can anyone explain how to make a box plot easily! I have a test otday and i still dont understand how to get q
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Your boxplot is based on the '5 number summary', that is... \n1. The minimum value\n2. The 1st quartile (25th percentile)\n3. The median (called the 2nd quartile)\n4. The 3rd quartile (75th percentile) and..\n5. Tthe maximum.\n\nThe 1st, 2nd and 3d quartiles make up the box, and the lines extend from the top and bottom ends of the box to the maximum and minimum.
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1. Start with a new workbook and enter your task data as in Sheet1. Column A contains the task descriptions, column B contains the beginning date for each task, and column C contains the number of days to complete the task. Column D contains formulas thhat determine the ending day for each task. For example, the formula in cell D2 is =B2-C2-1. Column D isn’t really necessary (and is not used in the chart), but including it will tell you end dates easily.\n\n2. Create a stacked horizontal chart from the data in A2:C13. The Chart Wizard will probably guess these series incorrectly, so you’ll need to set the category axis labels and data series manually. The category (x) axis labels should be range A2:A13, the Series 1 data should be B2:B13, and Series 2 data should be C2:C13. \n\n3. Remove the chart’s legend and adjust the height of the chart so that all of the category axis labels are visible (or, change the font to a smaller size). \n\n4. Access the Format Axis dialog box for the horizontal (value) axis. Set the Minimum and Maximum values to correspond to the earliest and latest dates in your project. Note that you can\nenter actual dates into this dialog box. To display the dates at weekly intervals, set the Minimum to a Monday, the Maximum to a Sunday, and the Major Unit to 7. \n\n5. Select the data series that corresponds to the data in Column B and access the Format Data Series dialog box. Set Borders to None and Area to None. This essentially makes the first data\nseries invisible – and makes the chart resemble a Gantt chart.\n\n6. Apply other formatting as desired. For example, you may want to add gridlines and a title. You’ll be able to adjust your project schedule and the chart will update automatically. However, if you use dates outside of the original date range, you’ll need to change the scaling for the horizontal (value) axis.
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In the given qts, the data provided is typical for using an orthodox yet nice method called L.P.P. (Linear programming) which is actually a graph method with the cartesian co-ordinates. I am sorry but i won't be able to provide you with the graph solution. Here is the theory of the required solution.\n\nLet the no of matches per box in the first case be x\nLet the no of matches per box in the second case be (x-y) where y is the number of matches decremented per box.\n\nNumber of matches in first case = 24x\nNumber of matches in second case = 28(x-y)\n\nHowever number of match boxes are equal\nHence, 24x = 28(x-y)\n       24x = 28x - 28y\n       28y = 4x\n        7y = x  ----------- (1)\n\nNote that since the total number of matches is not = 0\n hence,   x > y \n          x > 0\n          y > 0  \n and x,y are integers ----- (2)\n\nHence, x is minimum when y is minimum.\nMinimum integral value of y is 1\nHence minimum value of x = 7y = 7*1 = 7 ----- (3)\n\nFrom (1), (2) and (3), we conclude that...\n\nTotal number of matches required for the given qts should have a minimum value         \n   = 24x \n   = 24*7 \n   = 168   ........ (ANSWER)\n\nThe data provided by you is sufficent for finding minimum no of matches required since that's the limiting case. If you have a more precise condition in the given qts, a unique solution shall be obtained for the same. Graphs would have made the solution more interesting.\n\nI saw someone getting an answer as 672. It's possible if there is a condition in the given qts that "When the number of matches in each box are decremented, number of boxes required are 28. However, the number of matches placed in each box is found to be four less than the intial number of matches placed in each box". Then the solution would be like this :\n\nLet the no of matches per box in the first case be x\nHence, the no of matches per box in the second case be (x-4) \n\nNumber of matches in first case = 24x\nNumber of matches in second case = 28(x-4)\n\nHowever number of match boxes are equal\nHence, 24x = 28(x-4)\n       24x = 28x - 112\n        4x = 112\n         x = 28 \n\nClearly, like in the previous solution, in this case too, \n \n   x > 4\n  and its obvious that x > 0 and x is an integer.\n\nHence, the unique number of matches required in this case is \n         = 24x \n         = 24 * 28 \n         = 672 ... (ANSWER)\n\nI hope that this solution serves its purpose and helps you conquer this question satisfactorily.
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Try to fit your data with Excel’s exponential fitting curves,\nIt will give you Coefficients which make the exponential looks like a line.\n\nOr try the experiment more rapidly.
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Try this, http://www.collegeboard.com/appt/qotd/question
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Make one on a spreadsheet
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There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
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QUESTION1 -- Profits = Revenue - Cost, so you want to come up with an equation for profits using R(x) and C(x), then use differentiation to determine your maximum.  Let P(x) = profits.  So:\n\nP(x) = R(x) - C(x) =\n50x - 0.5x² - 9x - 3 =\n-0.5x² + 41x - 3\n\nP '(x) = -x + 41\nP '(41) = 0  << so you have a min or max at x=41\n\nP ''(x) = -1\nP ''(41) = -1  << tells us x=41 is a max\n\nSo, answer for #1 = Maximum profit at x=41\n\nQUESTION 2 -- Let x = side of base, so L and W each = x since the base is a square.  The height is unknown, but we can solve for it since we know the volume.\n\nV = 57 ft³ = x²h,\nso h = 57/x²\n\nWe want to maximize surface area, so let's first get our formula (all in terms of x) for the surface area:\n\nS(x) = x² + 4(x*57/x²) = x² + 228/x\n\nMaximize same way we did above using differentiation:\n\nS '(x) = 2x -228/x²\n\nS '(x) = 0 when 2x = 228/x², i.e., when x³ = 114; i.e., when x = cube.rt(114) ft.\n\nSo the dimensions of your box are L = W = x = cu.rt.(114), and h =  57/x² = 57/114^(2/3) ft.\n\n~ ♥ ~
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That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
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<SCRIPT>\nrnd.today=new Date();\nrnd.seed=rnd.today.getTime();\n\n    function rnd() {\n    rnd.seed = (rnd.seed*9301+49297) % 233280;\n    return rnd.seed/(233280.0);\n};\n\n\n    function rand(number) {\n    return Math.ceil(rnd()*number);\n};\n</SCRIPT>\n\nThis is alot more simple then using the IF statements. To create a random number between 1 and 10 use rand(10). To display an random image named banner1.gif to banner10.gif use: \n\n<SCRIPT LANGUAGE=JAVASCRIPT><!--\ndocument.write('<IMG SRC="banner' + rand(10) + '.gif" WIDTH="400" HEIGHT="40">');\n//--></SCRIPT>\n\nIf you have at least 10 or more banners... You shouldnt have a problem with dups. If it is essensial not to dup let me know and I'll update the code.
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Let me see if I can answer your question in English. :-)\n\nI'm sure you have an equation in your notes or book that you can just start plugging stuff into.  It probably looks like this:\n\ns = square root of (X - X bar)^2/(n-1)\n\nBut if you want to know why it works, keep reading.\n\nYou have a list of numbers.  First, you need to find the average, or mean, of those numbers, by adding them up and dividing by how many there were.\n\nNow, if you subtract each number from the mean, you get a new list of numbers.  These are the deviation scores, and tell you how far above or below average each number was.  If you add them up, you will get zero, because some of them are positive and some negative.  That's not very interesting, so we square them to make them all positive.  Then we add up all those squared deviation scores.\n\nYou probably got a pretty large number, and not the one we want, yet.  We want the Standard Deviation, which is sort of like the average deviation.  So just like you divide by the number of scores to get the average, you need to divide the sum of the squared deviation scores by something.  Sometimes we divide it by the number of scores, like you would think, but for complicated reasons it works better if you divide by one less than that.  So if you started with 5 numbers, divide by 4 now.  (Be sure to check your notes or your book to see what your teacher said you should divide by.)\n\nWe're still one step away.  The number you should have now is called the Variance.  It is also a measure of how spread out your numbers were, but it's bigger than we want.  Remember how we squared the deviation scores earlier?  Now we need to undo that by taking the square root.  What you get is the Standard Deviation you wanted.
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Try Chi square testing, it starts out complicated but once you understand it, it is way faster than punnett squares! You can find all kinds of stuff about it on the internet, it's a statistical analysis.
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I think all you have to do is plug in a few values for C to get a graph.  For example, try 0, 5, 10, 15, 20.  It's continuous so you can connect the dots.\n\n(0,0), (5,87.5), (10,350), (15,787.5), (20,1400)\n\nTotal labor for 20 is 1400.
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Use the pastel or whatever the kindergarden kids use.  It is like mud but comes in different colours.  The model should be a representation of the cross-section of your cell.  Use strings and other things to show different structures of the cell.  Some diagrams are drawn in 3D, and that's what your model should look like.  I wish they made us do that in university.  It would be fun.
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on a graph...when it's 1/2 or 0.5 then graph the point half way between the two lines (between 0 and 1)...if it's 1/3 then split the box into three sections...etc etc.
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are u kidding, just use the dam formulas for confidence interval and z scores it is not hard just plug them in
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Here's one that is free that I've been using in a college class.\n\nhttp://math.exeter.edu/rparris/winplot.html