Answer:
I only see one true equation. Please check the statements.
Step-by-step explanation:
О 0+451=0← for A <u><em>FALSE</em></u>
О 130+210=210+130← for B <u><em>FALSE</em></u>
О (123+57)+180=180+(87+180)←for C <u><em>FALSE</em></u>
√ for= 497+0=497←for D <u><em>TRUE</em></u>
The answer is 51/40=1 and 11/40
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:
14-5i
Step-by-step explanation:
Distribute using FOIL
12+3i-8i-2i^2
i^2 = -1
12+3i-8i-2(-1)
12+3i-8i+2
Combine like terms
14-5i