Answer:
-7zsquared/4+1
Step-by-step explanation:
It goes on and on..It's a repeating decimal. The answer rounded to the nearest hundredth is 1.23.
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
Answer:
Option D
Step-by-step explanation:
A type I error occurs when you reject the null hypothesis when it is actually true.
The null hypothesis in this case is minimum breaking strength is less than or equal to 0.5.
A type one error would be allowing the production process to continue when the true breaking strength is below specifications.
Answer:
10.63015
10.63015
Step-by-step explanation:
c=a2+b2=82+72