Answer:
x=3/8, y=-7/8. (3/8, -7/8).
Step-by-step explanation:
y=-5x+1
y=3x-2
----------
-5x+1=3x-2
-5x-3x+1=-2
-8x+1=-2
-8x=-2-1
-8x=-3
8x=3
x=3/8
y=3(3/8)-2=9/8-2=9/8-16/8=-7/8
If 
is the cumulative distribution function for 
, then
Then the probability density function for is 
:
The 
th moment of is
![E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5Enf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bn-1%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Let 
, so that 
and 
:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu%7De%5E%7B-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du)
Complete the square in the exponent:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B%5Cfrac12%28n%5E2-%28u-n%29%5E2%29%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac%7Be%5E%7B%5Cfrac12n%5E2%7D%7D%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du)
But 
is exactly the PDF of a normal distribution with mean and variance 1; in other words, the 0th moment of a random variable 
:
![E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1](https://tex.z-dn.net/?f=E%5BU%5E0%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du%3D1)
so we end up with
![E[Y^n]=e^{\frac12n^2}](https://tex.z-dn.net/?f=E%5BY%5En%5D%3De%5E%7B%5Cfrac12n%5E2%7D)
The answer would be 1. The commutative postulate for multiplication. Hope this helped! Good luck! :)
Answer:
m =7.
Step-by-step explanation:
Let the constant number = k.
g(m) = k/m,
3.5 = k/10,
k = 35.
Thus, when g(m) = 5 = 35/m, m = 7.
Answer:
10
Step-by-step explanation:
