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:
The common factor is b since they both share it.
Hope this helps
we have that
*-------------------------*--------------------------------*
E F G
EF=2x-12
FG= 3x-15
EG=23
we know that
EF + FG = EG
so
[2x - 12] + [3x - 15] = 23 simplify
5x - 27 = 23 add 27 to both sides
5x = 50 divide both sides by 5
x = 10
EF=2x-12-------> EF=2*10-12-------> EF=8
FG= 3x-15------> FG=3*10-15------> FG=15
therefore
the answer part a) is
the value of x is 10
the answer part b) is
the value of EF is 8
the answer part c) is
the value of FG is 15
Answer:Replacing x with -x will reflect a graph across the y-axis. Multiplying x by 3 before the function is applied will compress the graph horizontally. This means the correct choice is D.
