Answer:
<u>The ages of Paul and Selena are 13 and 7 years old.</u>
Step-by-step explanation:
Age of Paul = x
Age of Selena = x -6
Age of Paul and Selena in 15 years:
x + 15 = 4 (x - 6)
x + 15 = 4x - 24
x - 4x = -24 - 15
- 3x = -39
x = 13
<u>Paul now is 13 years old and Selena is 7 years old. In 15 years, Paul will be 28, that is 4 times the present age of Selena.</u>
Answer:
s = 4
Step-by-step explanation:
The equation for the line of best fit for the data is p=25s+40.
Where p is no. of points and s is no. fo strikes
We need to find the number of strikes made by a bowler with 140 points.
Put p = 140
p=25s+40
140 = 25s+40
Subtract 40 to both sides,
140-40 = 25s+40-40
100 = 25s
⇒ s = 4
So, there are 4 strikes made by a bowler with 140 points.
Answer: false
Step-by-step explanation:
Answer:
Rahul reads 20 pages of a story book everyday and finish it in 12 days Ram is a slow reader and work for two days he was then joined by b find the total time taken to finish the work?
Answer:
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)

Step-by-step explanation:
For this case we can use the moment generating function for the normal model given by:
![\phi(t) = E[e^{tX}]](https://tex.z-dn.net/?f=%20%5Cphi%28t%29%20%3D%20E%5Be%5E%7BtX%7D%5D)
And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

And we have that the moment generating function can be write like this:

And we can write this as an infinite series like this:

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:
![E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%20E%5B1%2B%20tX%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28tX%29%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn%21%7D%28tX%29%5En%20%2B....%5D)
![E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%201%2B%20E%5BX%5Dt%20%2B%5Cfrac%7B1%7D%7B2%7DE%5BX%5E2%5Dt%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn1%7DE%5BX%5En%5D%20t%5En%2B...)
and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:
![\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%282k%29%21%7D%20E%5BX%5E%7B2k%7D%5D%20t%5E%7B2k%7D%3D%5Cfrac%7B1%7D%7Bk%21%7D%20%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5Ek%20%3D%5Cfrac%7B1%7D%7B2%5Ek%20k%21%7D%20t%5E%7B2k%7D)
And then we have this:
![E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...](https://tex.z-dn.net/?f=E%5BX%5E%7B2k%7D%5D%3D%5Cfrac%7B%282k%29%21%7D%7B2%5Ek%20k%21%7D%2C%20k%3D0%2C1%2C2%2C...)
And then we can find the ![E[X^2]](https://tex.z-dn.net/?f=E%5BX%5E2%5D)
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)
And we can find the variance like this :
![Var(X^2) = E[X^4]-[E(X^2)]^2](https://tex.z-dn.net/?f=Var%28X%5E2%29%20%3D%20E%5BX%5E4%5D-%5BE%28X%5E2%29%5D%5E2)
And first we find:
![E[X^4]= \frac{4!}{2^2 2!}= 3](https://tex.z-dn.net/?f=E%5BX%5E4%5D%3D%20%5Cfrac%7B4%21%7D%7B2%5E2%202%21%7D%3D%203)
And then the variance is given by:
