Answer:
(9/2,0) and (0,-1)
Step-by-step explanation:
y(0) = log(1)-1 = -1
y=0 when log(2x+1) = 1
2x+1 = 10
x = 9/2
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
F - (g-2) should be the answer. In order to solve this, you would use P.E.M.D.A.S....Hopefully this helps!