The route used by a certain motorist in commuting to workcontains two intersections with traffic signals. The probabilitythat he must stop at the first signal is 0.39, the analogousprobability for the second signal is 0.54, and the probability thathe must stop at least one of the two signals is 0.64. What is theprobability that he must stop. a.) At both signals?
b.) At the first signal but not at the second one?
c.) At exactly on signal?

a) P(A∩B) = 0.29

b) P1 = 0.1

c) P = 0.35

Step-by-step explanation:

Let's call A the event that the motorist stop at the first signal, and B the event that the motorist stop at the second signal.

From the question we know:

P(A) = 0.39

P(B) = 0.54

P(A∪B) = 0.64

Where P(A∪B) is the probability that he stop in the first, the second or both signals. Additionally, P(A∪B) can be calculated as:

P(A∪B)  = P(A) + P(B) - P(A∩B)

Where P(A∩B) is the probability that he stops at both signals.

So, replacing the values and solving for P(A∩B), we get:

0.64 = 0.39 + 0.54 - P(A∩B)

P(A∩B) = 0.29

Then, the probability P1 that he just stop at the first signal can be calculated as:

P1 = P(A) - P(A∩B) = 0.39 - 0.29 = 0.1

At the same way, the probability P2 that he just stop at the second signal can be calculated as:

P2 = P(B) - P(A∩B) = 0.54 - 0.29 = 0.25

Finally, the probability P that he stops at exactly one signal is:

P = P1 + P2 = 0.1 + 0.25 = 0.35

answer: haha (: i’ll go see what it is

step-by-step explanation:

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x