Answer:
no solutions
Step-by-step explanation:
if you solve by elimination you will get 0=-4
That statement is false
Answer:24
xexponet
3
−
5
Step-by-step explanation:
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:
Okays so for 100 guests we can make the equation $5000 ÷ 100 = $50.00
For 100 guests at the wedding it would cost <u>$50.00</u> per plate
For 45 guests we can make the equation $2750 ÷ 45 = $61.111111111111, it would be a repeating decimal
For 45 guests at the wedding, it would cost <u>$61.111111111111</u>
I hope that this can help you! ^‿^
8.54. The last number is low so it can not round up to 5