Let 
denote the value on the 
-th drawn ball. We want to find the expectation of 
, which by linearity of expectation is
![E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i]](https://tex.z-dn.net/?f=E%5BS%5D%3DE%5Cleft%5B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E5B_i%5Cright%5D%3D%5Csum_%7Bi%3D1%7D%5E5E%5BB_i%5D)
(which is true regardless of whether the 
are independent!)
At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.
and so
![E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5](https://tex.z-dn.net/?f=E%5BX_i%5D%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B10%7Dx%5C%2CP%28X_i%3Dx%29%3D%5Cfrac1%7B10%7D%5Cfrac%7B10%2810%2B1%29%7D2%3D5.5)
Then the expected value of the total is
![E[S]=5(5.5)=\boxed{27.5}](https://tex.z-dn.net/?f=E%5BS%5D%3D5%285.5%29%3D%5Cboxed%7B27.5%7D)
Answer:
Step-by-step explanation:
So, we are given:
First, we should immediately rule out 0 as an answer. This is because the if 
, the equation would be undefined.
Now, cross multiply.
Divide everything by x (and we can do this safely because we already know x cannot be equal to zero).
We didn't run into any possibilities for extraneous solutions.
Answer:
B would be the answer.
Step-by-step explanation:
Because 1 x 35 is 35. 35 x 2 is 70. 35 x 3 is 105. And so on. :> Good luckk
Answer:
the answer is 35
Step-by-step explanation:
