4. 119*4 = $476 total dollars for four nights
5.
Answer:
0.6672 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8.4 minutes
Standard Deviation, σ = 3.5 minutes
We are given that the distribution of distribution of taxi and takeoff times is a bell shaped distribution that is a normal distribution.
According to central limit theorem the sum measurement of n is normal with mean
and standard deviation 
Sample size, n = 37
Standard Deviation =

P(taxi and takeoff time will be less than 320 minutes)

Calculation the value from standard normal z table, we have,

0.6672 is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
Just need to find out under early to add graduate and undergraduate together and then see if how much out of the total of those who registered early is undergraduate
Answer:
(2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×-(2×2×2)×(2×2×2)