Answer:
ABCD is not a parallelogram
Step-by-step explanation:
Use the distance formula to determine whether ABCD below is a parallelogram. A(-3,2) B(-3,3) C (5,-3) D (-1.-5)
We have to find the length of the sides of the parallelogram using the formula below
= √(x2 - x1)² + (y2 - y1)² when given vertices (x1, y1) and (x2, y2)
For side AB
A(-3,2) B(-3,3)
= √(-3 -(-3))² + (3 -2)²
= √0² + 1²
= √1
= 1 unit
For side BC
B(-3,3) C (5,-3)
= √(5 -(-3))² + (-3 -3)²
= √8² + -6²
= √64 + 36
= √100
= 10 units
For side CD
C (5,-3) D (-1.-5)
= √(-1 - 5)² + (-5 - (-3))²
= √-6² + -2²
= √36 + 4
= √40 units
For sides AD
A(-3,2) D (-1.-5)
= √(-1 - (-3))² + (-5 -2)²
= √(2² + -7²)
= √(4 + 49)
= √53 units
A parallelogram is a quadrilateral with it's opposite sides equal
From the above calculation
Side AB ≠ CD
BC ≠ AD
Therefore, ABCD is not a parallelogram
Answer:
Step-by-step explanation:
-3(y²+4y+4)-5+6y
=-3y²-12 y-12-5+6 y
=-3y²-6y-17
9514 1404 393
Answer:
D. 14
Step-by-step explanation:
Point G divides each median into parts that have the ratio ...
short : long = 1 : 2
Then ...
GD : GC = 1 : 2 = 7 : 14 . . . . . . . . multiply the ratio by 7
GC = 14
Answer:
23.32
Step-by-step explanation:
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)
