It depends on what you are rounding off to.
If you are round it off to the nearest ten, you must look at the unit number.
If you are rounding off to the nearest hundred, you must look at the ten number.
In this case I think they are asking to round to the nearest hundred. Now we must look at the ten number, which is the number after the 'hundred' number.
The ten number is '5' and the hundred number is '1'. If the ten number is 5 or above, it changes to 0 and it makes the hundred number one higher.
So because the ten number is 5, it changes to 0 and it makes the hundred number one higher, to become 2.
The number is now 200.
        
             
        
        
        
I'm going to assume the joint density function is

a. In order for  to be a proper probability density function, the integral over its support must be 1.
 to be a proper probability density function, the integral over its support must be 1.

b. You get the marginal density  by integrating the joint density over all possible values of
 by integrating the joint density over all possible values of  :
:

c. We have

d. We have

and by definition of conditional probability,


e. We can find the expectation of  using the marginal distribution found earlier.
 using the marginal distribution found earlier.
![E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac67%5Cint_0%5E1%282x%5E2%2Bx%29%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B%5Cfrac57%7D)
f. This part is cut off, but if you're supposed to find the expectation of  , there are several ways to do so.
, there are several ways to do so.
- Compute the marginal density of  , then directly compute the expected value. , then directly compute the expected value.

![\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac87)
- Compute the conditional density of  given given , then use the law of total expectation. , then use the law of total expectation.

The law of total expectation says
![E[Y]=E[E[Y\mid X]]](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5BE%5BY%5Cmid%20X%5D%5D)
We have
![E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}](https://tex.z-dn.net/?f=E%5BY%5Cmid%20X%3Dx%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_%7BY%5Cmid%20X%7D%28y%5Cmid%20x%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac%7B6x%2B4%7D%7B6x%2B3%7D%3D1%2B%5Cfrac1%7B6x%2B3%7D)
![\implies E[Y\mid X]=1+\dfrac1{6X+3}](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5Cmid%20X%5D%3D1%2B%5Cdfrac1%7B6X%2B3%7D)
This random variable is undefined only when  which is outside the support of
 which is outside the support of  , so we have
, so we have
![E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5Cleft%5B1%2B%5Cdfrac1%7B6X%2B3%7D%5Cright%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cleft%281%2B%5Cfrac1%7B6x%2B3%7D%5Cright%29f_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac87)
 
        
             
        
        
        
Answer:
I think that the answer is D (13,20)
 
        
                    
             
        
        
        
I believe it would be 360 students.