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:
4/8-3/8=1/8more
Therefore Keith at 1/8 more of the sandwich than Henry.
9g-45h, you have to distribute 9
The number lines should show the weights the car seats are designed for in comparison the the weight of the 32lb child. The car seat made for children 30lb and lighter would not work because this child is 32lbs. Model this by placing a filled in circle at 30 and the arrow should face left. The seat for children between 15lbs and 40lbs and the seat designed for a child that is between 30lbs and 85lbs would work because the child is within these weight ranges. The number line for the seat for a 15lb-40lb child would have open circles at 15 and 40 with a line connecting the two points. The number line for a 30lb-85lb child should have filled in circles at these two point with a line connecting them.
