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
Factor each
4k=2*2*k
18k⁴=2*3*3*k*k*k*k
12=2*2*3
GCF=2
the greatest common factor is 2
X must be greater than 0 and y must be less than 0.
I hope this helped.
Answer:
4x + 6y > 120
Step-by-step explanation:
i got the same question on my midterm exam.
