You should first make 7 3/7 and 3 3/5 into improper fractions that would be 52/7 and 18/5 then we can add but we have to make them equivelent in order to do so.
52/7 = 260/35
18/5 = 126/35
Add 260 and 126 and we will get 386/35 but remember we still have to make it a mixed number so we finally get the answer of 11 1/35
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.
The first one is the 5th tile and the second one is the 3rd tile
Answer:
you need a photo my dude
Step-by-step explanation:
Answer:
- 15
Step-by-step explanation: