Mathematics, 03.11.2020 17:20 shyyyy13

(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

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Answer from: chasadyyy

The answer is $\frac{7}{8}$ of the population.

Step-by-step explanation:

The question is wrong. The joint density function is $p(x,y)=(\frac{1}{16})xy$ in R

and p(x,y)=0 outside R.

R is defined as the rectangle $0\leq x\leq 4$ , $0\leq y\leq 2$

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

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 p(x,y)=0 outside ''R'', the integral will be equal to 0 (because of the joint density function).

Inside the rectangle ''R'' and given the constraint $x\geq y$ , 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 ⇒

$\int\int p(x,y) dx dy$ ⇒

$\int \int\limits_1 {p(x,y)} \, dx dy + \int \int\limits_2{p(x,y)} \, dx dy$ , where ''1'' is the green region and ''2'' is the blue region.

⇒ $\int\limits^2_0 \int\limits^x_0 (\frac{1}{16} xy) dy dx$ + $\int\limits^4_2\int\limits^2_0 (\frac{1}{16}xy) dydx$ = $\frac{1}{8}+\frac{3}{4}=\frac{7}{8}=0.875$

We find that $\frac{7}{8}$ of the population satisfy the constraint $x\geq y$ .

(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

Answer from: Quest

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

Answer from: Quest

answer: ay

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ya, ya me está gustando más de lo normal

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(1 point) Let p be the joint density function such that p(x, y)=116xy in R, the rectangle 0≤x≤4,0≤y≤...

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