Answer:
I think the answer is$12 - $3 = $9
the amount u need is $9 to have the complete amount u need to purchase the ticket
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

3. 10 because the opposite of a negative is a positive number
Step two is the answer i hope that helps