Answer:
a) P(X∩Y) = 0.2
b) 
= 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability 
that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is:
Answer:
x> 13
Step-by-step explanation:
x-4 >9
Add 4 to each side
x-4+4 >9+4
x> 13
I think the answer is D. But I am not sure if that's the right answer.
Think of it this way:
_ _ _ (3 digit number)
for the first blank you could have 7 possibilities
for the second blank you could have 6 possibilities
for the third blank you could have 5 possibilities
because each time you fill in a blank one number is removed from the list of possibilities
7×6×5=210. there are 210 possibilities
