Answer:
237.5
Step-by-step explanation:
First, you should solve for
, which equals
. Now, solve the integral of
=
, to get that
. You can check this by taking the integral of what you got. Now by the Fundamental Theorem
![\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B4x%7D%20%5C%2C%20dx%3D%5B2x%5E2%5D%20%5E%7B2%7D_%7B0%7D%3D2%282%29%5E%7B2%7D-2%280%29%5E2%3D8)
.
This should be the answer to your question, if I understood what you were asking correctly.
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
THE ANSWER IS NOT IN THE LINK THAY THAT PERSON GAVE
