Answer: √58 cm or 7.62 cm
Step-by-step explanation:
The Pythagorean Theorem is a²+b²=c². Since we are given the lengths of a and b, we can plug them into this formula to find the length of the hypotenuse.
7²+3²=c²
49+9=c²
58=c²
c=√58 or 7.62 cm
Answer:
1000 m³
Step-by-step explanation:
Put the number into the formula and do the arithmetic.
v = a^(3/2) = (100 m²)^(3/2) = (√100)³ m³ = 1000 m³
The volume is 1000 cubic meters.
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)
Answer:
84 is a solution.
Step-by-step explanation:
To find out if 84 is a solution or not, we plug in the value and simplify.
Therefore, 84 is a solution.
Given:
The two functions are:


To find:
The value of
.
Solution:
We have,


We know that,


![(h\circ g)(b)=[(5b-9)-1]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28b%29%3D%5B%285b-9%29-1%5D%5E2)
![(h\circ g)(b)=[5b-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28b%29%3D%5B5b-10%5D%5E2)
Putting
, we get
![(h\circ g)(-6)=[5(-6)-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B5%28-6%29-10%5D%5E2)
![(h\circ g)(-6)=[-39-10]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B-39-10%5D%5E2)
![(h\circ g)(-6)=[-49]^2](https://tex.z-dn.net/?f=%28h%5Ccirc%20g%29%28-6%29%3D%5B-49%5D%5E2)

Therefore, the value of
is 2401.