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:
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no because mathematically 30X10=300 not 900 its would have to be 30X30 to get the product of 900 so therefore no your statement is incorrect
Answer:
135
Step-by-step explanation:
I think you mean the LCM instead of LMC. LCM stands for Least Common Multiple. Anyway, some of the multiples of 27 are 27, 54, 81, 108, and 135. Some multiples for 45 are 45, 90, 135. We can see here that 27 and 45 share the mutiple of 135 which is the first number they have in common.