you can only see values of
Ranging from $-3$ to $3$ and they're included, so domain is $[-3,3]$
and $y$ values ranging from $-2$ to $4$ but $-2$ is not included so range is $(-2,4]$
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:
Look below at the image I have provided for you <3
Step-by-step explanation:
I hope this helps!
18/4ths, 27/6ths, and 36/8ths.
(0.35x + 0.14 * 18) / y = 0.23
x + 18 = y