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:

Check the picture below, notice those three quadrilaterals.
the square, is really a rectangle and is also a rhombus, but with right-angles and equal sides.
Answer:
x = 34
Step-by-step explanation:
Add together all three angles because together they equal 180 degrees
x + 12 + x + 100 = 180 Combine like terms
2x + 112 = 180
- 112 - 112 Subtract 112 from both sides
2x = 68 Divide both sides by 2
x = 34