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:
40,600x10=406,000 lol dude its easy
Answer: 0
Step-by-step explanation:
h ( t ) = − 2 t ^2 + 10 t
h = height of ball above the ground
t = time, seconds after ball was kicked
Height of ball at the time it is kicked?
At the time ball is kicked, time (t) = 0
Therefore,
h ( t ) = − 2 t ^2 + 10 t
h( 0) = - 2(0)^2 + 10(0)
h = 0 + 0
h = 0
The ball was the on the grounf at the time it was kicked.
Answer:
a=-7.275*967
Step-by-step explanation:
Answer:
(x,y) = (.5x,5y)
Step-by-step explanation:
