Answer:
the second option!
Step-by-step explanation:
hope this helps! will appreciate brainliest!
Answer:
2.14 cm^3
Step-by-step explanation:
V = 4/3 pi r^3
V = 4/3 pi 0.512
V = 2.14
The answer to our question is yes
Answer:
Step-by-step explanation:
Suppose the dimensions of the playground are x and y.
The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore, y= 780/2-3/2x. Now, the total area (A )to be fenced is
A=x*y= x*(390-3/2x)=-3/2 x^2+390x
Calculating the derivative of A and setting it equals to 0 to find the maximum
A'= -3x+390=0
This yields x=130.
Therefore y=780/2-3/2*130=195
Thus, the maximum area is 130*195=25,350ft^2
Answer:
1) 
2) 
3) 
And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
4) 
And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be:
Step-by-step explanation:
For this case we have the following distributions given:
Probability M J
0.3 14% 22%
0.4 10% 4%
0.3 19% 12%
Part 1
The expected value is given by this formula:

And replacing we got:

Part 2

Part 3
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
Part 4
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be: