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:
Step-by-step explanation:
AB and CD is not parallel because co-interior sum is 180 and another reason is that parallel sign isn't given.
Answer:
I gave you most of the answer. I'll let you check my work and find the point using the solution.
Step-by-step explanation:
The first thing we do is we divide by negative one in the first equation to get
y = -x.
-3x + 3y = -36
Plug in y = -x and get
-3x + 3(-x) = -36
= -3x - 3x = -36
This equals -6x = -36
divide both sides by -6 and you get 6. 6 is your x value
Plug 6 back in to the second equation.
-3x + 3y = -36
-3(6) + 3y = -36
-18 + 3y = -36
3y = -18
y = -6
Answer:
im in this right now, the answer is b
Step-by-step explanation:
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