Let X and Y denote the latitude and longitude of a person who is thought to be lost in a certain part of a forest. If we assume that the density of X is uniform for 2 < X < 8.2 and the density of Y is also uniform for 3.01 < Y < 6.3, and we assume that X and Y are independent, then the joint density of X and Y is

P(X, Y) = 1/(20.398 ) if X ∈ (2, 8.2) and Y ∈ (3.01, 6.3)

P(X, Y) = 0 if  X ∉ (2, 8.2) and Y ∉ (3.01, 6.3)

Step-by-step explanation:

We know that the possible values of X (all with the same probability) are:

2 < X < 8.2

And for Y we have the range (also uniform):

3.01 < Y < 6.3

Then we have a rectangle of area.

(8.2 - 2)*(6.3 - 3.01) = 20.398

Such that the probability for all the points inside the rectangle should be exactly equal, then we will have:

P(X, Y) = 1/(20.398 ) if X ∈ (2, 8.2) and Y ∈ (3.01, 6.3)

Such that if we integrate this in the given region, the integral will be equal to 1, as we should expect.

And zero if the point is outside that region, this is:

P(X, Y) = 0 if  X ∉ (2, 8.2) and Y ∉ (3.01, 6.3)

you should pay $0.88

what is the image?

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x