Answer:go watch beluga
Step-by-step explanation:
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Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )

E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
Answer:
I do pls
Step-by-step explanation:
Answer:
x(6 + 15y)
Step-by-step explanation:
In this instance, you can divide by a factor of x.
6x/x = 6
15xy/x = 15y
Therefore, we are left with 6 + 15y.
Answer:
the answer is
, and -
which is C.
Step-by-step explanation:
well I got that answer by doing this
we have
= 5
For
= f (a) the solutions are x =
, - 
so in this case the solutions are
, and -
HOPE THIS HELPS :)