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Answer:
2nd car is running faster than the first car by 2.01 units.
Step-by-step explanation:
Let's assume that
- the velocity of first car,
- and the velocity of second car
=> speed of first car,
= 32.01 units
and speed of second car,
= 30 units
= 2.01 units
Hence, 2nd car is running faster than the first car by 2.01 units.
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: