and 9 are the squares of, respectively,
and 3.
So, using

we have

Answer:
Step-by-step explanation:
Given that:
X(t) = be the number of customers that have arrived up to time t.
... = the successive arrival times of the customers.
(a)
Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;




Now 
(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

Now; 
(c) Determine the conditional probability density function for W2, given that X(t)=5.
So ; the conditional probability density function of
given that X(t)=5 is:

Answer:
x = -8
Step-by-step explanation:
-2.5x - 2 = 18.
Add 2 to both sides.
-2.5x = 20.
Divide both sides by -2.5.
x = -8
Answer:
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7ft = 84in 84 ÷ 15 = 5.6 pieces he can cut from each board. Since .6 is not a whole piece measuring 15 inches long, he can only cut 5 pieces from each board.
So, 5 x 2 = 10
He would be able to cut 10 pieces measuring 15 inches long.
Hope this helps :)