I'm assuming
is the shape parameter and
is the scale parameter. Then the PDF is

a. The expectation is
![E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac29%5Cint_0%5E%5Cinfty%20x%5E2e%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20dx)
To compute this integral, recall the definition of the Gamma function,

For this particular integral, first integrate by parts, taking


![E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle-xe%5E%7B-x%5E2%2F9%7D%5Cbigg%7C_0%5E%5Cinfty%2B%5Cint_0%5E%5Cinfty%20e%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20x)
![E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E%5Cinfty%20e%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20dx)
Substitute
, so that
:
![E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cfrac32%5Cint_0%5E%5Cinfty%20y%5E%7B-1%2F2%7De%5E%7B-y%7D%5C%2C%5Cmathrm%20dy)
![\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}](https://tex.z-dn.net/?f=%5Cboxed%7BE%5BX%5D%3D%5Cdfrac32%5CGamma%5Cleft%28%5Cdfrac12%5Cright%29%3D%5Cdfrac%7B3%5Csqrt%5Cpi%7D2%5Capprox2.659%7D)
The variance is
![\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2-2XE%5BX%5D%2BE%5BX%5D%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
The second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2f_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac29%5Cint_0%5E%5Cinfty%20x%5E3e%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20dx)
Integrate by parts, taking


![E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle-x%5E2e%5E%7B-x%5E2%2F9%7D%5Cbigg%7C_0%5E%5Cinfty%2B2%5Cint_0%5E%5Cinfty%20xe%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20dx)
![E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle2%5Cint_0%5E%5Cinfty%20xe%5E%7B-x%5E2%2F9%7D%5C%2C%5Cmathrm%20dx)
Substitute
again to get
![E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle9%5Cint_0%5E%5Cinfty%20e%5E%7B-y%7D%5C%2C%5Cmathrm%20dy%3D9)
Then the variance is
![\mathrm{Var}[X]=9-E[X]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3D9-E%5BX%5D%5E2)
![\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cmathrm%7BVar%7D%5BX%5D%3D9-%5Cdfrac94%5Cpi%5Capprox1.931%7D)
b. The probability that
is

which can be handled with the same substitution used in part (a). We get

c. Same procedure as in (b). We have

and

Then

The volume of cube and rectangular prism are same. Option B.
Step-by-step explanation:
Given,
The length of the edge of the cube (a) = 5 cm
The dimension of rectangular prism (l×b×h) = 5 cm×25 cm×1 cm
To find the relation between the volume of cube and rectangular prism.
Formula
The volume of a cube = a³ cube cm
The volume of rectangular prism = l×b×h cube cm
Now,
The volume of a cube = 5³ cube cm = 125 cube cm
The volume of rectangular prism = 5×25×1 cube cm = 125 cube cm
Hence,
The volume of cube and rectangular prism are same.
Answer: X = 9
Show Your Work:
All Angles & Sides = 60.
All Triangles = 180.
60 x 3 = 180.
Angle N:
7x - 3 = 60
7 * 9 - 3 = 60.
Answer:
50
Step-by-step explanation:
6/0.12=50
<h3>
Answer: 35 inches</h3>
Each side of the pentagon is multiplied by 5/3 to get the length of each new side, so the perimeter is multiplied by 5/3 to get the new perimeter.
New perimeter = (scale factor)*(old perimeter)
New perimeter = (5/3)*(old perimeter)
New Perimeter = (5/3)*(21)
New Perimeter = (5/3)*(21/1)
New Perimeter = (5*21)/(3*1)
New Perimeter = 105/3
New Perimeter = 35 inches