Answer:
Step-by-step explanation:
- given λ =one for every 20minutes = 60/20 = 3customers/hr
- μ = average of 15minutes = 60/15 = 4customers/hr
a) percentage when judy was idle = 1- λ/μ = 1- 0.75 = 0.25
%service time = 0.75
%idle time = 0.25
b) How much time, on average, does a student spend waiting in line;
= λ/ μ( μ- λ)
= 0.75hrs = 0.75 x60 = 45minutes
c) How long is the waiting line on average;
= average waiting time x arrival rate = 0.75hrs x 3 customers/hr
= 2.25customers
d) What is the probability that an arriving student will find at least one other student waiting in line ; Po( probability of idle time i.e no customer to attend to) = 0.25
P1( Probability of having a customer to attend to) = 0.25 x 0.75 = 0.1875
P2( Probability of having 2 customer to attend to) = o.25 x 0.75x0.75 = 0.14
Hence, probability of finding at least one customer = 1 -[ po + p1]
= 1 - 0.25 - 0.1875 = 0.5625
Step-by-step explanation:
C. yes sure qamar would have a lacquer to cover it .
D. justice League of the mathematical model..
E.she needs about 45ft
I think it is B. Hope this helps!
Answer:
child tickets: 115
adult tickets: 175
Step-by-step explanation:
number of child tickets = x
number of adult tickets = y
(value of child ticket [8] * number of child tickets [x]) + (value of adult ticket [8.5] * number of adult ticket [y]) = total value (2407.50)
8x + 8.5y= 2407.50
290 (total number of tickets) = x (number of child tickets) + y (number of adult tickets)
290 = x+y
system of equations: 8x + 8.5y= 2407.50
290 = x + y
solve equation:
x = 115
y = 175