B}$ relative to P(B), the probability that B has occurred. • This results in ${\textstyle P(A\|B)=P(A\cap B)/P(B)}$ whenever P(B) > 0 and 0 otherwise. This approach results in a probability measure that is consistent with the original probability measure and satisfies all the Kolmogorov axioms. This conditional probability measure also could have resulted by assuming that the relative magnitude of the probability of A with respect to X will be preserved with respect to B (cf. a Formal Derivation below). The wording "evidence" or "information" is generally used in the Bayesian interpretation of probability. The conditioning event is interpreted as evidence for the conditioned event. That is, P(A) is the probability of A before accounting for evidence E, and P(A\|E) is the probability of A after having accounted for evidence E or after having updated P(A). This is consistent with the frequentist interpretation, which is the first definition given above. ## Statistical independence Events A and B are defined to be statistically independent if ${\displaystyle P(A\cap B)=P(A)P(B).}$ If P(B) is not zero, then this is equivalent to the statement that ${\displaystyle P(A\mid B)=P(A).}$ Similarly, if P(A) is not zero, then ${\displaystyle P(B\mid A)=P(B)}$ is also equivalent. Although the derived forms may seem more intuitive, they are not the preferred definition as the conditional probabilities may be undefined, and the preferred definition is symmetrical in A and B. Independent events vs. mutually exclusive events The concepts of mutually independent events and mutually exclusive events are separate and distinct. The following table contrasts results for the two cases (provided the probability of the conditioning event is not zero). If statistically independent If mutually exclusive ${\displaystyle P(A\mid B)=}$ ${\displaystyle P(A)}$ 0 ${\displaystyle P(B\mid A)=}$ ${\displaystyle P(B)}$ 0 ${\displaystyle P(A\cap B)=}$ ${\displaystyle P(A)P(B)}$ 0 In fact, mutually exclusive events cannot be statistically independent (unless they both are impossible), since knowing that one occurs gives information about the other (specifically, that it certainly does not occur). ## Common fallacies These fallacies should not be confused with Robert K. Shope's 1978 "conditional fallacy", which deals with counterfactual examples that beg the question. ### Assuming conditional probability is of similar size to its inverse In general, it cannot be assumed that P(A\|B) P(B\|A). This can be an insidious error, even for those who are highly conversant with statistics.[10] The relationship between P(A\|B) and P(B\|A) is given by Bayes' theorem: {\displaystyle {\begin{aligned}P(B\mid A)&={\frac {P(A\mid B)P(B)}{P(A)}}\\\Leftrightarrow {\frac {P(B\mid A)}{P(A\mid B)}}&={\frac {P(B)}{P(A)}}\end{aligned}}} That is, P(A\|B) P(B\|A) only if P(B)/P(A) 1, or equivalently, P(A) P(B). ### Assuming marginal and conditional probabilities are of similar size In general, it cannot be assumed that P(A) P(A\|B). These probabilities are linked through the law of total probability: ${\displaystyle P(A)=\sum _{n}P(A\cap B_{n})=\sum _{n}P(A\mid B_{n})P(B_{n}).}$ where the events ${\displaystyle (B_{n})}$ form a countable partition of ${\displaystyle \Omega }$. This fallacy may arise through selection bias.[11] For example, in the context of a medical claim, let SC be the event that a sequela (chronic disease) S occurs as a consequence of circumstance (acute condition) C. Let H be the event that an individual seeks medical help. Suppose that in most cases, C does not cause S so P(SC) is low. Suppose also that medical attention is only sought if S has occurred due to C. From experience of patients, a doctor may therefore erroneously conclude that P(SC) is high. The actual probability observed by the doctor is P(SC\|H). ### Over- or under-weighting priors Not taking prior probability into account partially or completely is called base rate neglect. The reverse, insufficient adjustment from the prior probability is conservatism. ## Formal derivation Formally, P(A \| B) is defined as the probability of A according to a new probability function on the sample space, such that outcomes not in B have probability 0 and that it is consistent with all original probability measures.[12][13] Let Ω be a sample space with elementary events {ω}. Suppose we are told the event B Ω has occurred. A new probability distribution (denoted by the conditional notation) is to be assigned on {ω} to reflect this. For events in B, it is reasonable to assume that the relative magnitudes of the probabilities will be preserved. For some constant scale factor α, the new distribution will therefore satisfy: {\displaystyle {\begin{aligned}&{\text{1. }}\omega \in B:P(\omega \mid B)=\alpha P(\omega )\\&{\text{2. }}\omega \notin B:P(\omega \mid B)=0\\&{\text{3. }}\sum _{\omega \in \Omega }{P(\omega \mid B)}=1.\end{aligned}}} Substituting 1 and 2 into 3 to select α: {\displaystyle {\begin{aligned}1&=\sum _{\omega \in \Omega }{P(\omega \mid B)}\\&=\sum _{\omega \in B}{P(\omega \mid B)}+{\cancelto {0}{\sum _{\omega \notin B}P(\omega \mid B)}}\\&=\alpha \sum _{\omega \in B}{P(\omega )}\\[5pt]&=\alpha \cdot P(B)\\[5pt]\Rightarrow \alpha &={\frac {1}{P(B)}}\end{aligned}}} So the new probability distribution is {\displaystyle {\begin{aligned}{\text{1. }}\omega \in B&:P(\omega \mid B)={\frac {P(\omega )}{P(B)}}\\{\text{2. }}\omega \notin B&:P(\omega \mid B)=0\end{aligned}}} Now for a general event A, {\displaystyle {\begin{aligned}P(A\mid B)&=\sum _{\omega \in A\cap B}{P(\omega \mid B)}+{\cancelto {0}{\sum _{\omega \in A\cap B^{c}}P(\omega \mid B)}}\\&=\sum _{\omega \in A\cap B}{\frac {P(\omega )}{P(B)}}\\[5pt]&={\frac {P(A\cap B)}{P(B)}}\end{aligned}}} ## References 1. Gut, Allan (2013). Probability: A Graduate Course (Second ed.). New York, NY: Springer. ISBN 978-1-4614-4707-8. 2. Ross, Sheldon (2010). A First Course in Probability (8th ed.). Pearson Prentice Hall. ISBN 978-0-13-603313-4. 3. Casella, George; Berger, Roger L. (2002). Statistical Inference. Duxbury Press. ISBN 0-534-24312-6. 4. Kolmogorov, Andrey (1956), Foundations of the Theory of Probability, Chelsea 5. Gillies, Donald (2000); "Philosophical Theories of Probability"; Routledge; Chapter 4 "The subjective theory" 6. Draheim, Dirk (2017). "An Operational Semantics of Conditional Probabilities that Fully Adheres to Kolmogorov's Explication of Probability Theory". doi:10.13140/RG.2.2.10050.48323/3. 7. Draheim, Dirk (2017). "Generalized Jeffrey Conditionalization (A Frequentist Semantics of Partial Conditionalization)". Springer. Retrieved December 19, 2017. 8. Jeffrey, Richard C. (1983), The Logic of Decision, 2nd edition, University of Chicago Press 9. "Bayesian Epistemology". Stanford Encyclopedia of Philosophy. 2017. Retrieved December 29, 2017. 10. Paulos, J.A. (1988) Innumeracy: Mathematical Illiteracy and its Consequences, Hill and Wang. ISBN 0-8090-7447-8 (p. 63 et seq.) 11. Thomas Bruss, F; Der Wyatt Earp Effekt; Spektrum der Wissenschaft; March 2007 12. George Casella and Roger L. Berger (1990), Statistical Inference, Duxbury Press, ISBN 0-534-11958-1 (p. 18 et seq.)	5,114	18,220	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 57, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2020-40	longest	en	0.952523
https://mathzsolution.com/how-to-find-perpendicular-vector-to-another-vector/	1,675,667,414,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00359.warc.gz	400,907,454	18,167	How to find perpendicular vector to another vector? How do I find a vector perpendicular to a vector like this: Could anyone explain this to me, please? I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components… When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks. For finding all of them, just choose 2 perpendicular vectors, like $v_1=(4\mathbf{i}-3\mathbf{j})$ and $v_2=(2\mathbf{i}+3\mathbf{k})$ and any linear combination of them is also perpendicular to the original vector:	168	618	{"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2023-06	latest	en	0.899176
https://kidsworksheetfun.com/factoring-polynomials-worksheet-grade-8/	1,716,498,667,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00514.warc.gz	301,209,286	25,128	# Factoring Polynomials Worksheet Grade 8 Factoring polynomials 1 first determine if a common monomial factor greatest common factor exists. Showing top 8 worksheets in the category factoring of polynomials. Factoring Polynomials Coloring Activity Factoring Polynomials Polynomials Factoring Polynomials Activity ### Apply factoring techniques to solve problems involving area and volume. Factoring polynomials worksheet grade 8. Unit 2 worksheet 8 factoring polynomials factor each of the following polynomials. Displaying top 8 worksheets found for problems involving factors of polynomials. Some of the worksheets displayed are factoring polynomials addition and subtraction when adding multiplying polynomials date period factoring quadratic expressions work 2 6 factorizing algebraic expressions factoring trinomials a 1 date period factoring polynomials gcf and quadratic expressions. Showing top 8 worksheets in the category factoring polynomial word problems in grade 8. Walk through these factoring polynomials worksheets to acquire abundant practice in factoring linear expressions quadratic expressions monomials binomials and polynomials using a variety of methods like grouping synthetic division and box method. Vocabulary match each term on the left with a definition on the right. A x x2 3 28 b x x2 5 6 c x x2 6 5 d x x2 8 16 e x x2 9 22 f x2 25 g x2 9 h 2 7 6x x2 i 2 13 21x x2 j 3 10 8x x2 k 2 3 35x x2 l 5 13 6x x2n n. Printable worksheets on the greatest common factors of whole numbers and polynomials are presented on this page 6th grade 7th grade and 8th grade students. Levels from easy to difficult are discussed. Some of the worksheets for this concept are factoring solvingequations and problem solving 5 factoring polynomials factoring polynomials factoring polynomials work and answers 5 4 factoring polynomials work answers factoring polynomials gina wilson work the remainder and factor synthetic division special. A b and b a these may become the same by factoring 1 from one of them. 540 chapter 8 factoring polynomials factor polynomials. A11nls c08 0540 0543 indd 540 8 18 09 10 18 19 am. Some of the worksheets displayed are factoring trinomials a 1 date period factoring polynomials gcf and quadratic expressions factoring polynomials factoring practice factoring quadratic expressions factoring polynomials 1 algebra 1 factoring polynomials name. From 8th grade algebra worksheets to course syllabus we have got all of it covered. Displaying top 8 worksheets found for factoring polynomial word problems in grade 8. Factor trees may be used to find the gcf of difficult numbers. Some of the worksheets for this concept are grade 9 math polynomial word problems midgrp factoring polynomials factoring polynomials factoring solvingequations and problem solving 5 lake washington institute of technology lake washington factoring quadratic expressions polynomials word problems work. Hone your skills in. Gcf of polynomials worksheets. Be aware of opposites. Education Factoring Polynomials Gcf Worksheet Factoring Polynomials Algebra Worksheets Polynomials Algebra 1 Worksheets Monomials And Polynomials Worksheets Algebra Worksheets Factoring Polynomials Polynomials Factoring Easy Trinomials Pic High School Algebra School Algebra Algebra Worksheets This Worksheet Includes 15 Practice With Factoring Trinomials As Well As Special Cases Such As Difference Factoring Polynomials Polynomials Writing Equations Factorize Each Polynomial Using Algebraic Identities Polynomials Algebra Worksheets Factoring Polynomials Multiply Polynomials Worksheet 4 Polynomials Multiplying Polynomials Solving Linear Equations Precised Worksheets On Factors Monomials Factoring Polynomials Pre Algebra Worksheets Chemistry Worksheets 30 Factorisation Class 8 Worksheets Factoring Polynomials Using The Gcf Factoring Polynomials Polynomials Quadratics Discovery Lesson Factoring Trinomials Math In The Middle Factor Trinomials Common Core Math Middle School Polynomials Factoring Polynomials Gcf Factoring Polynomials Algebra Polynomials Factoring Polynomials Trinomials Activity Beginner Factoring Trinomials Activity Factoring Polynomials Education Math Factoring Trinomials Worksheet Answers Luxury Factoring General Trinomials In 2020 Factoring Polynomials Factor Trinomials Algebra Worksheets Algebraic Expressions Pdf Printable Worksheets With Integers Simplifying Algebraic Expressions Algebraic Expressions Polynomials 5 Year 8 Algebra Worksheets In 2020 Algebra Worksheets 9th Grade Math Factoring Polynomials Multiplying And Factoring Polynomials Card Sort Factoring Polynomials Teaching Algebra Teaching Math Factoring Polynomials Free Worksheet Factoring Polynomials Teaching Algebra School Algebra Unit 2 Worksheet 8 Factoring Polynomials Answer Key Teaching Algebra School Algebra Math Interactive Notebook	1,042	4,866	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.15625	3	CC-MAIN-2024-22	latest	en	0.854844
http://forums.autodesk.com/t5/Robot-Structural-Analysis/Plate-and-Shell-Deflections-Calculations/m-p/3343691	1,394,251,182,000,000,000	text/html	crawl-data/CC-MAIN-2014-10/segments/1393999652934/warc/CC-MAIN-20140305060732-00019-ip-10-183-142-35.ec2.internal.warc.gz	67,287,249	19,621	Discussion Groups ## Robot Structural Analysis Mentor Posts: 160 Registered: ‎01-24-2009 # Plate and Shell Deflections - Calculations 691 Views, 4 Replies 02-22-2012 07:28 PM Hi All Robot help says: "If properties of materials used during design are identical to those used in a model, then the value of the coefficient D / B > 1.0. This coefficient can be interpreted (mainly for slabs subjected to unidirectional bending) as an elastic deflection multiplier. If different materials are used in a model and calculations (for example, with different classes such as concretes with different Young's modulus or Poisson's ratio), the coefficient value is corrected automatically. It may result in disturbing the the earlier mentioned inequality." 1) This coefficient can be interpreted (mainly for slabs subjected to unidirectional bending) as an elastic deflection multiplier: • Which is the meaning of "elastic" in this context ? The coefficient D/B take in account of "element cracking, rheological effects, adherence of calculated reinforcement, etc. and averaged for both directions" so it is, The intent is to simulate a inherently inelastic situation • "rheological effects" means long term delayed strain estimation as in UNI EN 1992-2 5.8.4 - Creep Annex KK ? • "mainly for slabs subjected to unidirectional bending" I think that for bidirectional slabs should be the same thing 2) it may result in disturbing the the earlier mentioned inequality. Which is the meaning ?. If in the model I am using i.e.. RCK 25 and in RCK 35 for the calculation off shell deflection D/B may be less than 1 ? 3) If I am using reduced stiffness of elements like http://forums.autodesk.com/t5/Autodesk-Robot-Structural/EC8-4-3-1-7-Reduced-moment-of-Inertia/td-p/3... the D/B calculation uses reduced or full inertia of elements ? Please use plain text. Product Support Posts: 4,021 Registered: ‎12-17-2010 # Re: Plate and Shell Deflections - Calculations 02-23-2012 01:01 AM in reply to: GabrieleNovembri1027 1) This coefficient can be interpreted (mainly for slabs subjected to unidirectional bending) as an elastic deflection multiplier: • Which is the meaning of "elastic" in this context ? The one you get after running the static analysis of the model. • The coefficient D/B take in account of "element cracking, rheological effects, adherence of calculated reinforcement, etc. and averaged for both directions" so it is, The intent is to simulate a inherently inelastic situation? Correct. • "rheological effects" means long term delayed strain estimation as in UNI EN 1992-2 5.8.4 - Creep Annex KK ? I'm not sure which effects you mean as I'm don't have what is included in the UNI EN 1992-2 5.8.4 - Creep Annex KK provisions • "mainly for slabs subjected to unidirectional bending" I think that for bidirectional slabs should be the same thing Yes, the scaling as described in the description of the method you refer to is done for all shapes of panels. The main intention of this statement is to inform that the influence of in-plane forces is disregarded. 2) it may result in disturbing the the earlier mentioned inequality. Which is the meaning ?. If in the model I am using i.e.. RCK 25 and in RCK 35 for the calculation off shell deflection D/B may be less than 1 ? I'm not sure if more or larger than 1 for this particular case but theoretically you can use 10MPa concrete in the model and decide on setting e.g. 50MPa concrete with creep coefficient set as 0 in the panel design parameters 3) If I am using reduced stiffness of elements like http://forums.autodesk.com/t5/Autodesk-Robot-Structural/EC8-4-3-1-7-Reduced-moment-of-Inertia/td-p/3... the D/B calculation uses reduced or full inertia of elements ? Reduced (for the value of D) If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you. Artur Kosakowski Please use plain text. Mentor Posts: 160 Registered: ‎01-24-2009 # Re: Plate and Shell Deflections - Calculations 02-24-2012 10:33 AM in reply to: Artur.Kosakowski Hi Artur, You are right. "UNI EN 1992-2 5.8.4 - Creep Annex KK provisions" is related to bridges instead of buildings. Anyway I am looking for a detailed description of the method used by Robot to take into account the concrete creep in the D/B coefficient calculation. Thanks Please use plain text. Product Support Posts: 4,021 Registered: ‎12-17-2010 # Re: Plate and Shell Deflections - Calculations 02-26-2012 11:24 PM in reply to: GabrieleNovembri1027 Robot uses the formula 7.20 from 7.4.3.(5) of EC2. If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you. Artur Kosakowski Please use plain text. Mentor Posts: 160 Registered: ‎01-24-2009 # Re: Plate and Shell Deflections - Calculations 02-26-2012 11:56 PM in reply to: Artur.Kosakowski Thanks Artur Please use plain text.	1,300	4,990	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2014-10	latest	en	0.837373
https://ojs.bonviewpress.com/index.php/AAES/article/view/1322	1,718,936,897,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00786.warc.gz	374,613,853	9,305	# Statistical Study of Bisection Method for Cubic Equations with Random Coefficients ## Authors • Shikha Kumari Department of Mathematics, Birla Institute of Technology, India • Soubhik Chakraborty Department of Mathematics, Birla Institute of Technology, India ## Keywords: bisection method, uniform distribution, normal distribution, cubic equation, statistical analysis ## Abstract The bisection method is an iterative approach used in numerical analysis to find solutions to nonlinear equations. The main purpose of this paper is to study how the parameters of a probability distribution characterizing the coefficients of a cubic polynomial can influence the convergence of the bisection method. The study covers discrete and continuous distributions, including discrete uniform, continuous uniform, and normal distributions. It was found that for both types of uniform distribution inputs, a second-degree polynomial equation can predict the average iteration for a given parameter r, where r indicates the distribution interval [−r, r]. Interestingly, the coefficients of the second-degree polynomial are nearly identical for discrete and continuous uniform distributions. For normal distribution input, the average iteration does not depend upon the standard deviation when the mean is fixed and the standard deviation is varying. But when the standard deviation is fixed and the mean is varying, the second-degree polynomial is still the best fit. This means the average iteration depends upon the mean of the normal distribution. Overall, our paper concludes that: I. For uniform distribution input, the average iteration does not depend on whether the distribution is discrete or continuous but rather depends on the range of the distribution which is its parameter. II. For non-uniform distribution input, the average iteration depends on the mean of the distribution (location parameter) but not on the standard deviation (scale parameter). Finally, a curtain is raised in the future direction of research in which we propose to combine the bisection method with the regula falsi and Newton-Raphson methods to increase the rate of convergence. Received: 5 July 2023 \| Revised: 9 August 2023 \| Accepted: 11 August 2023 Conflicts of Interest The authors declare that they have no conflicts of interest to this work. Data Availability Statement Data sharing is not applicable to this article as no new data were created or analyzed in this study. 2023-08-14 ## How to Cite Kumari, S., & Chakraborty, S. (2023). Statistical Study of Bisection Method for Cubic Equations with Random Coefficients. Archives of Advanced Engineering Science, 2(1), 37–52. https://doi.org/10.47852/bonviewAAES32021322 Articles	557	2,728	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2024-26	latest	en	0.868581
https://www.santaclaritahomeguide.com/what-is-the-rate/	1,696,470,363,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00296.warc.gz	1,052,026,937	12,386	# What Is The Rate Your basal metabolic rate (BMR) is the number of calories your body needs to accomplish its most basic (basal) life-sustaining functions, such as breathing, circulation, nutrient processing, and. Rate (mathematics) A rate defined using two numbers of the same units (such as tax rates) or counts (such as literacy rate) will result in a dimensionless quantity, which can be expressed as a percentage (for example, the global literacy rate in 1998 was 80%) or fraction or as a multiple . A normal resting heart rate for adults ranges from 60 to 100 beats per minute. Generally, a lower heart rate at rest implies more efficient heart function and better cardiovascular fitness. For example, a well-trained athlete might have a normal resting heart rate closer to 40 beats per minute. Rates of severe coal workers’ pneumoconiosis-also known as black lung disease-among coal miners have been on the rise. Definition: Base rate is the minimum rate set by the Reserve Bank of India below which banks are not allowed to lend to its customers. Description: Base rate is. Mathematics and science. Rate (mathematics), a specific kind of ratio, in which two measurements are related to each other (often with respect to time) Rate function, a function used to quantify the probabilities of a rare event; rate (particle physics), rate at which a specific subatomic particle reaction occurs 7 rows · Define rate. rate synonyms, rate pronunciation, rate translation, English dictionary definition of rate. n. 1. A quantity measured with respect to another measured quantity: a rate of speed of 60 miles an hour. 2. A measure of a part with respect to a whole; a. A rate is a little bit different than the ratio, it is a special ratio. It is a comparison of measurements that have different units, like cents and grams. A unit rate is a rate with a denominator of 1. Mortgage interest rates determine your monthly payments over the life of the loan. Even a slight difference in rates can drive your monthly payments up or down, and you could pay thousands of. For years, universities across the nation have workshopped ideas for improving graduation rates among minority students. UNC. sitemap	469	2,210	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.96875	3	CC-MAIN-2023-40	latest	en	0.927114
http://excel.bigresource.com/Fiscal-Week-in-any-month-date-VYlfr1mt.html	1,555,690,678,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578527865.32/warc/CC-MAIN-20190419161226-20190419183226-00429.warc.gz	65,784,424	12,570	# Fiscal Week In Any Month Date Jul 24, 2013 I have data for 6 monthsone column has all the dates 1/1/2013 1/2//2013 1/5/2013 1/8/2013 1/10/2013 1/15/2013 1/21/2013 1/31/2013 I have for 6 months like these dates I need to convert all these dates to week 1- week 5 week 1: 1-7, week 2: 8-14, week3: 15-21, week4: 22-28, week5: 29-31 Instead of 1/1/2013 it should show week1 instead 1/10/2013 week2 like this ADVERTISEMENT ## Macro To Create Fiscal Week To Month And Date Dec 2, 2012 I have to generate daily report in which data is fiscal week format. We need to show Fiscal week in actual month and day format. create a macro so that for every fiscal week anywhere in the sheet, it automatically converts fiscal week to month and days of the week For example: WK 1 OCT FY2013 get converted to Oct 1-7 WK 2 OCT FY2013 coverted to Oct 8 - 14 WK 3 OCT FY2013 get converted to Oct 15 -21 WK 4 OCT FY2013 to Oct 22-28 WK 5 OCT FY2013 to Oct 29- Nov4 ## Convert Fiscal Date To Corresponding Week Number Nov 12, 2009 I have a spreadsheet that I use to convert a purchase order ship date from the actual date to the corresponding week it falls out on. The fiscal year always starts on February 1 regardless of the day of the week. The problem i am encountering is when the year changes. As soon as I enter 01/01/2010, the response I get is -4, where as 12/31/2009 is 48. I am using the following formula that I found somewhere, where R2 = 02/01/2009 (02/01/2009 falls out on a Sunday). =INT((R2-DATE(YEAR(R2),2,1)-WEEKDAY(R2,1))/7)+2. I need to make the formula "not care about" the day of the week. ## Get Date From Week Number And Year ( Fiscal Calendar) Jan 22, 2013 I'm having a data only pull week number and year. We are using Fiscal calendar starting in July. For example, A1 = Week number and A2= Year. How to set up a formula to retrieve a date for this? If A1 = 2 , A2 = 2013, the date will be 07/14/2012. I want the date pull of on Saturday every week. ## Enter A Date And Return The Fiscal Month Dec 15, 2008 Is there any way to defined our own fiscal month. I have a fiscal year where the start & end date is different from the normal calendar. I have store the start & end date on the different column. What i need is that when people enter a date, it will look up to the table and return the fiscal month.It's something like If the value is >=column A and <=column B, then the fiscal month=column C(refer to the below table). But i have no idea how to make comparison on date value. what kind of formula i should use? And also how to write the code if i want to use macro to implement this? Start End Month 09/21/08 10/18/08 Oct 10/19/08 11/22/08 Nov 11/23/08 12/27/08 Dec 12/28/08 01/24/09 Jan 01/25/09 02/21/09 Feb 02/22/09 03/28/09 Mar 03/29/09 04/25/09 Apr 04/26/09 05/23/09 May 05/24/09 06/27/09 Jun 06/28/09 07/25/09 Jul 07/26/09 08/22/09 Aug 08/23/09 09/26/09 Sep ## Formatting Date: Add The Month And The Text "Week" Before The Week Number Nov 24, 2009 I have a column where I am convering the Date into a Fiscal week number. For example 10/6/2009 is Work week 41 Now I want to show October Week 41 I need to add the month and the text "Week" before the week number. what is the formula I use. ## Check If Final Week Of Fiscal Period Oct 8, 2008 I am attempting to check to see if the current week is the end of a period or not. If it is, I need to display some text, and if it is not, I need to display nothing. Since I am Computer Science major, the best way I can describe this is through and if structure from a low-level programming lanuage (like C++ or Java): ## Formula To Get Date From Month And / Or Week May 30, 2014 I uploaded an example file. Now, what I need to accomplish is that the D1 and D3's in sheet 2 need to result in a date next to the correct country (the date (in full) must be the first monday of the correct week). I find it quit difficult to do this because in sheet 2 you have once the country name, but several possible dates. So in sheet 1 there must be a date for every D1 or D3 but under each other. The second problem is that I need to accomplish to get a "x" in sheet 3 under the correct month where there is an D1 or D3 in sheet 2 (week). So I need to go from a week to a month and this can be for one country 1, 2, 3 or even more months (it depends from the D1 and D3's in sheet 2). ## Vba To Find The Month And Week Number From Given Date Apr 22, 2009 In the attch file i have the date coulumn from this date column i need to calulate the month & week no. (like WEEK1,WEEK2..) The Week ( Monday to sunday) which need to be calculated is the week no. in the given month like for month of April the week1 is print in the week column for 6april to 12 april date and Week2 print for 13 April to 19 april ## Update Day Of Week And Date Based On Month Sep 3, 2007 I'm attempting to force excel to auto update the day of the week, and the date in a spreadsheet. The date isn't as important, since it can be hard coded. The only problem there is some months have 31 days, some 30, and another with 28. I've uploaded an image of the spreadsheet, and you can see in field A1 the date/year is input. I'm wanting to find a way to force the days/dates in fields 2E and 3E to update based on the month. ## Calculate Fiscal Month Oct 7, 2008 I'm looking for a way to calculate (name) a fiscal month based on a date range. Example attached. Columns A & B are a date range and column D is the desired outcome. I'd prefer a formula solution, but am open to anything as I haven't a clue how to begin with this. ## How To Calculate The Week Number Of The Month Based On A Date Dec 30, 2013 I would like to calculate the week number of the month based on a date. Now my days would only include working weeks (Monday - Friday). Supposed the date is 12/31/2012: M 31-Dec T 1-Jan W 2-Jan TH 3-Jan F 4-Jan Since it only occupies 1 day of the workweek, then it will be considered as Week 1 of January. If the date is 1/28/2012: M 28-Jan T 29-Jan W 30-Jan TH 31-Jan F 1-Feb It will be considered as Week 5 of January since it occupies 4 days of the working week. If the date is 4/29/2013: M 29-Apr T 30-Apr W 1-May TH 2-May F 3-May It will be considered as Week 1 of May since it occupies only 2 days of the working week. Basically if the date's month occupies 3 or more of the working days of the workweek then it will be considered as part of that month's working week. Is this possible with formulas? I tried to explain it the best I can. ## Date Transformation Formula - Convert Particular Week To Day And Month Jul 20, 2012 This might be a simple date transformation formula that I need Column A has numbers like: 200517, 200530, 200544, 201036, 201043, etc I'd like to get a formula in Col B in corresponding rows that would show date as dd/mm/yy I know the numbers above are yyyyww = yyyy is year, and ww is the week of that year. So the formula would have to convert that particular week to the day and month. I understand that are 7 days in a week, but if the formula would reflect Monday of that week, it would suffice. ## Excel Formula To Calculate Fiscal Month Dec 19, 2012 I have a large data sheet with dates in column B - in column A, I'm trying to write a formula that will determine what fiscal month it should be mapped to. I have the calendar listed on a different tab. Fiscal Month Start Date End Date January 1/1/2012 1/31/2012 [Code] ......... The formula I tried using is: =LOOKUP(B2,Calendar!\$B\$2:\$C\$25) ## Formula To Tell When Its Week 1 Or Week 2 Of A Any Given Month Jun 16, 2014 I'm trying to write a formula that will tell me when its week one or week two, week three and week 4 based on a given date of any month. I'm using weekday formula but no luck. ## Add Fiscal Quarter To Fiscal Macro? May 13, 2014 I have a Fiscal Date macro that will looks at a date to give me the current fiscal month or week, but I would also like to add the functionality for it to give me the quarter. In my business, the fiscal year starts in October. The quarters of course are as follows: Oct-Dec = Q1 Jan-Mar = Q2 Apr-Jun = Q3 Jul-Sep = Q4 I would like for the quarter to be formatted with the year ex. FY14Q1 Here is the Code, but I have also attached an Excel file with the .bas file. [Code] ..... Attached File : Fiscal.xlsm‎ ## Auto Open Macro To Find Correct Week Tab & Day/date In 52 Week Worksheets Jan 19, 2010 I am new to VBA & not sure of the full understanding of code copied from a workbook which worked on the same principle but with Monthly (12) tabs. I thought if modified to show weeks, the macro would be able to locate the current week tab & day/date within - but upon opening, the cell stops at WK19 & column O - rather than WK43, Column N (which changes daily). Sub Auto_Open() week(1) = "WK1" week(2) = "WK2" week(3) = "WK3" week(4) = "WK4" week(5) = "WK5" week(6) = "WK6" week(7) = "WK7" week(8) = "WK8" week(9) = "WK9" week(10) = "WK10" week(11) = "WK11" week(12) = "WK12" week(13) = "WK13" week(14) = "WK14" week(15) = "WK15" week(16) = "WK16" week(17) = "WK17" week(18) = "WK18" week(19) = "WK19" week(20) = "WK20" week(21) = "WK21" week(22) = "WK22" week(23) = "WK23" week(24) = "WK24"...................................... ## Results By Month And Week Of Month Mar 26, 2008 I have a range of data which is as follows: Week in month: 1 1 1 5 Site: 01/03 02/03 03/03 etc 30/03 etc Leeds 10 9 15 20 Manchester 8 5 1 2 Etc Here's what I need to produce: March 08 April 08 Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Leeds Manchester I need to sum week 1 to 6 for each month Mar, Apr and so on. The different sites are in the same order so that doesn't matter too much. ## User Defined Function To Return Fiscal Year Of A Date May 2, 2012 I am working with a fiscal year that starts in December and ends in November. I want to make a user defined function that will return the fiscal year of a date. I've created the below code, but it returns a zero. Function FiscalYear(DateFY) If Month(DateFY) = 12 Then Year (DateFY) + 1 Else Year (DateFY) End If End Function ## Formula To Output Upcoming Quarter End Date (for A Broken Fiscal Year) Jan 20, 2014 I am looking to create a function that outputs the upcoming quarter end date based on a specified start date, for which the quarter end is based on a broken fiscal year ending december 15. As an example, say that you sign up as a customer with an internet provider on 2014-01-01. The internet provider charges all their clients on a quarterly basis and have a broken fiscal year ending on december 15. Hence, as you signed up on 2014-01-01 you will be charged on 2014-03-15, which is the date of the company's first quarter end. So what I would like to do is to set up a function that outputs the first date I will be charged based on the date that I sign up. If I sign up between december 2013-12-16 and 2014-03-15, the formula should output 2014-03-15. If I sign up between 2014-03-16 and 2014-06-15, the formula should output 2014-06-15 etc. etc. ## Month Day Day Of Week Nov 2, 2008 1 2 3 4 5 6 11/1 11/2 11/3 11/4 11/5 11/6 sat sun mon tues wed thur what i am trying to do is to be able to put a date in a cell and it would fille in the month day and day of week =\$D\$1 =\$D\$1+E5-1 =TEXT(WEEKDAY(D6),"ddd") cells are not correct in the formulas just copied them. i got this to work in a set up xls file when used it in the real setting i get ###. ## Number Of The Week In Month Mar 26, 2008 I want to use a formula to calculate what week number in the month it is (i.e 1 to 6) from a particular date. I know how to calculate this on an annual basis (i.e. 1 to 52) but not within the month. This is what I have so far... =IF(OR(D58=1, D58>=D57), ROUNDUP(DAY(D61)/7,0),ROUNDUP(DAY(D61)/7,0)+1) d58 is a Weeday formula looking at d61 which is the date i want to look at. d57 is the weekday number of the first day of the month in cell d61. ## Converting Week To Month May 10, 2012 Assuming all my date inputs are in column A, week in B, and month in C. I found a formula from a forum in converting my date inputs to week number, and here is the formula in column B: =IF(A40,"WEEK"&INT((A4+1-DATE(YEAR(A4),1,1))/7)+1,"") - INT(bold) formula part, as those part I just copy paste it, add this and that, and it works - the reason why I add the "+1" on the very last part is because on my weekly salary payment, every Saturday salary will be paid on the week after. (ex: 04/28/12 is supposed to be WEEK17, but then I added +1 so it became WEEK18. The main reason is 04/28 is Saturday and like I said I need every Saturday salary to be paid on the week after) So, I've got the solution to convert my date inputs to Week#, but then I also need this week# to be converted into "MONTH" in a way that the last day/s of any months, but still in the same week of the new month, to be grouped together. - For an example, 04/28/12 is supposed to be considered as April month, but I need that day to be on the May month. ## Formula For Week # Of Month Mar 12, 2008 what is the equivalent command to WEEKNUM if I want to properly calculate Week # of Month? For example (Sunday being the first day of the week): January 5th 2008 = Week 1 of January January 6th 2008 = Week 2 of January February 2nd 2008 = Week 1 of February February 3rd 2008 = Week 2 of February WEEKNUM perfectly calculates this, but it is applicable for the whole year. ## Averages By Week By Month May 5, 2009 I have the following data by week: ------------------------------------------------------------------------ 29-Mar 5-Apr 12-Apr 19-Apr 26-Apr 3-May 10-May # of work days 4 4 4 4 5 5 5 Manpower 106 85 115 115 120 121 87 ------------------------------------------------------------------------ On a second tab, I want to show the data by month: ---------------------------------------------------------------- Apr May Jun # of workdays 21 (4+4+4+4+5) 21 20 Avg. manpower 108 104 .... --------------------------------------------------------------- ## Get Which Week Number Will Falls In Which Month? Nov 15, 2013 I have week numbers from 1 to 52, now i want to get which week number will falls in which month, is there any formula in excel for eg. Week 01 - 05 will fall in January month (2014), likewise.. ## Days Of Month And Day Of Week Automatically Set? Feb 25, 2013 I currently am trying to refine some spreadsheets at work (hospital setting). The type of files im working with are medication sheets where on the left it states the medication and to the right of it, the cells have the days of the month(1-31) but I need them to change depending on the day they come into our facility. Above the numbers i would also like it to say the day of week with the first initial (M, T, W, T, F, S, S) in the cells are the top. It is something that we have to make for each day it it gets really annoying and is a waste of time moving the dates over for every day. find a way where I can open the file and the numbers and letters are all in the right place without having to change it for the day that the patients are coming in. ## Visualizing The Data For Per Month And Per Week May 1, 2009 I have 15 projects which are currently been taken, i need to record how much time each staff member takes each week/ per month on the projects, so at the end of week 5 i require a total column, and i require a totals columns at the end of the data. How can i lay this out cos i'm having trouble visualing the data for per month and per week? ## COUNTA And Sum: Counts How Many Week Days There Are In A Particular Month Feb 4, 2010 I have using the following Formula: =COUNTA(A3:A7,A10:A14,A17:A21,A24:A28,A31:A35) Basically is counts how many week days there are in a particular month. Now I have a cell (B47) that counts how many Bank Holidays are in that month. I am having trouble using the formula then minus B47. I must be missing something really simple. ## Day Of Week -day Of Month List Auto Setup Feb 18, 2009 What I'm after is a sheet that self generates the day of the week in column A and the day of the month in column B. I have a month long sheet where daily entries get made in the DOW row, the day of the month is a reference. I have a macro to generate a new sheet for the next month and would like to auto populate the DOW and DOM. This typically gets done on the second day of the month (data from the first day is entered on the second) Copyrights 2005-15 www.BigResource.com, All rights reserved	4,760	16,591	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.078125	3	CC-MAIN-2019-18	latest	en	0.896855
http://www.eduplace.com/kids/mw/bt/3/3_01-2q.html	1,440,905,665,000,000,000	text/html	crawl-data/CC-MAIN-2015-35/segments/1440644064869.18/warc/CC-MAIN-20150827025424-00283-ip-10-171-96-226.ec2.internal.warc.gz	421,170,900	2,690	# Change Range ## Question Rosa, Glen, Cathy, and Errol each bought snacks from several vending machines. Each person got back 6 coins, but they were different combinations of coins. Each person received less than \$1 in change. The machines returned only nickels, dimes, and quarters. Each person had at least one of each coin. • Rosa had the fewest dimes but the same number of quarters as Errol. • Glen had an equal number of nickels, dimes, and quarters. • Cathy had the same number of quarters as Glen. • Errol had more dimes than Cathy. • Cathy had more dimes than quarters. • Rosa had fewer quarters than Cathy. How much money did each person get back from the machines? How much money did they get back in all?	169	724	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.890625	3	CC-MAIN-2015-35	longest	en	0.989276
http://eranraviv.com/modern-statistical-discoveries/	1,498,380,998,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320476.39/warc/CC-MAIN-20170625083108-20170625103108-00542.warc.gz	135,907,381	19,189	# Present-day great statistical discoveries Some time during the 18th century the biologist and geologist Louis Agassiz said: “Every great scientific truth goes through three stages. First, people say it conflicts with the Bible. Next they say it has been discovered before. Lastly they say they always believed it”. Nowadays I am not sure about the Bible but yeah, it happens. I express here my long-standing and long-lasting admiration for the following triplet of present-day great discoveries. The authors of all three papers had initially struggled to advance their ideas, which echos the quote above. Here they are, in no particular order. 1. The Bootstrap (1979), by Bradly Efron. 2. If you study or studied Applied Statistics, there is simply no getting around it, in a good way. It is nice to know that some quantity is asymptotically distributed F. However, the word asymptotically means approaching. Sometimes we are not sure how good is the approximation, how quickly is it approaching? Is it ‘Japanese high-speed train’ approaching? or is it ‘my grandpa goes to the toilet’ approaching? Bootstrapping relies on sampling from finite samples. So the number of data points is accounted for. More often than not, we simply sample from our empirical distribution, the one that we observe, or some variant thereof. At the time of this writing, the original paper alone is cited over 12,500 times. And you can now find a great many bootstrap techniques solving many different complications: Bayesian bootstrap, smooth bootstrap, parametric bootstrap, residual sampling, usual regression bootstrap, Wild bootstrap to handle Heteroscedasticity, block bootstrap to handle time-dependencies and panel bootstrap to handle panel data (some references are also below, more are welcome). In this interview, Prof. Efron reveals that the paper was practically rejected on first submission. 3. Comparing Predictive Accuracy (1995), by Francis X. Diebold and Roberto S. Mariano (DM henceforth) 4. Circulated first in 1991, so far with 4658 citations. I would thoughtfully argue that this is the most common way to statistically compare the accuracy of two forecasts. Looking at two money managers, one of them is going to do better than the other. The paper helps to reply on the question: “is he really better?” You can see where it comes in handy.. I can only do worse than simply citing Prof. Francis Diebold himself: If the need for predictive accuracy tests seems obvious ex post, it was not at all obvious to empirical econometricians circa 1991. Bobby Mariano and I simply noticed the defective situation, called attention to it, and proposed a rather general yet trivially-simple approach to the testing problem. And then – boom! – use of predictive accuracy tests exploded. The paper was submitted to the highly prestigious Econometrica journal and was rejected. A referee expressed bewilderment as to why anyone would care about the subject it addressed. “.. Lastly they say they always believed it.” Around a year later, an extension was published: “Asymptotic Inference About Predictive Ability” by K.D West. Eleven years after the DM paper, Econometrica published another important extension: “Tests of Conditional Predictive Ability” (2006) by Giacomini, R. and H. White (GW henceforth). For quite some time, I did not really understand the difference between DM test and GW test. Perhaps because computationally there is no difference. In a nutshell, the biggest difference between the DM and the GW tests for predictive ability is in the way they arrive to the asymptotic normal distribution. The DM proof relies on population parameters, while the GW proof lets us stay with our inaccurate sample estimates for those parameters. Practically, this means that if your forecasts are based on expanding windows, the two tests are asymptotically equivalent. However, if your forecasts are based on rolling estimation window, the DM test is not valid. Being that by using rolling window (with a constant size), you can’t assume anymore convergence to the population parameters which the proof is based on. If you are using rolling window, computation is the same but you are relying on the GW proof. 5. Controlling the False Discovery Rate: A Practical and Powerful Approach to Multiple Testing (1995), by Yoav Benjamini and Yosef Hochberg 6. This ingenious paper concerns with multiple testing. If you have 10 possible independent regressors, none of which matter, you have a good chance to find at least one is important. A good chance being 40%: prob(one or more looks important) = 1 – prob(non looks important) = Not really adequate. You generally would like to keep your error rate at 5% regardless of how many tests you make. So, over the years various suggestions have been made. A first and intuitive solution was given by Bonferroni (1936), just use 5%/10=0.005 (0.5%) for each test. This will ensure that the chance you will wrongly decide to reject the null is less than 5%. You are now controlling what they call the ‘Family-wise error rate’ (FWER). This is very conservative solution. Drawing from genetics, you will never decide a gene is important since you have so many of them (0.05/(~20,000) is a very small number). Benjamini and Hochberg (1995) suggest to relax that. They come up with a different quantity to control. where V is the number of false rejections and S is the number of correct rejections. Hence, instead of controlling the proportion of false rejections relative to the total number of tests (as in the traditional procedure), we aim to control the proportion of false rejections relative to the total number of rejections. Why is this quantity more relevant nowadays? I understand why you don’t want to be wrong when you check a couple of things, but presently you probably check more than a couple, simply because you can. The idea here is that you should perform, but you don’t have to be perfect. As long as you discover correct rejections (discover real effects, e.g. discover real market anomalies), you can make a mistake every-once-in-a-while. Every-once-in-a-while being not higher than say 1 unreal discovery for every 20 real discoveries (so confusingly, again 5%). Nice eh? Evidently 29785 citations nice. To be fair, there are a great deal more fields where multiple testing is applied, so many of those citations come from biology, medicine and so forth. Here is the link to the friendly original paper. Also in this case, the paper was not an overnight success. Far from it actually, you will be surprised to hear that it took 5 years and three journals for the paper to see the light of day. As a final word, Journals’ editors are the gate-keepers for all kinds of papers coming our way. Those are only few examples of “true negatives”, I hope and believe that we are in good shape, with filtering process that albeit often painstakingly slow, works. For those of you out there who had the courage to explore some out-of-the-box original ideas, these three examples should provide some inspiration and comfort. Original ideas may encounter strong opposition. REFERENCES (1) • Peter BÜhlmann. Bootstraps for time series. Statistical Science, pages 52-72, 2002. • M.P. Clements and J.H. Kim. Bootstrap prediction intervals for autoregressive time series. Computational statistics & data analysis, 51(7):3580-3594, 2007. • Russell Davidson and Emmanuel Flachaire. The wild bootstrap, tamed at last. Journal of Econometrics, 146(1):162-169, 2008. • B. Efron. Model selection, estimation, and bootstrap smoothing. 2012. • Bradley Efron. The bootstrap and markov-chain monte carlo. Journal of biopharmaceutical statistics, 21(6):1052–1062, 2011 • B.E. Hansen. The grid bootstrap and the autoregressive model. Review of Economics and Statistics, 81(4):594-607, 1999. • Regina Y. Liu. Bootstrap procedures under some non-iid models. The Annals of Statistics, pages 1696 – 1708, 1988. • D.B. Rubin. The bayesian bootstrap. The annals of statistics, 9(1):130-134, 1981. • Xiaofeng Shao. The dependent wild bootstrap. Journal of the American Statistical Association, 105(489):218-235, 2010. • Marco Meyer and Jens-Peter Kreiss. On the vector autoregressive sieve bootstrap. Journal of Time Series Analysis, 2014 • Joseph P Romano, Azeem M Shaikh, and Michael Wolf. Control of the false discovery rate under dependence using the bootstrap and subsampling. Test, 17(3):417–442, 2008 REFERENCES (2) • Francis X. Diebold. Comparing predictive accuracy, twenty years later: A personal perspective on the use and abuse of diebold-mariano tests. Working Paper 18391, National Bureau of Economic Research, September 2012. • Raffaella Giacomini and Halbert White. Tests of Conditional Predictive Ability, Econometrica Vol. 74, No. 6 (Nov., 2006), pp. 1545-1578 • Diebold, F.X. and R.S. Mariano (1991). Comparing Predictive Accuracy I: An Asymptotic Test. Discussion Paper 52, Institute for Empirical Macroeconomics, Federal Reserve Bank of Minneapolis • West, K.D. (1996), Asymptotic Inference About Predictive Ability, Econometrica, 64, 1067–1084. REFERENCES (3) • Benjamini, Y. and Hochberg, Y. (1995) Controlling the false discovery rate: a practical and powerful approach to multiple testing. J. R. Statist. Soc. B, 57, 289–300. • Yoav Benjamini (2010) Discovering the false discovery rate Econometrica, J. R. Statist. Soc. B, 72, Part 4, pp. 405–416	2,180	9,412	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2017-26	latest	en	0.938821
https://fr.scribd.com/document/378930800/Fisher-s-Test	1,563,928,877,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195530246.91/warc/CC-MAIN-20190723235815-20190724021815-00408.warc.gz	409,401,528	77,199	Vous êtes sur la page 1sur 42 # STAT 135 Lab 11 ## Tests for Categorical Data (Fisher’s Exact test, χ2 tests for Homogeneity and Independence) and Linear Regression ## April 20, 2015 Fisher’s Exact Test Fisher’s Exact Test ## A categorical variable is a variable that can take a finite number of possible values, thus assigning each individual to a particular group or “category”. ## We have seen categorical variables before, for example, our ANOVA factors: I pen color I gender I beer type However, for ANOVA we also had a continuous response variable. Fisher’s Exact Test ## Purely categorical data are summarized in the form of a contingency table, where the entries in the table describe how many observations fall into each grouping. For example Males 43 9 52 Females 44 4 48 Totals 87 13 100 ## The total number of individuals represented in the contingency table, is the number in the bottom right corner. Fisher’s Exact Test Males 43 9 52 Females 44 4 48 Totals 87 13 100 ## We might want to know whether the proportion of right-handed-ness is significantly different between males and females. ## We can use Fisher’s exact test to test this hypothesis. I It is called an exact test because we know the distribution of the test statistic exactly rather than just approximately or asymptotically. Fisher’s Exact Test ## Suppose that we have two categorical factors, A and B, each of which has two-levels, and the number of observations in each group is given as follows: A1 A2 Total B1 N11 N12 n1· B2 N21 N22 n2· Totals n·1 n·2 n·· ## Then under the null hypothesis that there is no difference between the proportions of the levels for factor A or B, i.e. that the two factors are independent, then the distribution of N11 is hypergeometric. Fisher’s Exact Test ## The Hypergeometric(N, K, n) distribution describes the probability of k successes in n draws, without replacement, from a finite population of size N that contains exactly K successes, wherein each draw is either a success or failure. K N −K k n−k P (X = k) = N n ## and we note that nK E(X) = N Fisher’s Exact Test A1 A2 Total B1 N11 N12 n1· B2 N21 N22 n2· Totals n·1 n·2 n·· ## N = n·· , K = n1· , n = n·1 So n1· n2· n11 n21 P (N11 = n11 ) = n·· n·1 and n·1 n1· E(N11 ) = n·· Note that under the symmetry of the problem, I could transpose the matrix and get the same answer (I usually just think about what do I want to be considered a “success”). Fisher’s Exact Test A1 A2 Total B1 N11 N12 n1· B2 N21 N22 n2· Totals n·1 n·2 n·· ## To conduct the hypothesis test, we can think about rejecting H0 for extreme values of N11 I N11 is our test statistic Fisher’s Exact Test ## A two-sided alternative hypothesis can be stated in several equivalent ways, for example: I the proportion of right-handedness differs between men and women I the proportion of women differs between right-handed people and left-handed people I right/left handedness and gender are independent ## P (\|N11 − E(N11 )\| ≥ \|n11 − E(N11 \|) Fisher’s Exact Test ## A one-sided alternative hypothesis can also be stated in several equivalent ways, for example: I the proportion of right-handedness is higher (lower) for men than for women I the proportion of men is higher (lower) for right-handed people than for left-handed people ## P (N11 ≥ n11 ) or P (N11 ≤ n11 ) Fisher’s Exact Test For our example Males 43 9 52 Females 44 4 48 Totals 87 13 100 ## N11 ∼ Hypergeoetric(100, 87, 52) n·1 n1· 52 × 87 n11 = 43 E(N11 ) = = = 45.24 n·· 100 So that our two-sided p-value is given by ## P (\|N11 −E(N11 )\| ≥ \|n11 − E(N11 \|) = P (\|N11 − 45.24\| ≥ 2.24) = P (N11 − 45.24 ≥ 2.24) + P (N11 − 45.24 ≤ −2.24) = P (N11 ≥ 47.48) + P (N11 ≤ 43) = 0.24 Exercise Exercise: Fisher’s Exact Test (Rice 13.8.19) ## I A psychological experiment was done to investigate the effect of anxiety on a person’s desire to be alone or in company. I A group of 30 subjects was randomly divided into two groups of sizes 13 and 17. I The subjects were told that they would be subject to electric shocks. I The “high anxiety” group was told that the shocks would be quite painful I The “low-anxiety” group was told that they would be mild and painless. I Both groups were told that there would be a 10-min wait before the experiment began and each subject was given the choice of waiting alone or with other subjects. The results are as follows: ## Wait Together Wait Alone Total High-Anxiety 12 5 17 Low-Anxiety 4 9 13 Total 16 14 30 ## Test whether there is a significant difference between the high- and low-anxiety groups. χ2 Test for Homogeneity χ2 Test for Homogeneity ## Suppose now that instead of a 2 × 2 table, we have an arbitrary I × J table, and we want to see if the count proportions are differently distributed across different populations. ## For example, suppose we are interested in whether TV show preference differs significantly between different age groups. 18 to 30 30 to 45 Total Game of Thrones 40 31 71 House of Cards 25 35 60 Orange is the New Black 30 23 53 Total 95 89 184 χ2 Test for Homogeneity Suppose that we have J multinomial distributions, each having I categories. If the probability of the ith category of the jth multinomial is denoted πij , the null hypothesis to be tested is ## However, we can view this as a goodness-of-fit test: Does the model prescribed by the null hypothesis fit the data? Recall that Pearson’s χ2 statistic is given by I X J 2 X (Oij − Eij )2 X = Eij i=1 j=1 I X J X (nij − ni· n·j /n·· )2 = ni· nj· /n·· i=1 j=1 χ2 Test for Homogeneity ## The degrees of freedom for the χ2 statistic are the number of independent counts minus the number of independent parameters estimated from the data. Under the assumption of homogeneity, we have I J(I − 1) independent counts, since each of the J multinomials has I − 1 independent counts. I (I − 1) independent parameters that have been estimated since the totals for each multinomial are fixed The degrees of freedom for our test statistic is thus df = J(I − 1) − (I − 1) = (I − 1)(J − 1) χ2 Test for Homogeneity ## our p-value is given by p-value = P χ2(I−1)(J−1) ≥ X 2 χ2 Test for Homogeneity ## So for our example, our observed counts are given by 18 to 30 30 to 45 Total Game of Thrones O11 = 40 O12 = 31 71 House of Cards O21 = 25 O22 = 35 60 Orange is the New Black O31 = 30 O32 = 23 53 Total 95 89 184 ## and our expected counts under H0 are given by 18 to 30 30 to 45 95×71 89×71 Game of Thrones E11 = 184 = 36.7 E12 = 184 = 34.3 95×60 89×60 House of Cards E21 = 184 = 31.0 E22 = 184 = 29.0 95×53 89×53 Orange is the New Black E31 = 184 = 27.4 E32 = 184 = 25.6 χ2 Test for Homogeneity So for our example, our observed (expected) counts are given by 18 to 30 30 to 45 Total Game of Thrones 40 (36.7) 31 (34.3) 71 House of Cards 25 (31.0) 35 (29.0) 60 Orange is the New Black 30 (27.4) 23 (25.6) 53 Total 95 89 184 ## so our test statistic is given by J I X X (Oij − Eij )2 (40 − 36.7)2 (25 − 31)2 X2 = = + Eij 36.7 31 i=1 j=1 (30 − 27.4)2 (31 − 34.3)2 + + 27.4 34.3 (35 − 29)2 (23 − 25.6)2 + + 29 25.6 ≈ 3.528 χ2 Test for Homogeneity I X J X (Oij − Eij )2 X2 = = 3.528 Eij i=1 j=1 ## and thus the p-value for the test is P χ2(I−1)(J−1) ≥ X 2 = P χ22 ≥ 3.528 = 0.17 ## which is larger than 0.05, so we fail to reject our null hypothesis and conclude that there is no significant difference between TV preferences for the different age groups. χ2 Test for Independence χ2 Test for Independence The χ2 test for independence is eerily similar to (read exactly the same as) the χ2 test for homogeneity, but is aimed at ## Suppose that a psychologist wants to test whether there is a relationship between personality and color preference. Then they are testing the null hypothesis that color preference and personality are independent. They have observed the following counts ## Blue Red Yellow Total Extroverted 5 20 5 30 Introverted 10 5 5 20 Total 15 25 10 50 ## We again use the X 2 test statistic. χ2 Test for Independence ## The degrees of freedom for the χ2 statistic are the number of independent counts minus the number of independent parameters estimated from the data. Under the assumption of independence, we have I IJ − 1 independent counts, since we have IJ cells, with any one cell entirely determined by the sum of the others. I (I − 1) + (J − 1) independent parameters that have been estimated to give the marginal probabilities that determine the expected counts The degrees of freedom for our test statistic is thus ## df = (IJ − 1) − [(I − 1) + (J − 1)] = (I − 1)(J − 1) χ2 Test for Independence ## our p-value is given by p-value = P χ2(I−1)(J−1) ≥ X 2 ## which is exactly the same as the χ2 test for homogeneity. Exercise Exercise: χ2 Test for Independence ## Determine whether there is a relationship between color preference and personality in the psychologists’s experiment ## Blue Red Yellow Total Extroverted 5 20 5 30 Introverted 10 5 5 20 Total 15 25 10 50 Simple Linear Regression Simple Linear Regression ## We have spent the last couple of weeks identifying if variables of interest have an effect on some response. Now we turn to asking the question of how our variables of interest affect the response. ## For example, we might ask “if I increase the value of variable x by one unit, what happens to the response, y?”. ## Simple linear regression refers to the case when we have only one explanatory variable, x. Simple Linear Regression The idea behind linear regression by least squares is to identify the line of best fit when we plot y against x by minimizing the sum of the squared distances from the points to the line. y = β0 + β1 x ## and our goal is to estimate the intercept β0 and the slope β1 using our observed data. Simple Linear Regression The statistical model corresponding to our data is given by yi = β0 + β1 xi + i ## where i corresponds to the unobserved random noise which explains the deviation from the line β0 + β1 xi . i satisfies E(i ) = 0 and V ar(i ) = σ 2 (xi is considered fixed). ## We estimate β0 and β1 by finding the values, β̂0 and β̂1 , that minimize the following equation n X S(β0 , β1 ) = (yi − β0 − β1 xi )2 i=1 ## which corresponds to minimizing the vertical distance from each observed point, yi , to the corresponding point on the fitted line β̂0 + β̂1 xi (i.e. we want to minimize the residuals). Exercise Exercise: Simple Linear Regression (Rice 14.9.10) ## Show that the values of β0 and β1 that minimize n X S(β0 , β1 ) = (yi − β0 − β1 xi )2 i=1 are given by β̂0 = y − β̂1 x Pn (x − x)(yi − y) βˆ1 = Pn i i=1 2 i=1 (xi − x) Simple Linear Regression ## Note that these estimators can be written as β̂0 = y − β̂1 x Pn (xi − x)(yi − y) r Cov(x, y) sxy sy βˆ1 = i=1 Pn 2 = = =ρ i=1 (xi − x) V ar(x) sx sx s where ρ = √sxy x sy is the correlation coefficient between x and y. Note that \|ρ\| ≤ 1 The correlation describes how much of a linear relationship there is between x and y. A strong linear relationship will have \|ρ\| close to 1 (however the converse is not necessarily true – always make plots to check!) Simple Linear Regression Some theoretical results about these estimators: β̂0 and β̂1 are unbiased: E(β̂j ) = βj ## Moreover, we can calculate the variance and covariance of these estimators to find that σ 2 ni=1 x2i P V ar(β̂0 ) = Pn n i=1 x2i − ( ni=1 xi )2 P nσ 2 V ar(β̂1 ) = − ( ni=1 xi )2 Pn 2 P n i=1 xi −σ 2 ni=1 xi P Cov(β̂0 , β̂1 ) = Pn n i=1 x2i − ( ni=1 xi )2 P ## Question: where does the randomness come from? Simple Linear Regression ## To estimate the variance of the estimators, we need to know σ 2 , which we rarely do. We can, however, estimate σ 2 . Recall that our model is yi = β0 + β1 xi + i where i ∼ N (0, σ 2 ). ## Then we can rearrange as follows: i = yi − β0 − β1 xi ## so perhaps we can estimate σ 2 by looking at the variances of the i ... but we don’t observe i ! Simple Linear Regression i = yi − β0 − β1 xi We can estimate i by substituting βb0 and βb1 , so we define the ith residual to be ei = yi − βb0 − βb1 xi Pn 2 σ b = = i=1 i n−p n−p ## where p is the number of variables we have (p = 2 for simple linear regression) Exercise Exercise: Simple Linear Regression The faithful dataset in R contains data on (1) the waiting time (in mins) between eruptions and the duration of the eruption (in mins) for the Old Faithful geyser in Yellowstone National Park, Wyoming, USA. We have 272 observations. Exercise: Simple Linear Regression ## The faithful dataset in R contains data on (1) the waiting time (in mins) between eruptions and the duration of the eruption (in mins) for the Old Faithful geyser in Yellowstone National Park, Wyoming, USA. We have 272 observations. I Plot duration versus waiting time. I Apply the simple linear regression model where we are regressing duration on waiting time. I Report the fitted regression line, as well as the variance and covariance of the parameter estimates. I Estimate the next eruption duration if the waiting time since the last eruption has been 80 minutes ago.	4,041	13,120	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2019-30	latest	en	0.859912
http://forums.wolfram.com/mathgroup/archive/2010/Oct/msg00076.html	1,590,751,011,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00330.warc.gz	48,196,668	7,861	What assumptions to use to check for orthogonality under integration? • To: mathgroup at smc.vnet.net • Subject: [mg112868] What assumptions to use to check for orthogonality under integration? • From: "Nasser M. Abbasi" <nma at 12000.org> • Date: Mon, 4 Oct 2010 06:06:03 -0400 (EDT) ```This is basic thing, and I remember doing this or reading about it before. I am trying to show that Cos[m Pi x], Cos[n Pi x] are orthogonal functions, m,n are integers, i.e. using the inner product definition Integrate[Cos[nPix]Cos[mPix], {x, 0, 1}]; So, the above is ZERO when n not equal to m and 1/2 when n=m. hence orthogonal functions. This is what I tried: ------ case 1 ------------- Clear[n, m, x] r = Integrate[Cos[nPix]Cos[mPix], {x, 0, 1}]; Assuming[Element[{n, m}, Integers], Simplify[r]] Out[167]= 0 ---------------- I was expecting to get a result with conditional on it using Piecewise notation. Then I tried ---------case 2 ------------ Clear[n, m, x] r = Integrate[Cos[nPix]Cos[mPix], {x, 0, 1}]; Assuming[Element[{n, m}, Integers] && n != m, Simplify[r]] Out[140]= 0 Assuming[Element[{n, m}, Integers] && n == m, Simplify[r]] Out[184]= Indeterminate ---------------- So, it looks like one has to do the limit by 'hand' to see that for n=m we get non-zero? ------------------- Clear[n, m, x] r = Integrate[Cos[nPix]Cos[mPix], {x, 0, 1}]; Limit[Limit[r, n -> m], m -> 1] Out[155]= 1/2 Limit[Limit[r, n -> 1], m -> 99] Out[187]= 0 ---------------------------- So, is there a way to get Mathematica to tell me that the integral is zero for m!=n and 1/2 when n=m? (tried Reduce, Refine). It seems the problem is that the Integrate is not taking the limit automatically to determine what happens when n=m? Should it at least in case have told me that when n!=m it is zero, and when n=m it is Indeterminate? It just said zero which is not correct when n=m and I did say n,m are integers. thanks --Nasser ``` • Prev by Date: double loop do • Next by Date: Re: NMINIMIZATION WITH COMPLEX CONJUGATE • Previous by thread: Re: double loop do • Next by thread: How to interpolate vectors?	667	2,125	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2020-24	longest	en	0.777022
https://cses.fi/241/result/142587	1,701,463,554,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00119.warc.gz	229,986,957	6,055	CSES - NOI 2019 Open - Results Task: Distance Code Sender: knutalos Submission time: 2019-03-07 19:20:09 +0200 Language: Python3 Status: READY Result: 0 Feedback groupverdictscore #10 #20 #30 Test results testverdicttimegroup #10.04 s1, 2, 3details #20.04 s1, 2, 3details #30.05 s1, 2, 3details #40.04 s1, 2, 3details #50.04 s1, 2, 3details #60.04 s1, 2, 3details #70.04 s1, 2, 3details #80.05 s1, 2, 3details #90.05 s1, 2, 3details #100.04 s1, 2, 3details #110.04 s1, 2, 3details #120.05 s2, 3details #130.07 s2, 3details #140.04 s2, 3details #150.05 s2, 3details #16--3details #17--3details #180.75 s3details #19--3details #200.04 s1, 2, 3details ### Code ```import sys encode = int(sys.stdin.readline()) == 1 for i in range(n): if encode: for i in range(n - 1): toblader = [int(x) - 1 for x in sys.stdin.readline().split()] while len(blader) != n: for i, verdi in enumerate(superblader): if verdi == None: continue if len(verdi) == 0: elif len(verdi) == 1: """for key, value in motsattblad.items(): if len(value) <= 1: for verdi in value: motsattblad[verdi] = [x for x in motsattblad[verdi] if x != key]""" #print(key,value, len(value)) else: exit("Decode")``` ### Test details Group: 1, 2, 3 Verdict: input 1 2 2 1 correct output (empty) user output [0, 1] Group: 1, 2, 3 Verdict: input 1 3 3 1 2 1 correct output (empty) user output [1, 2, 0] Group: 1, 2, 3 Verdict: input 1 4 3 2 2 1 4 1 correct output (empty) user output [2, 3, 0, 1] Group: 1, 2, 3 Verdict: input 1 4 2 3 3 4 1 3 correct output (empty) user output [0, 1, 2, 3] Group: 1, 2, 3 Verdict: input 1 5 3 5 4 1 1 3 ... correct output (empty) user output [1, 3, 4, 0, 2] Group: 1, 2, 3 Verdict: input 1 5 3 2 3 4 5 1 ... correct output (empty) user output [1, 3, 4, 0, 2] Group: 1, 2, 3 Verdict: input 1 5 4 3 1 4 4 2 ... correct output (empty) user output [0, 1, 2, 3, 4] #### Test 8 Group: 1, 2, 3 Verdict: input 1 10 9 3 8 9 2 9 ... correct output (empty) user output [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] #### Test 9 Group: 1, 2, 3 Verdict: input 1 10 9 2 5 8 7 1 ... correct output (empty) user output [3, 4, 7, 9, 2, 5, 0, 1, 6, 8] #### Test 10 Group: 1, 2, 3 Verdict: input 1 10 10 4 9 1 4 7 ... correct output (empty) user output [2, 4, 7, 8, 9, 0, 1, 3, 5, 6] #### Test 11 Group: 1, 2, 3 Verdict: input 1 10 2 6 4 3 3 5 ... correct output (empty) user output [3, 8, 9, 2, 4, 5, 6, 7, 0, 1] #### Test 12 Group: 2, 3 Verdict: input 1 500 10 6 6 255 6 428 ... correct output (empty) user output [0, 1, 2, 3, 4, 6, 7, 8, 9, 10... #### Test 13 Group: 2, 3 Verdict: input 1 500 152 466 451 313 158 479 ... correct output (empty) user output [71, 105, 188, 378, 425, 36, 3... #### Test 14 Group: 2, 3 Verdict: input 1 500 109 440 330 190 443 161 ... correct output (empty) user output [1, 2, 4, 5, 6, 7, 9, 11, 12, ... #### Test 15 Group: 2, 3 Verdict: input 1 500 144 373 257 233 341 318 ... correct output (empty) user output [24, 27, 32, 38, 51, 73, 77, 1... Group: 3 Verdict: input 1 100000 54983 75172 93807 75172 44082 75172 ... correct output (empty) user output (empty) Group: 3 Verdict: input 1 100000 88863 19059 86423 76688 98536 95984 ... correct output (empty) user output (empty) #### Test 18 Group: 3 Verdict: input 1 100000 59979 6389 19097 24999 27846 82330 ... correct output (empty) user output [1, 3, 4, 5, 6, 8, 9, 10, 11, ... Group: 3 Verdict: input 1 100000 58761 66001 25102 51081 98625 67861 ... correct output (empty) user output (empty) #### Test 20 Group: 1, 2, 3 Verdict: input 1 6 2 1 3 2 4 2 ... correct output (empty) user output [0, 3, 4, 5, 1, 2]	1,696	3,664	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2023-50	latest	en	0.273536
https://concensure.com/excel-functions-and-formulas/	1,611,421,179,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00695.warc.gz	268,214,826	15,035	Select Page ### Working with Basic Functions Figuring out formulas for calculations you want to make in Excel can be tedious and complicated. Fortunately, Excel has an entire library of functions or predefined formulas that you can take advantage of. You may be familiar with common functions like sum, average, product or count, but there are hundreds of functions in Excel, even for things like formatting text, referencing cells, calculating financial rates, analyzing statistics, and more. In this lesson, you will learn the basics of inserting common functions into your worksheet by utilizing the AutoSum and Insert Functions commands. You will also become familiar with how to search and find various functions, including exploring Excel’s Functions Library A function is a predefined formula that performs calculations using specific values in a particular order. One of the key benefits of functions is that they can save you time since you do not have to write the formula yourself. Excel has hundreds of different functions to assist with your calculations. In order to use these functions correctly, you need to understand the different parts of a function and how to create arguments in functions to calculate values and cell references. ### The Parts of a Function The order in which you insert a function is important. Each function has a specific order, called syntax, which must be followed for the function to work correctly. The basic syntax to create a formula with a function is to insert anequal sign (=), a function name (SUM, for example, is the function name for addition), and an argument. Arguments contain the information you want the formula to calculate, such as a range of cell references. ### Working with Basic Arguments Arguments must be enclosed in parentheses. Individual values or cell references inside the parentheses are separated by either colons or commas. Colons create a reference to a range of cells. For example, =AVERAGE(E19:E23) would calculate the average of the cell range E19 through E23. Commas separate individual values, cell references, and cell ranges in the parentheses. If there is more than one argument, you must separate each argument by a comma. For example, =COUNT(C6:C14,C19:C23,C28) will count all the cells in the three arguments that are included in parentheses. ### Create a Basic Function in Excel To Create a Basic Function in Excel: 1. Select the cell where the answer will appear (F15, for example) 2. Type the equal sign (=) and enter the function name (SUM, for example). 3. Enter the cells for the argument inside the parenthesis. 4. Press Enter and the result will appear. Result	572	2,671	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2021-04	latest	en	0.811472
https://user.services.openoffice.org/en/forum/viewtopic.php?f=75&t=98212	1,610,745,730,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00259.warc.gz	614,937,593	8,583	## [Tutorial] Calc formula terms Forum rules No question in this forum please For any question related to a topic, create a new thread in the relevant section. ### [Tutorial] Calc formula terms Note that questions are not allowed in the Tutorials section of the forum. Ask them in the Applications → Calc section. In this post the word apostrophe always means Unicode character U+0027: ' In this post the word quote always means Unicode character U+0022: " ‘ ↖ Those ↗ are not apostrophes: U+2018/U+2019 (Left/Right Single Quotation Mark) ↖ Those ↗ are not quotes: U+201C/U+201D (Left/Right Double Quotation Mark) OpenOffice Calc formulas can contain the following terms: • A numeric constant A sequence of digits with at most one decimal separator, optionally followed by this sequence: E or e, an optional + or -, and at least one digit If a decimal separator is used — period for English (USA) locale — the constant is locale dependent The decimal separator for a locale is normally a period or a comma All numeric constants are positive; one uses the negation operator to obtain negative numbers Numeric constants can be written to be locale-independent by using exponential notation d…Ennn (this represents d… times 10ⁿⁿⁿ) or d…E-nnn (this represents d… divided by 10ⁿⁿⁿ) The following constants are written differently but all have the same value `Example: 3.14159265358979 (for locale using period for decimal separator)Example: 3,14159265358979 (for locale using comma for decimal separator)Example: 314159265358979E-14 (locale independent)` • A text constant A quote, followed by zero or more Unicode characters, followed by a quote The value of the constant does not include those two surrounding quotes A quote within the constant is represented as two quotes: "" `Example: "Hello world!" (12 characters)Example: "" (0 characters: the empty string)Example: """" (1 character: a quote)Example: """\""" (3 characters: quote backslash quote)` • A boolean constant (locale dependent) Calc converts these locale dependent constants to 0 (for FALSE) and 1 (for TRUE) `Example: TRUE [English (USA) locale]Example: FALSE [English (USA) locale]` • An array constant A {, followed by zero or more of the terms above separated by ; or \|, followed by } ; is used to separate terms in a row \| is used to begin a new row All rows must have the same number of columns All columns must have the same number of rows `Example: {0;1;2;3;4;5;6;7;8;9} (1 row, 10 columns)Example: {0\|1\|2\|3\|4\|5\|6\|7\|8\|9} (10 rows, 1 column)Example: {"Nineteen";19\|"Fifty two";52} (2 rows, 2 columns)` • A cell reference An optional sheet reference terminated by a period, an optional \$, a column name, an optional \$, a row number Column names are A through AMJ Row numbers are 1 through 1048576, or 1 through 65536 before OpenOffice 3.3 Use of the \$ is explained in 8. Using formulas and cell references in Ten concepts that every Calc user should know. Cell references are to single cells; cell ranges are created by using the : operator In the sentences below, Σ represents a sheet name References to sheets in the same spreadsheet use an optional \$ followed by a sheet name Σ If the sheet name contains other than letters, digits, or underscores, surround the name with apostrophes If the sheet name begins with a digit, surround the name with apostrophes If the sheet name contains a an apostrophe, use two apostrophes to represent it `Example: \$A\$1 (upper left cell)Example: \$AMJ\$65536 (lower right cell, before OpenOffice 3.3)Example: \$AMJ\$1048576 (lower right cell, OpenOffice 3.3 and later)Example: \$Sheet1.A1 (reference to cell on Sheet1)Example: Sheet1.\$A1 (reference to cell on previous sheet when used on Sheet2)Example: 'Sheet One'.A\$1 (sheet name containing a space)Example: 'Dad''s Sheet'.A1 (sheet name containing an apostrophe)` References to sheets in other spreadsheets are 'file://Φ'#\$Σ where Φ is a file name If the file name contains a an apostrophe, use two apostrophes to represent it `Example: 'file:///Users/Guest User/Desktop/Data.ods'#\$Sheet1.A1Example: 'file:///Users/Guest User/Desktop/Dad''s Data.ods'#\$Sheet1.A1Example: 'file:///Users/Guest User/Desktop/Dad''s Data.ods'#\$'Dad''s Sheet'.A1Example: 'file:///C:/Users/Bozo/Desktop/Data.ods'#\$Sheet1.A1` References to "cells" in CSV files always use Sheet1 for the sheet name `Example: 'file:///Users/Guest User/Desktop/Data.csv'#\$Sheet1.A1` • A defined name (created with Insert → Names → Define) A letter or underscore followed by a sequence of letters, digits, underscores, and periods The name may not end in a period nor look like a cell reference nor be R, RC, or C The name may not begin with R followed by digits followed by C `Example: InventoryExample: March.InventoryExample: _` • A row/column label range An apostrophe, followed by Unicode characters (except apostrophe), followed by an apostrophe The option Calc → Calculate → Automatically find row and column labels must be enabled Defined names offer much of the same function as row/column label ranges but with additional features `Example: 'Expenses'Example: 'My Expenses'` • Functions (locale dependent) A letter, optionally followed by letters, digits, underscores, and periods, a (, zero or more parameters separated by ;, a ) Parameters can be functions or any of the terms above Entering the function's name in the wrong language for the locale will produce a #NAME? error `Example: TODAY() [English (USA) locale, current date]Example: AUJOURDHUI() [French (France) locale, current date]Example: HEUTE() [German (Germany) locale, current date]Example: HOY() [Spanish (Spain) locale, current date]Example: IF(A2=B2;"Equal";IF(A2<B2;"A<B";"A>B")) [English (USA) locale, nested IF functions]` • Calc's Operators: +, :, !, ~, -, %, ^, /, *, +, -, &, =, >, >=, <>, <, <= See [Tutorial] Order of Operations in Calc • Parentheses ( and ) are used to modify the standard order of operations A parenthesized expression is treated as a single term of a formula Warning: LibreOffice provides options to change the array constant separators and the function parameter separators, so syntax in LibreOffice formulas may differ from what is shown here. Note that questions are not allowed in the Tutorials section of the forum. Ask them in the Applications → Calc section. Mr. Programmer AOO 4.1.7 Build 9800 on MacOS 10.14.6. The locale for any menus or Calc formulas in my posts is English (USA). MrProgrammer Moderator Posts: 3854 Joined: Fri Jun 04, 2010 7:57 pm Location: Wisconsin, USA	1,717	6,813	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.03125	3	CC-MAIN-2021-04	latest	en	0.796767
https://www.physicsforums.com/threads/relative-velocity.43438/	1,524,311,597,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125945143.71/warc/CC-MAIN-20180421110245-20180421130245-00309.warc.gz	862,138,021	16,006	Relative velocity 1. Sep 17, 2004 FateMaster Hmm... This is a chp in a-maths... I appear to be having a mental block regarding the vector diagram thingy.... Help me plz ??? Thx.... 2. Sep 17, 2004 Integral Staff Emeritus 3. Sep 18, 2004 FateMaster Realli ??? Hmm... Well... We can start with the basics... i dun even understand the basics of this topic... So... i guess you can start from there..... ^^ 4. Sep 18, 2004 recon For those of you not in the know, A-maths stands for Additional Mathematics, which is an additional subject you can take in Singapore/Brunei if you're really good at maths. FateMaster is having trouble understanding vector diagrams constructed for relative velocity. FateMaster, you really need to be more specific in posting your problems. Understand that most people in PF (Physics Forums) are not from Singapore, or any South East Asian country. I'm from Brunei, though, so I perfectly understand your post. Unfortunately, we've yet to cover that topic in our A-Maths class. We might be using the same textbook. The textbook that I'm using is "New Additional Mathematics" by Ho Soo Thong and Khor Nyak Hiong. Is there something in the book that is badly explained? If so, let me know. I might be able to help. EDIT: I'm just curious. Are you in Secondary 3 or 4? Last edited: Sep 18, 2004 5. Sep 18, 2004 PRodQuanta x-----> x-----> x-----> x-----> This shows a car (x) with a constant velocity using vector diagrams. x x-> x--> x---> x----> x----------> This shows a car (x) accelerating using a vector diagram. EDIT: maybe try this website to refresh your memmory. http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/kinema/avd.html [Broken] Last edited by a moderator: May 1, 2017 6. Sep 18, 2004 recon PRodQuanta, FateMaster is talking about vector diagrams used in finding relative velocity. FateMaster, it may be helpful to look around in the homework forums. As I recall, there were quite a lot of problems being posted regarding this topic not too long ago. 7. Sep 18, 2004 PRodQuanta AH, i see. Misunderstanding. 8. Sep 18, 2004 PRodQuanta This should refresh your memmory then: http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l1f.html [Broken]	610	2,222	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2018-17	longest	en	0.90657
http://riverware.org/HelpSystem/Objects/AppendixTableInterpolation.html	1,563,463,829,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00001.warc.gz	122,312,508	5,454	Table Interpolation Objects and Methods Table Interpolation This document describes the algorithms employed by RiverWare for approximating continuous functions using linear interpolation between a sample of points defining the function. These data are input by the user into data tables (Table Slots), thus we refer to this approach to function approximation as table interpolation. Table interpolation occurs in the following contexts: • Execution of some user methods during simulation. For example, the Plant Efficiency Curve method in the Power Calculation category on power reservoirs performs a three-dimensional table interpolation on the Plant Power Table. • An Optimization run, as part of the process of approximating nonlinear functions as one or more linear functions. For example, linearizing which used the Stage Flow Tailwater Table on power reservoirs. • A Rules or Accounting run in which a RPL block executes the interpolation function. For example, the user could create a table on a data object representing evaporation as a function of temperature. Within a rule one could use two-dimensional interpolation to find the evaporation corresponding to any temperature in the range of this table. RiverWare supports table interpolation for functions of two and three dimensions. For functions of two variables it is useful to refer to the table columns as x and y, where x is the independent variable and y is the dependent variable: y = f(x). For functions of three variables, we refer to the columns as x, y, and z, where y = f(x, z). We denote a particular approximation using the table by an asterisk: y* = f(x) or y = f(x, z) Two-dimensional data format For two-dimensional interpolation RiverWare assumes that the values in the x column of the data table are increasing. Table B.1 is an example of the proper way to formulate a table for two-dimensional interpolation. Table B.1 Elevation volume table for a reservoir Pool Elevation (ft) Storage (acre-ft) 440 439,400 441 455,900 442 472,600 443 489,600 445 507,000 Two-dimensional table interpolation For two-dimensional functions, we apply linear interpolation between data points. Figure B.1 illustrates this approach. Figure B.1 Two-dimensional linear interpolation The following types of errors may be reported during two-dimensional table interpolation: • Invalid value (data error): an x or y value is invalid (xi = NaN or yi = NaN, for some i). • Non-increasing x (data error): the x values are not increasing (xi >= xi-1, for some i). • Out of range (interpolation error): the x value being interpolated is out of the range of the table, that is, the domain of the function being approximated (x* < xmin or x* > xmax). Three-dimensional data format For three-dimensional interpolation, the z values define blocks: each block has a constant z value and increasing x values, and the blocks are arranged in order of increasing z value. In other words, the three dimensional surface is represented by multiple slices or contours in the x-y plane, each of which may be represented by any arbitrary number of data points, just as with ordinary two-dimensional curves. Table B.2 is an example of the proper way to formulate a table for three-dimensional interpolation. Table B.2 Plant power table for a power reservoir Turbine Release (cfs) Power (kW) 100 0 0 100 10 2000 100 20 3000 100 30 4000 200 0 0 200 10 2500 200 20 3500 200 25 3800 200 30 4500 300 0 0 300 10 3000 300 25 5000 Three-dimensional table interpolation For three-dimensional functions, the algorithm for interpolation has two basic cases. If the z value being interpolated is equal to the z value for one of the blocks in the table, then we just perform a two-dimensional interpolation along the curve represented by that block. When the z value is not exactly equal to any of the z values found in the table, RiverWare first identifies the constant z-blocks whose values bound the z value being interpolated and performs a two-dimensional interpolation along these curves. This yields two points, one on each bounding constant z-curve, and the final answer is computed by a linear interpolation between these two points. Figure B.2 illustrates this case. Figure B.2 Three-dimensional linear interpolation There is one special case in which the interpolation behavior is slightly different: when the x value being interpolated is within the domain of one of the bracketing constant z curves but not the other. In this case, we interpolate between the encompassing curve and the extrapolation of the other (shorter) curve. We extrapolate this curve with either the slope of its last segment or the slope of the corresponding segment of the encompassing curve, as appropriate for the particular table. To avoid over-ambitious extrapolation, RiverWare requires that the answer lie in the region bounded by the constant-z curves (i.e., their convex hull). Figure B.3 illustrates this case, where the short curve is extrapolated with the slope of its last segment. Figure B.3 Three-dimensional linear interpolation The following types of errors may be reported during three-dimensional table interpolation: • Invalid value (data error): an x, y, or z value is invalid (xi = NaN, yi = NaN, or zi = NaN for some i). • Non-increasing z (data error): the z values are not increasing for one block to another (zi >= zi-1, for some i). • Non-increasing x (data error): the x values are not increasing (xi >= xi-1, for some i). • z value out of range (interpolation error): the z value being interpolated is out of the range of the table (z* < zmin or z* > zmax). • x value out of range (interpolation error): the x value being interpolated is out of the domain of both of the two bounding constant z-curves. Revised: 06/03/2019	1,286	5,771	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2019-30	latest	en	0.859839
https://www.ademcetinkaya.com/2023/02/saia-saia-inc-common-stock.html	1,679,913,948,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00000.warc.gz	697,621,812	60,738	Outlook: Saia Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Dominant Strategy : Sell Time series to forecast n: 21 Feb 2023 for (n+6 month) Methodology : Deductive Inference (ML) ## Abstract Saia Inc. Common Stock prediction model is evaluated with Deductive Inference (ML) and Wilcoxon Rank-Sum Test1,2,3,4 and it is concluded that the SAIA stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell ## Key Points 1. What statistical methods are used to analyze data? 2. What is prediction in deep learning? 3. How accurate is machine learning in stock market? ## SAIA Target Price Prediction Modeling Methodology We consider Saia Inc. Common Stock Decision Process with Deductive Inference (ML) where A is the set of discrete actions of SAIA stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4 F(Wilcoxon Rank-Sum Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Deductive Inference (ML)) X S(n):→ (n+6 month) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$ n:Time series to forecast p:Price signals of SAIA stock j:Nash equilibria (Neural Network) k:Dominated move a:Best response for target price For further technical information as per how our model work we invite you to visit the article below: How do AC Investment Research machine learning (predictive) algorithms actually work? ## SAIA Stock Forecast (Buy or Sell) for (n+6 month) Sample Set: Neural Network Stock/Index: SAIA Saia Inc. Common Stock Time series to forecast n: 21 Feb 2023 for (n+6 month) According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell X axis: Likelihood% (The higher the percentage value, the more likely the event will occur.) Y axis: Potential Impact% (The higher the percentage value, the more likely the price will deviate.) Z axis (Grey to Black): Technical Analysis% ## IFRS Reconciliation Adjustments for Saia Inc. Common Stock 1. A hedge of a firm commitment (for example, a hedge of the change in fuel price relating to an unrecognised contractual commitment by an electric utility to purchase fuel at a fixed price) is a hedge of an exposure to a change in fair value. Accordingly, such a hedge is a fair value hedge. However, in accordance with paragraph 6.5.4, a hedge of the foreign currency risk of a firm commitment could alternatively be accounted for as a cash flow hedge. 2. Paragraph 5.7.5 permits an entity to make an irrevocable election to present in other comprehensive income subsequent changes in the fair value of particular investments in equity instruments. Such an investment is not a monetary item. Accordingly, the gain or loss that is presented in other comprehensive income in accordance with paragraph 5.7.5 includes any related foreign exchange component. 3. If such a mismatch would be created or enlarged, the entity is required to present all changes in fair value (including the effects of changes in the credit risk of the liability) in profit or loss. If such a mismatch would not be created or enlarged, the entity is required to present the effects of changes in the liability's credit risk in other comprehensive income. 4. For a discontinued hedging relationship, when the interest rate benchmark on which the hedged future cash flows had been based is changed as required by interest rate benchmark reform, for the purpose of applying paragraph 6.5.12 in order to determine whether the hedged future cash flows are expected to occur, the amount accumulated in the cash flow hedge reserve for that hedging relationship shall be deemed to be based on the alternative benchmark rate on which the hedged future cash flows will be based. International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS. ## Conclusions Saia Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Saia Inc. Common Stock prediction model is evaluated with Deductive Inference (ML) and Wilcoxon Rank-Sum Test1,2,3,4 and it is concluded that the SAIA stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell ### SAIA Saia Inc. Common Stock Financial Analysis* Rating Short-Term Long-Term Senior OutlookBa1Ba1 Income StatementB3Baa2 Balance SheetBaa2Baa2 Leverage RatiosBaa2Ba2 Cash FlowCB3 Rates of Return and ProfitabilityBaa2Baa2 Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents. How does neural network examine financial reports and understand financial state of the company? ### Prediction Confidence Score Trust metric by Neural Network: 82 out of 100 with 631 signals. ## References 1. O. Bardou, N. Frikha, and G. Pag`es. Computing VaR and CVaR using stochastic approximation and adaptive unconstrained importance sampling. Monte Carlo Methods and Applications, 15(3):173–210, 2009. 2. J. Baxter and P. Bartlett. Infinite-horizon policy-gradient estimation. Journal of Artificial Intelligence Re- search, 15:319–350, 2001. 3. Christou, C., P. A. V. B. Swamy G. S. Tavlas (1996), "Modelling optimal strategies for the allocation of wealth in multicurrency investments," International Journal of Forecasting, 12, 483–493. 4. J. Ott. A Markov decision model for a surveillance application and risk-sensitive Markov decision processes. PhD thesis, Karlsruhe Institute of Technology, 2010. 5. Bengio Y, Schwenk H, Senécal JS, Morin F, Gauvain JL. 2006. Neural probabilistic language models. In Innovations in Machine Learning: Theory and Applications, ed. DE Holmes, pp. 137–86. Berlin: Springer 6. Bengio Y, Schwenk H, Senécal JS, Morin F, Gauvain JL. 2006. Neural probabilistic language models. In Innovations in Machine Learning: Theory and Applications, ed. DE Holmes, pp. 137–86. Berlin: Springer 7. Friedman JH. 2002. Stochastic gradient boosting. Comput. Stat. Data Anal. 38:367–78 Frequently Asked QuestionsQ: What is the prediction methodology for SAIA stock? A: SAIA stock prediction methodology: We evaluate the prediction models Deductive Inference (ML) and Wilcoxon Rank-Sum Test Q: Is SAIA stock a buy or sell? A: The dominant strategy among neural network is to Sell SAIA Stock. Q: Is Saia Inc. Common Stock stock a good investment? A: The consensus rating for Saia Inc. Common Stock is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating. Q: What is the consensus rating of SAIA stock? A: The consensus rating for SAIA is Sell. Q: What is the prediction period for SAIA stock? A: The prediction period for SAIA is (n+6 month)	1,848	7,485	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2023-14	longest	en	0.824395

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