I HAVE A QUIZ TMRW PLEASE HELP SITH 12 or 13
1 answer:
You might be interested in
Answer:
The graph of the rectangle missing in the question is shown in the figure attached.
Coordinates of the points are:
A(1,8)
B(4,5)
C(14,21)
D(17,18)
The area of the rectangle is AB*BD.
The length of a segment given two points (x1, y1) and (x2, y2) is computed as follows:
√[(y2 - y1)^2 + (x2 - x1)^2]
For segment AB:
√[(5 - 8)^2 + (4 - 1)^2] = √18
For segment BD:
√[(18 - 5)^2 + (17- 4)^2] = √338
Then, the area is:
AB*BD = (√18)*(√338) = √(18*338) = √6084 = 78
Answer:
A. p.d.f(x,y)= 1/2
B. P(X>Y)= 3/4
Step-by-step explanation:
A. We are told that p.d.f(x,y) is constant on the rectangle 0<x<2, 0<y<1. To find that constant the property of probabilities that is useful here is the following:
We know that f(x,y)=c (c is a constant) in the rectangle and outside of it f(x,y)=0, so the integral above could be rewritten easily as:
Solving the integral it´s possible to find the value c:
This is the value of f(x,y) in the rectangle
B. Now that we know our p.d.f(x,y), we are asked to find P(X>Y) in the rectangle. There´s a graph of the area (probability) that we are looking for, it´ll make easy the understanding of the move that we are going to make.
We are asked for the probability that X>Y, i.e the part of the rectangle below the graph of X=Y. Because x is between 0 and 2, whenever x>1, it will immediately be greater than y. If x<1, we just need to assure that 0<y<x. With this data we can find the probability this way:
The first integral represents the area (that´s equal to the probability) of the triangle at the left of the blue line, and the other integral it´s related to the square that´s right of the blue line.
And now we are ready to solve the problem:
We conclude that P(X>Y)=3/4
