What is the variance in completion time for the critical path?
105317
it depends on the familiarity of time study of the maker of the critical path
282592
d + 4 = distance of downward trail\nd = distance of upward trail\nt = 4.5 hours\nv = 2.5 km/hr (average)\n\ntotal distance = 4.5*2.5 = 11.25 km\n\nd + (d + 4) = 11.25\n2d + 4 = 11.25\n2d = 7.25\nd = 3.625 km
1185986
how ever long it takes to get it done
742680
You said that your timing equipment has an accuracy of +- 33 milliseconds. In statistics, I can assume that the +- 33 milliseconds is a 95% confidence interval of the mean period of the pendulum. So, since, the critical value in a normal distribution for a 95% confidence interval is 1.96, therefore, your standard deviation is about 0.0168367 s. Working shown below,\n\n1.96 * stddev = 0.033\nstddev = 0.033/1.96 = 0.0168367\n\nSo, using the Central Limit Theorem, the standard deviation of the estimated mean is stddev/sqrt(n), where n is the number of measurements and sqrt is the square root.\n\nSo the accuracy (95% confidence interval) of the period would be\n1.96 * stddev/sqrt(n)\n= 1.96 * 0.0168367 / sqrt(n)\n= 0.033 / sqrt(n)\n\nSo as you can see the accuracy drops by a factor of 1/sqrt(n)\n\nSo, if you do 2 measurements, you get an accuracy of plus minus 0.033/sqrt(2) = 0.023 s, ie plus minus 23 milliseconds\n\nIf you do 3 measuremenets, you  get an accuracy of plus minus 0.033/sqrt(3) = 0.019 s, ie plus minus 19 milliseconds
764277
during and distructive test, danger exists. knowlege of the equiptment and proceedures are a must. if the exact failure point is to be measured, then at yield the test should be slowed. It will also minimise flying samples.
141128
My calculation says it's about 4.35 minutes.
1062962
so Joe is taking 20 hrs and Fred is taking 32 hours. If these two are to be given work, then 20/52% should be given to Fred and 32/52% should be given to Joe to get things done in time ( I mean both finishes at the same time)\n\nSo Joe's portion of work would be 64% and Fred wil do 36% work...\n\nsimple !
43752
Figure an hour to an hour and a half if there are no complications for the actual procedure. You'll probably make at least one, if not two, trips back for a crown. All told, 2 1/2-3 hours. The first one is the tough one. Good luck...you have my sympathies. Been there, done that...more than once. :)
877147
(A) The probability that a wait will be necessary is the same as the probability that a car comes within the next T seconds: 1 - exp(-lambda T). So the probability of no wait is:\n\nP = exp(-lambda*T)\n\n(B) Let Q be the expected waiting time. If t is the time until the next car arrives (t < T), the expected additional wait time will be Q, giving an expected total wait time of t + Q. In other words the expectation value of the total wait time is:\n\nQ = &#92;integral_0^T (lambda exp(-lambda*t))(t + Q) dt\n= [-exp(-lambda t)(Q + 1/lambda + t]_0^T\n= Q + 1/lambda - exp(-lambda T)(Q + 1/lambda + T)\n\nSolving for Q gives the answer:\n\nQ = (exp(+lambda T) - 1)/lambda - T
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.
796175
By plain, I have no idea.  By plane depends on the route but direct would be about 11 hours.
29755
it depends on the servers the sender & receiver are using and how large the file is.  usually under 5 minutes.
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
391404
Your final project is due when?\n\nServes you right for waiting until the last minute!
1016234
This could take some time to do right...when is this due...and how long should it be?
812753
a.  820 is 10 min lates sooo... 25% prob\nb. 800 is 10 min early sooo..... 15%\n\nc.  and.. then hmmm i gues you need to find out what "mutauilly exclusive means and compare the two answers\nd. same for d
149109
...Please complete it