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I might misinterpret the question.Brent Hanneson wrote:Rectangular region R lies on the coordinate plane and has vertices at (0,0), (0,3), (4,3) and (4,0). If a point in region R is randomly selected, what is the probability that the point’s x-coordinate will be greater than the y-coordinate?
(A) 5/16
(B) 5/8
(C) 7/12
(D) 2/3
(E) 3/4
No Brent, I actually dont thats why I also have those GREEN areas. But I am still missing something :roll:Brent Hanneson wrote:This is a good start, Logitech, but I believe that you are assuming that the x- and y-coordinates must be integers.
Try drawing a straight line from (0,1) to (4,3)logitech wrote:No Brent, I actually dont thats why I also have those GREEN areas. But I am still missing something :roll:Brent Hanneson wrote:This is a good start, Logitech, but I believe that you are assuming that the x- and y-coordinates must be integers.
Good point, Moshum. My bad.moshum wrote:Brent, shouldn't be the straight line from (0,0) to (3,3) to separate the 2 regions?
Because the "switching" points where the y coordinate is getting bigger than the x coordinate is at (0.5,0.5) ; (1,1) ; (2,2) etc...
Can you explain the 7.5? Im getting 1/2. I see where you are talking about not being able to assume positive integers but that would inversely work for Y as well wouldn't it?Brent Hanneson wrote:Good point, Moshum. My bad.moshum wrote:Brent, shouldn't be the straight line from (0,0) to (3,3) to separate the 2 regions?
Because the "switching" points where the y coordinate is getting bigger than the x coordinate is at (0.5,0.5) ; (1,1) ; (2,2) etc...
Yes, once we've drawn the line from (0,0) to (3,3) one region consists of points where the x-coordinate is greater than the y-coordinate and the other region consists of points where the y-coordinate is greater than the x-coordinate.
The total area is 12.
The region where the x-coordinate is greater than the y-coordinate has an area of 7.5
So, the probability is 7.5/12 = 5/8 (answer choice B)
