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anonymous
 5 years ago
question in the reply :)
anonymous
 5 years ago
question in the reply :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Continuous uniform distribution? Many students find that sitting through an entire lecture is difficult to do without falling asleep. Suppose the number of minutes a single, randomly selected student is asleep during a lecture is uniformly distributed between 7 and 14 minutes. B) Over the course of the semester (say, 50 lectures), what is the probability that a randomly selected student will sleep a total of between 130 and 140 minutes? Assume that the minutes slept in any lecture is independent of the number of minutes slept in any other lectures. (C) How many lectures must a student attend to be 95% sure that they will have slept at least 28 minutes in total? Use the quadratic equation to solve for the quadratic in this question. Hint: replace x with sqrt(n) to solve for n.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0need more information. need equation to work with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not sure what the equation is, thats the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you can't solve those questions unfortunately

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ye i tried. not sure how tho.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0formula is \[1/(\beta  \alpha)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r u sure? thats what i used for the first part of this question. this part seems more complicated?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you integrate that formula i gave you then it becomes 1/7 (x) you follow up to that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean 1/7(x) from 130 to 140 mins

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the sample size 50 goes where?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have the final answer by any chance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the number of lecture should matter because it's uniform random variable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so answer should be approx. 1.4286

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you integrate 1/7 you get (1/7)x then you take the limit from 140 to 130

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure it's 140 and 130 mins?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm maybe you're supposed to divide the minutes by 50 lectures

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think you're supposed to do that then you get .02857 = 2.857%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. wud u be able to tell me how u got the limit form 130 to 140?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you first divide 140/50 and 130/50 to get 2.8 and 2.6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you (1/7)(2.8)  (1/7)(2/6) = .0287

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1/7)(2.8)  (1/7)(2.6) = .0287

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh okay.thanks :) do u kno how to do parrt C?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure since i didn't learn this using quadratic only hypergeometric :( sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0look at your notes and set it equal to .95

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kays np. thanks for ur help :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol theres a medal for u :)
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