Of the people who fished at Clearwater Park today, 81 had a fishing license, and 9 did not. Of the people who fished at Mountain View Park today, 60 had a license, and 40 did not. (No one fished at both parks.) Suppose that one fisher from each park is chosen at random. What is the probability that the fisher chosen from Clearwater had a license and the fisher chosen from Mountain View did not have a license? Do not round your answer.

9/25 = the probability that the fisher chosen from Clearwater had a license and the fisher chosen from Mountain View did not have a license

Step-by-step explanation:

There are 90 fishers at Clearwater and 100 fishers at Mountain View

The probability of a fisher at Clearwater had a license = 81/90 = 9/10

and the probability of a fisher at Mountain View

did not have a license = 40/100 = 2/5

the probability that the fisher chosen from Clearwater had a license and the fisher chosen from Mountain View did not have a license = 9/10(2/5) = 9/25

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.