Assume the arrival in a parking facility follows a Poisson process with mean arrival of 4.2 veh/min. The processing time is exponentially distributed with a mean processing rate of 5 veh/min. Assume there is only 1 processing booth. Calculate the average length in queue, average time spent

Average number of vehicles in the queue is 4.4 ≅ 4 veh/min

Average waiting time of a vehicles in the queue = 1.05 veh/min

Step-by-step explanation:

Step1 :- Given λ = 4.2 veh/min

and μ= 5 veh/min

a) Average number of customers in the queue

E(m) =

on simplification, we get

Average number of vehicles in the queue is 4.4 ≅ 4 veh/min

b) Average waiting time of a customer in the queue

on simplification, we get

Average waiting time of a vehicles in the queue = 1.05 veh/min

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.