Answer:
- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that
P(8 arrive) = (0.56)^8 = 0.00967
- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.
Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be
(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).
So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079
- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,
P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)
= 1 - P(overbooked) - P(exactly booked)
= 1 - 0.00967 - 0.06079
= 0.9295.
Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.
Hope that helps!
Have a great day :P
Answer:
<em>24</em>
Step-by-step explanation:
1 hour and 36 minutes = 60 minutes + 36 minutes = 96 minutes
6/24 = x/96
24x = 6 * 96
24x = 6 * 4 * 24
x = 24
Answer:
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds
Step-by-step explanation:
The given table is presented as follows;
The number of laps in the range 82 to 84 seconds = 1
The number of laps in the range 84 to 86 seconds = 4
The number of laps in the range 86 to 88 seconds = 2
The number of laps in the range 88 to 90 seconds = 4
The number of laps in the range 90 to 92 seconds = 6
The number of laps in the range 92 to 94 seconds = 5
The number of laps in the range 94 to 96 seconds = 2
The number of laps in the range 96 to 98 seconds = 0
Therefore, the histogram that represents Blanca's lap times for the three days of practice is described as follows;
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds