Answer:
0.79
Step-by-step explanation:
Here,
Let X be the event that the flights depart on time
Let Y be the event that flights arrive on time
So,
X∩Y will denote the event that the flights departing on time also arrive on time.
Let P be the probability
P(X∩Y)=0.65
And
P(X)=0.82
We have to find P((Y│X)
We know that
P((Y│X)=P(X∩Y)/P(X) )
=0.65/0.82
=0.79
So the probability that a flight that departs on schedule also arrives on schedule is: 0.79
Answer:
2/25
Step-by-step explanation:
We can form a system of equations:
x*y=90
and
x=2y+3.
Solving gives dimensions of 15 feet by 6 feet.
Answer:
<RQP = 70°
Step-by-step explanation:
<RQP = 1/2 (140)
<RQP = 70°
