If the joint density function of x and y is f(x, y)=c(x2−y2)e−4x, with 0≤x< [infinity] and −x≤y≤x, find each of the following.

(a) the conditional probability density of x, given y=y> 0. conditional density fx|y(x, y)= (x^2-y^2)e^(4y-4x)/(y/8 + 1/32) (enter your answer as a function of x, with y as a parameter.)
(b) the conditional probability distribution of y, given x=x. conditional distribution fy|x(y|x)= (x^2 - y^2)/(4/3 x^3) (for −x≤y≤x). (enter your answer as a function of y, with x as a parameter.)

168

step-by-step explanation:

i am pretty sure that the answer is 168

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11

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step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0