At Jaxlyn’s Garage, 200 cars were inspected and worked on. There were 92 cars that needed new brakes, 136 cars that needed a new alternator, and 24 cars that needed neither repair. Part A: Determine the probability of a car needing both repairs. Part B: Determine the probability of a car needing an alternator, but not new brakes.

A. car needing both repairs is the intersection, which is P(A) * P(B).

    92/200 * 136/200 = 391/1250 = .3128

B. car needing an alternator and not needing new brakes is the union, which is P(A) + P(notB) - P(A and B). We solved for P(A and B) in part A.

P(A) + P(notB) - P(A and B)

136/200 + 108/200 - .3128 = .9072

(a) 13/50

(b) 21/50

Step-by-step explanation:

We need to make a Venn Diagram (see the attachment).

Notice that the two circles are "need a brake" and "need an alternator". The intersection is simply the cars that need both repairs.

(a) Let's say b is the number of cars that need both repairs. 92 is the number of cars that need a new brake, but that also includes the cars that need both repairs. The number of cars that only need brakes is 92 - b. Similarly, the number of cars that only need an alternator is 136 - b.

We add up 92 - b, b, and 136 - b to get the number of cars that need repairs, and we set this equal to 200 - 24 (because 24 cars don't need repairs, 200 - 24 are the total number of cars that do):

(92 - b) + b + (136 - b) = 176

228 - b = 176

b = 52

52 cars need both repairs, so the probability that a car needs both repairs will be 52/200 = 13/50.

(b) The number of cars that need an alternator but not new brakes is 136 - b = 136 - 52 = 84 cars. Then the probability that a car needs an alternator but not brakes is 84/200 = 21/50.

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had