The answer is "b"
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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

Attached is the solution.
The following log properties are used:

Answer: 313% 79% 3%
Step-by-step explanation:
Using a system of equations, it is found that Ajay is 25 years old today.
<h3>What is a system of equations?</h3>
A system of equations is when two or more variables are related, and equations are built to find the values of each variable.
In this problem, the variables are:
Wale was twice as old as Ajay 10 years back, hence:



Wale will be 40 years old in 10 years, hence:


Then, Ajay's present age is found as follows.




You can learn more about system of equations at brainly.com/question/14183076