Question

The tool crib at a large manufacturing company is responsible for providing tools to the factory workers on demand. The tool crib has a variable demand. Historically, its demand has ranged from 150 to 250 small tools per day with an average of 200. Diane, the tool crib attendant, works eight hours a day, five days a week. Each order is for one small tool and each small tool takes Diane 2 minutes to retrieve from the bins.


Diane has been asked to consider plans to add the retrieval of larger tooling fixtures to her duties. She anticipates that there would be an average of 12 tooling fixtures per day requested. Each tooling fixture would take Diane 4 minutes to retrieve.

What is the average waiting time, in minutes, if Diane continues to be the only worker that would retrieve the small tools as well as the larger tooling fixtures?

Select one:

A. 5.0 minutes

B. 10.0 minutes

C. 15.5 minutes

D. 18.5 minutes

E. 31.0 minutes

2.. M. Chain was to manufacture 1,000 chain saws next month. Its accountant has provided the following analysis of the total manufacturing costs.

Variable	Coefficient	Standard Error	t-Value
Constant	200	143.88	1.39
Predictor Variable	400	183.49	2.18

r² = 0.71

What is the estimated cost of producing the 1,000 chain saws?

Select one:

A. $18,000

B. $284,142

C. $400,200

D. $9,000

E. $200,400

3.The demand for Ballard's Glass Company's products varies, ranging from 10 to 20 windows a day with an average of 15. John Ballard is the only worker and he works a maximum of eight hours a day, five days a week. Each order is one window, and each window takes 26 minutes to install. The company plans to add doors to its product line and anticipates selling an average of 5 doors per day. Each door takes 12 minutes to install. What is the average waiting time if Ballard is the only worker?

Select one:

A. 410.0 minutes

B. 38.0 minutes

C. 181.0 minutes

D. 112.4 minutes

E. 523.4 minutes3.

Hello,

Are you able to show the work for all three?

Thanks,

Answer 1

1) Total Minutes per day = 8 hours = 8 *60 = 480 minutes

Average Waiting time ={ [200 * (2)^2] + [12 * 4^2] / {2*[480-(200*2) - (12*4)]}

= {[200 * 4] + [12*16]} / {2*[480-400-48]}

= {800+192} / {2*32}

= 992 / 64

= 15.5 Minutes

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