Answer: The correct option is (A). angle 1.
Step-by-step explanation: In the given diagram, the measure of ∠3 is 105°.
We are to find the angle that must also measure 120°.
We know that the measures of two vertically opposite angles are equal.
In the given diagram, since ∠1 and ∠3 are vertically opposite angles.
So, the measures of ∠1 and ∠3 are equal.
That is,
m∠1 = m∠3.
Also, m∠3 = 105°.
Therefore, m∠1 = 105°.
Thus, the measure of angle 1 must be 105°.
Option (A) is correct.
The answer is 2.686, and i am writing these extra characters to pass the 20 character qualifications.
Answer:
30
Step-by-step explanation:
Substitue 3 for every x in the equation
10(3)-(3*3)(2*3)+(3)^2(2*3)
Multiply
30-(9)(6)+9(6)
30-54+54=30
Answer:
mathematics is the study of numbers shapes and patterns
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)
