(1 point) Let p be the joint density function such that p(x, y)=116xy in R, the rectangle 0≤x≤4,0≤y≤2, and p(x, y)=0 outside R. Find the fraction of the population satisfying the constraint x≥y fraction

The answer is of the population.

Step-by-step explanation:

The question is wrong. The joint density function is in R

and outside R.

R is defined as the rectangle ,

In order to find the fraction of the population satisfying the constraint , we will need to integrate the joint density function over the region defined by the constraint. It is very convenient to draw the region ''R'' and the new region define by the constraint

I will attach a drawing with the region ''R'' and the new region where we need to apply the integral.

If we integrate outside ''R'', given that outside ''R'', the integral will be equal to 0 (because of the joint density function).

Inside the rectangle ''R'' and given the constraint , we define two new regions : the green region (I) and the blue region (II).

The final step is to integrate in (I) and in (II) and sum ⇒

⇒

, where ''1'' is the green region and ''2'' is the blue region.

⇒   +    =

We find that of the population satisfy the constraint .

step-by-step explanation:

f(x)=2x^2 -1 

translation is a function transformation that transform the function in a way of translating. by translating the function its graph is moved  up an down. moving the function f(x) 2 units to the left will be the new function g(x) with the equation: g(x)=f(x+2)=2(x+2)^2 -1

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation: