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:
224
Step-by-step explanation:
polynomial degree:224
leading term:7x^223
leading coefficient:7
42 ÷ 63
63 -> 420
63x6=378
420-378=42
63->420
So, how many times does 63 go into 42? Well, it doesn't. So put down a zero on your paper, and then a decimal. So if we add a zero onto 42, it becomes 420. Well, 420 is divisible by 63. In fact, 63 goes into 420 6 times, making a total of 378. 420-378 = 42. Then the process begins again. So you've got a 0.6, and that six just keeps on repeating. On paper, you're gonna wanna put a dash over the six to show that it's repeating.
Anyways, the answer is .66 repeating.
Answer:algebra ok
Step-by-step explanation:
Answer:
32 siblings from the class theres a possibility that some students had more siblings and not everyone has one
Step-by-step explanation:
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