Does it the answer will be 144 :)
Answer: a) 4.6798, and b) 19.8%.
Step-by-step explanation:
Since we have given that
P(n) = 
As we know the poisson process, we get that

So, for exactly one car would be
P(n=1) is given by

Hence, our required probability is 0.2599.
a. Approximate the number of these intervals in which exactly one car arrives
Number of these intervals in which exactly one car arrives is given by

We will find the traffic flow q such that

b. Estimate the percentage of time headways that will be 14 seconds or greater.
so, it becomes,

Hence, a) 4.6798, and b) 19.8%.
Answer:
- 9 outfits
- 6
- 6
Step-by-step explanation:
6600 /minute so...
396,000 /hour