Answer:
a) 0.125
b) 7
c) 0.875 hr
d) 1 hr
e) 0.875
Step-by-step explanation:l
Given:
Arrival rate, λ = 7
Service rate, μ = 8
a) probability that no requests for assistance are in the system (system is idle).
Let's first find p.
a) ρ = λ/μ

Probability that the system is idle =
1 - p
= 1 - 0.875
=0.125
probability that no requests for assistance are in the system is 0.125
b) average number of requests that will be waiting for service will be given as:
λ/(μ - λ)
= 7
(c) Average time in minutes before service
= λ/[μ(μ - λ)]
= 0.875 hour
(d) average time at the reference desk in minutes.
Average time in the system js given as: 1/(μ - λ)

= 1 hour
(e) Probability that a new arrival has to wait for service will be:
λ/μ =
= 0.875
The choice D is in scientific notation
7.8*10^5
Using the t-distribution, it is found that the p-value of the test is 0.007.
At the null hypothesis, it is <u>tested if the mean lifetime is not greater than 220,000 miles</u>, that is:

At the alternative hypothesis, it is <u>tested if the mean lifetime is greater than 220,000 miles</u>, that is:
.
We have the <u>standard deviation for the sample</u>, thus, the t-distribution is used. The test statistic is given by:
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem:

Then, the value of the test statistic is:



We have a right-tailed test(test if the mean is greater than a value), with <u>t = 2.69</u> and 23 - 1 = <u>22 df.</u>
Using a t-distribution calculator, the p-value of the test is of 0.007.
A similar problem is given at brainly.com/question/13873630
Answer:
c = 5
Step-by-step explanation:
a^2 + b^2 = c^2
3^2 + 4^2 = c^2
9 + 16 = 25
c^2 = 25
sqrt of 25 = 5
c= 5