Answer:

Step-by-step explanation:
Y/22 would be the answer to your question.
Answer:a
Step-by-step explanation:hope this helps
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)
The volume of the composite figure is the third option 385.17 cubic centimeters.
Step-by-step explanation:
Step 1:
The composite figure consists of a cone and a half-sphere on top.
We will have to calculate the volumes of the cone and the half-sphere separately and then add them to obtain the total volume.
Step 2:
The volume of a cone is determined by multiplying
with π, the square of the radius (r²) and height (h). Here we substitute π as 3.1415.
The radius is 4 cm and the height is 15 cm.
The volume of the cone :
cubic cm.
Step 3:
The area of a half-sphere is half of a full sphere.
The volume of a sphere is given by multiplying
with π and the cube of the radius (r³).
Here the radius is 4 cm. We take π as 3.1415.
The volume of a full sphere
cubic cm.
The volume of the half-sphere
cubic cm.
Step 4:
The total volume = The volume of the cone + The volume of the half sphere,
The total volume
cub cm. This is closest to the third option 385.17 cubic centimeters.