I know you said "without making any assumptions," but this one is pretty important. Assuming you mean
are shape/rate parameters (as opposed to shape/scale), the PDF of
is

if
, and 0 otherwise.
The MGF of
is given by
![\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20M_X%28t%29%20%3D%20%5CBbb%20E%5Cleft%5Be%5E%7BtX%7D%5Cright%5D%20%3D%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Btx%7D%20f_X%28x%29%20%5C%2C%20dx%20%3D%20%5Cfrac%7B2%5E8%7D%7B%5CGamma%288%29%7D%20%5Cint_0%5E%5Cinfty%20x%5E7%20e%5E%7B%28t-2%29%20x%7D%20%5C%2C%20dx)
Note that the integral converges only when
.
Define

Integrate by parts, with


so that

Note that

By substitution, we have

and so on, down to

The integral of interest then evaluates to

so the MGF is

The first moment/expectation is given by the first derivative of
at
.
![\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}](https://tex.z-dn.net/?f=%5CBbb%20E%5BX%5D%20%3D%20M_x%27%280%29%20%3D%20%5Cdfrac%7B8%5Ctimes%5Cfrac12%7D%7B%5Cleft%281-%5Cfrac%20t2%5Cright%29%5E9%7D%5Cbigg%7C_%7Bt%3D0%7D%20%3D%20%5Cboxed%7B4%7D)
Variance is defined by
![\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2](https://tex.z-dn.net/?f=%5CBbb%20V%5BX%5D%20%3D%20%5CBbb%20E%5Cleft%5B%28X%20-%20%5CBbb%20E%5BX%5D%29%5E2%5Cright%5D%20%3D%20%5CBbb%20E%5BX%5E2%5D%20-%20%5CBbb%20E%5BX%5D%5E2)
The second moment is given by the second derivative of the MGF at
.
![\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18](https://tex.z-dn.net/?f=%5CBbb%20E%5BX%5E2%5D%20%3D%20M_x%27%27%280%29%20%3D%20%5Cdfrac%7B8%5Ctimes9%5Ctimes%5Cfrac1%7B2%5E2%7D%7D%7B%5Cleft%281-%5Cfrac%20t2%5Cright%29%5E%7B10%7D%7D%20%3D%2018)
Then the variance is
![\Bbb V[X] = 18 - 4^2 = \boxed{2}](https://tex.z-dn.net/?f=%5CBbb%20V%5BX%5D%20%3D%2018%20-%204%5E2%20%3D%20%5Cboxed%7B2%7D)
Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

where
is the
-derivative of the MGF evaluated at
. This is also the
-th moment of
.
Recall that for
,

By differentiating both sides 7 times, we get

Then the
-th moment of
is

and we obtain the same results as before,
![\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4](https://tex.z-dn.net/?f=%5CBbb%20E%5BX%5D%20%3D%20%5Cdfrac%7B%28k%2B7%29%21%7D%7B7%21%5C%2C2%5Ek%7D%5Cbigg%7C_%7Bk%3D1%7D%20%3D%204)
![\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18](https://tex.z-dn.net/?f=%5CBbb%20E%5BX%5E2%5D%20%3D%20%5Cdfrac%7B%28k%2B7%29%21%7D%7B7%21%5C%2C2%5Ek%7D%5Cbigg%7C_%7Bk%3D2%7D%20%3D%2018)
and the same variance follows.
Answer: Yes 18/30 is equivalent to 3/5 heres the picture
Answer:
133,300
Step-by-step explanation:
3100 x 43 = 133,300
Answer:
She will have 142 watermelons
Step-by-step explanation:
First let's put our ratio
10 : 71
Since our lemons are now 20 instead of 10 we will divide 20 by 10 to see how many time we multiplied it by.

which equals 2
To find our water melon we will multiply it by 2

which equals 142
She will have 142 watermelons