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.

b. You get the marginal density
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.
![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.
- Compute the marginal density of
, 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
, 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
, 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)
Hey there!
To solve this, divide the numbers 54 and -9 . . .
54 divided by -9 = -6 <------------
(-6) x -9 = 54
_________________
d = -6
_________________
Hope this helps you.
Have a great day!
Answer:
Option (3)
Step-by-step explanation:
w = ![\frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5B%5Ctext%7Bcos%7D%28225%29%20%2B%20i%5Ctext%7Bsin%7D%28225%29%5D)
Since, cos(225) = cos(180 + 45)
= -cos(45) [Since, cos(180 + θ) = -cosθ]
= -
sin(225) = sin(180 + 45)
= -sin(45)
= -
Therefore, w = ![\frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5B-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%28-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%5D)
= 
= 
z = 1[cos(60) + i(sin(60)]
= 
= 
Now (w + z) = 
= 
= 
= 
Therefore, Option (3) will be the correct option.
Hey there!
3.14(a^2 + ab)
= 3.14(3^2 + 3 *4)
3^2
= 3 * 3
= 9
3 * 4
= 12
3.14(9 + 12)
9 + 12
= 21
3.14(21)
= 65.94
Answer: 65.94
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)