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3 years ago

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It is to be noted that the following question below is about a Trial Balance. This is a type of reconciliation book in Financial Accounting.

<h3>What is a Trial Balance?</h3>

Please, note that the original worksheet is not attached, hence the general answer.

This, in financial accounting, refers to the statements or records of all credits and debits in a double-entry accounting book which includes all errors or disagreements between figures and accounts.

Usually, all debit and credit columns sums must and should be equal to show that the account has been balanced.

See the link below about Trial Balance:

brainly.com/question/24914390

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If i filed my taxes on jan 31 when should i get them back.

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February 25th(direct deposit),March 4th(mailed check)

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All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxyg

OleMash [197]

Answer:

D. the high partial pressure of oxygen in the lungs

Explanation:

As the cells of the human body are made of the Hemoglobin molecules that transport this red blood to the bloodstreams.

Due to the presence of the dissolved oxygen concentration in the body that these molecules in the lungs are unloaded and then merges with the tissues.

The higher pressure is responsible for the flow of the oxygen in the lungs and the thus formation of the RBC in hemoglobin.

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4 years ago

An office has 5 copy machines that need to be serviced approximately once an hour either for paper, staples, or toner or repair.

Anton [14]

Answer:

a. Average number of copiers in line is 0.275

b. Average number of copiers still in operation is 4.23

c. Average number being serviced is 0.496

d. No, the office should not do it

Explanation:

N = number of copy machines = 5

U = average time between unit service requirements = 1 hour = 60 minutes

T = average service time = 7 minutes

M = number of servers = 1

Cost of copier downtime = $20

Cost of attendant = $15

(a) Service factor, X = T / (T+U) = 7 / (60+7) = 0.105

From the Finite Queuing Tables for a Population of N = 5,

For X = 0.105 and M=1, the efficiency factor, F = 0.945

So, The average number waiting in line, L = N × (1 - F) = 5 × (1 - 0.945) = 0.275

(b) Average number in operation, J = N×F×(1 - X) = 5×0.945×(1 - 0.105) = 4.23

(d) For M=1

The average number of copier down = N - J = 5 - 4.23 = 0.77

So, cost of downtime per hour = $20×0.77 = $15.4

Also, the cost of the server per hour = $15×M = $15×1 = $15

So, total cost = 15.4 + 15 = $30.4 per hour (i)

For M=2

From the Finite Queuing Tables for a Population of N = 5, with X = 0.105 and M=2, the efficiency factor, F = 0.997

J = N×F×(1 - X) = 5×0.997×(1 - 0.105) = 4.46

The average number of copier down = N - J = 5 - 4.46 = 0.54

So, cost of downtime per hour = $20×0.54 = $10.8

Also, the cost of the server per hour = $15×M = $15×2 = $30

So, total cost = 10.8 + 30 = $40.8 per hour (ii)

Comparing (i) and (ii), we can say that having another attendant is not cost-effective.

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