Answer:
Hence, the probability is:
0.82
Step-by-step explanation:
Let A denote the event that the flight departs on time.
Let B denote the event that the flight arrives on schedule.
Also, A∩B denotes the event that the flight that departs on time also arrives on schedule.
Let P denotes the probability of an event.
We are asked to find:
P(B|A)
We know that:
Hence, we have:
P(A)=0.83 x.
P(A∩B)=0.68 x.
where x are the total number of flights.
Hence,
Hence, the probability that a flight that departs on schedule also arrives on schedule is:
500(x - 3) =5000
x - 3 = 5000/500
x - 3 = 10
x = 10 + 3
x = 13
1 and 1
degree of 1 and 1 term.
m∠AOB+m∠BOC=180°
m∠AOB=8x+51°
m∠BOC=6x-25°
~
m∠AOB:-
Therefore, m∠AOB=139°
49.
35+35-15-10-19+23 is 49.