Rolle's & Mean Value thm question?
954215
1.\nSince f(0) and f(pi) are both equal to 0 and f is continuous over the interval [0,pi] and differentiable over the interval (0,pi), the conditions of Rolle's theorem are satisfied. Then there are values c, for which f'(c)=0, where f' is the first derivative of f(x) with respect to f.\n\nAll you have to do is find the number of solutions to 0=f'(x)=2cosx + 4cos4x. Using the Pythagorean identity, you can change this into a quartic equation in cosx, so there are at most 4 solutions. To be certain, graph the function on your calculator and count the number of turning points (minima/maxima or extrema) in the interval.\n\n2.\nf'(c)=(f(b)-f(a))/(b-a)\nPlugging in b=5 and a=0, we need to find the value of c satisfying:\nf'(c)=95/5=19\nSince f(x)=x^3 - 6x, f'(x)=3x^2 - 6.\nThen f'(c)=3c^2 - 6=19.\n3c^2=25\nc=5/sqrt{3}
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
884290
Ho: mean=40 hours/week\nHa: mean > 40 hours/week\nalpha: 95%\nn = 30\ndf = 30-1\n\nOne sample t-test:\n\nt=(mean-assumed mean)/(sample standard deviation/square root of n)\n\nt=(40-45)/(7.5/sqrt 30)\nt=-5/1.369\nt=-3.65\n\nFrom the tables with alpha = 0.05 and df = 29 the t value is 2.045.\n\nSince our calculated t-value is outside of the t value from the table, there is little statistical significance to conclude that the average work week has increased.
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
906033
first fin the mean with the sample of 280 \n\nmultiply 280 to .184 to get the mean \n\nmultiply 280 to .184 and to .816 (from 1-.184) then take the \n\nsquare root that will be your standard deviation \n\n\ntake the z score you will get   ( 56 - 51.52 )/6.483851\n\n=.6909   find teh valu in the standard normal distribution table \n\nyou will get 0.2549    subtract it from .5  because you need only \n\nthe probability higher than of 56  you will get \n\n24.51%
529075
Hey seriously I dont't understand "t0.05(9)=2.26"\n\nHowever I did some of the calculation based on what I remember.\nMean of the sample = 2.2\nVariance if the sample = 9.56\n\nTo find the t-value, using the t formula,\n\nt = (2.2 - 0) / [9.56/ sq root(10)]\n= 0.7277\n\nWhy 0? Because you want to test whether the sample mean is 0. O meaning there is no effect on the B.P.\n\nSince sample size is 10, degrees of freedom is 9.\nLooking at the t-table, at 5% sig, the t-value is 2.262\nat 10% sig, t- value is 1.833.\n\nNote that we are actually looking at the t-value where the p-values are 2.5% and 5% respectively. Here we divide the level of sig by 2 because the null hypothesis is an equal sign.\nNull Hypo: sample mean = 0.\nSince it is 'equal' sign, we will divide level of significace by 2. Remember this.\n\nSo you can see that actually our t-value is smaller than the 2 t-values and we do not rej the null hypo. \n\nConclusion: We do not have enough evidence to say that the stimulus have a significant effect on the B.P. Not being able to rej null hypo doesn't mean that we accept null hypo. So we cannot claim that the stimulus doesn't have an effect on the B.P. This point is impt. \n\nHope you understand my workings. Any mistakes pls point out to me. :) I did this based on what I remember. :)
189376
Hi,\n\nA fitted value is the mean (Average)\n\nOf a statistics, I know it's confusing sometimes!!\n\nCheers!!\nHope that helps!!
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
34804
what is that value equal to?
919154
It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
1062399
Mean is the average of the data.\nMedian is the middle number (if placed in ascending/descending order)\nMode is the number that appears the most.\n\nExamples:\n\nLet's say you are trying to discover the force of gravity by measuring the time it takes a car to roll down a ramp.  You know the length of the ramp and the angle of the ramp and therefore, need only a good approximation of the time that it takes for the car to go down the ramp (your number will be low do to friction).\n\nLet's say that these are the times that you got from 10 trials:\n\n\n9.3\n9.3\n9.9\n8.8\n9.5\n9.7\n15.4  (the car came off the track and was slowed down a lot)\n9.6\n8.9\n9.2\n\nThen you would have had a mean of 9.96\na median of  9.4 (line them up in ascending oder 8.9, 8.9, 9.2, 9.3, 9.3, 9.5, 9.6, 9.7, 9.9, 15.4 and since there is an even number of data points, take the mean of the 5th and 6th)\nand a mode of 9.3\n\nNow 15.4 is obviously a bad point.  Removing it we get:\na mean of about 9.7\na median of  9.3\nand a mode of 9.3\n\nSo only the mean really changes, this is because the mode and median are designed to get rid of "bad" data points.
360360
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b.\n\nSo for this question, the average will be given by\n\n(1/(4-0)) S 8+4e^-2t+te^-2t dt = (1/4)[8t - 2e^(-2t) + S te^-2t dt]\n\nLet u = e^(-t)\ndu =  -e^(-t) dt\n\nSo, S te^-2t dt = - S u^2 du\n= (-u^3)/3\n= -e^(-3t)/3\n\nSo your answer then becomes\n\n(1/4)[8t - 2e^(-2t) - e^(-3t)/3] = 2t - (1/2)e^(-2t) - e^(-3t)/12\n\nEvaluating from 0 to 4,\n= 2(4) - (1/2)e^(-2(4)) - e^(-3(4))/12 - [2(0) - (1/2)e^(-2(0)) - e^(-3(0))/12]\n= 8 - (1/2)e^(-8) - e^(-12)/12 + 1/2 + 1/12\n= 8 5/12 - (1/2)e^(-8) - e^(-12)/12\n~ $8.42
323008
By calculationg it's mean. Add all the values of observations and then divide by the number of observations.\nThat's all!
920953
1) X/25\n2) X/25\n\nExplanation:\n1) If you divide a set of values by a constant, then both the mean and the standard deviation are divided by that constant.\nSince your question is about the variance, which is the SQUARE of the standard deviation, the variance is divided by the SQUARE of the constant (5, in this case).\n\n2) When you add a constant to a set of values, the mean is increased by the amount of that constant, but the standard deviation is unchanged. This is true because the group of values has all moved up by a constant amount, but their relation to each other is unchanged, and their variation from each other is unchanged. Since the standard deviation is unchanged, the variance (the square of the standard deviation) is also unchanged.
502657
Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.
973127
Since the mean is 10, what does that mean the total is? 24.3*10 = 243.\nNow, how much has the total gone down by? (31-18) and (28-19) make 22. So the new total is 223. So the new mean is 22.3.\n\nLets list the numbers in order:\na b c 13 d e f.\nWe want to make the last one as big as possible, so lets make the others as small as possible:\n1 2 3 13 14 15 f\nNow we just have to find f which makes the total 13*7; which gives 43.\n\nAgain, this just comes down to working out totals. Boy total = 123*178.4. Girl total is 237*168.3. Total total = 61830.3.\nSo we divide that by the total number of people (360) to get 171.75, approx.
508493
I've found that the Wikipedia articles on probability distributions are top notch and contain enough information for you to solve this problem on your own.\n\nHint: solve the algebraic expressions that relate mean and variance so you can express p or n as algebraic relationship of one another and then do some substitions.\n\nAt first I thought you needed to know p or n up-front however I realize that you have all the info you need to solve this problem algebraically.\n\nGood luck with your homework.