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aliya0001 [1]
1 year ago
8

An office that dispenses automotive license plates has divided its customers into categories to level the office workload. Custo

mers arrive and enter one of three lines based on their residence location. Model this arrival activity as three independent arrival streams using an exponential interarrival distribution with mean 10 minutes for each stream, and an arrival at time 0 for each stream. Each customer type is assigned a single separate clerk to process the application forms and accept payment, with a separate queue for each. The service time is UNIF(8, 10) minutes for all customer types. After the completion of this step, all customers are sent to a single, second clerk who checks the forms and issues the plates (this clerk serves all three customer types, who merge into a single first come, first serve queue for this clerk). The service time for this activity is:_______
Mathematics
1 answer:
grigory [225]1 year ago
6 0

Answer:

UNIF(2.66,3.33) minutes for all customer types.

Step-by-step explanation:

In the problem above, it was stated that the office arranged its customers into different sections to ensure optimum performance and minimize workload. Furthermore, there was a service time of UNIF(8,10) minutes for everyone. Since there are only three different types of customers, the service time can be estimated as UNIF(8/3,10/3) minutes = UNIF(2.66,3.33) minutes.

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Aloiza [94]
1. 30 ÷ 50 = 0.6 = 6%
2. 20 ÷ 50 = 0.4 = 4%
29 ÷ 50 = 0.58 = 58%
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I can't say for sure if these are the corrects answers, as math is not one of my better suits. But, I hope that it can somewhat help you.
4 0
1 year ago
Write a program that prompts the user to read two integers and displays their sum . if anything but an integer is passed as inpu
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From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri
Leviafan [203]

Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

4 0
1 year ago
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murzikaleks [220]

Answer:

A

Step-by-step explanation:

Im good at math

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Help me with this question!!!
qwelly [4]

Step-by-step explanation:

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2

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