Answer:


Step-by-step explanation:
Given

Solving (a):
Find k
To solve for k, we use the definition of joint probability function:

Where

Substitute values for the interval of x and y respectively
So, we have:

Isolate k

Integrate y, leave x:
![k \int\limits^2_{0} y {dx} \, [0,x/2]= 1](https://tex.z-dn.net/?f=k%20%5Cint%5Climits%5E2_%7B0%7D%20y%20%7Bdx%7D%20%5C%2C%20%5B0%2Cx%2F2%5D%3D%201)
Substitute 0 and x/2 for y


Integrate x
![k * \frac{x^2}{2*2} [0,2]= 1](https://tex.z-dn.net/?f=k%20%2A%20%5Cfrac%7Bx%5E2%7D%7B2%2A2%7D%20%5B0%2C2%5D%3D%201)
![k * \frac{x^2}{4} [0,2]= 1](https://tex.z-dn.net/?f=k%20%2A%20%5Cfrac%7Bx%5E2%7D%7B4%7D%20%5B0%2C2%5D%3D%201)
Substitute 0 and 2 for x
![k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%20%5Cfrac%7B2%5E2%7D%7B4%7D%20-%20%5Cfrac%7B0%5E2%7D%7B4%7D%20%5D%3D%201)
![k *[ \frac{4}{4} - \frac{0}{4} ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%20%5Cfrac%7B4%7D%7B4%7D%20-%20%5Cfrac%7B0%7D%7B4%7D%20%5D%3D%201)
![k *[ 1-0 ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%201-0%20%5D%3D%201)
![k *[ 1]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%201%5D%3D%201)

Solving (b): 
We have:

Where 

To find
, we use:

So, we have:



Integrate x leave y
![P(x > 3y) = \int\limits^2_0 x [0,y/3]dy](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cint%5Climits%5E2_0%20%20x%20%5B0%2Cy%2F3%5Ddy)
Substitute 0 and y/3 for x
![P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cint%5Climits%5E2_0%20%20%5By%2F3%20-%200%5Ddy)

Integrate
![P(x > 3y) = \frac{y^2}{2*3} [0,2]](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cfrac%7By%5E2%7D%7B2%2A3%7D%20%5B0%2C2%5D)
![P(x > 3y) = \frac{y^2}{6} [0,2]\\](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cfrac%7By%5E2%7D%7B6%7D%20%5B0%2C2%5D%5C%5C)
Substitute 0 and 2 for y




Solve for x:
x^2 + 10 x + 12 = 36
36 = 36:
x^2 + 10 x + 12 = 36
Subtract 12 from both sides:
x^2 + 10 x = 24
Add 25 to both sides:
x^2 + 10 x + 25 = 49
Write the left hand side as a square:
(x + 5)^2 = 49
Take the square root of both sides:
x + 5 = 7 or x + 5 = -7
Subtract 5 from both sides:
x = 2 or x + 5 = -7
Subtract 5 from both sides:
Answer: x = 2 or x = -12 Thus the Answer is A.
Answer:
I believe it's 40/ 20 I'm sorry if it's wrong..
=Y2-10Y
We move all terms to the left:
-(Y2-10Y)=0
We add all the numbers together, and all the variables
-(+Y^2-10Y)=0
We get rid of parentheses
-Y^2+10Y=0
We add all the numbers together, and all the variables
-1Y^2+10Y=0
a = -1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-1)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
Y1=−b−Δ√2aY2=−b+Δ√2a
Δ‾‾√=100‾‾‾‾√=10
Y1=−b−Δ√2a=−(10)−102∗−1=−20−2=+10
Y2=−b+Δ√2a=−(10)+102∗−1=0−2=0