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Artyom0805 [142]
4 years ago
13

Martin Company applies manufacturing overhead based on direct labor hours. Information concerning manufacturing overhead and lab

or for the year follows:
Actual manufacturing overhead $150,000
Estimated manufacturing overhead $145,000
Direct labor hours incurred 4,800
Direct labor hours estimated 5,000

Instructions

Compute the predetermined overhead rate.
Business
1 answer:
bixtya [17]4 years ago
5 0

Answer:

POAR = $29 per hour

Explanation:

<em>The overhead absorption is a per-determined rate which is used to charge overheads to production units. Note that this rate is computed using estimated figures</em>

The rate is computed as follows:

Pre-determined overhead absorption rate (POAR)

POAR = Budgeted overhead for the period/Budgeted direct labour hours

          = $145,000/5,000 labour hours

          = $29 per hour

         

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Greger Peterson is a senior manager at a public accounting firm making a base salary of $180,000 a year ($15,000 per month). Emp
Rudik [331]

Answer:

$10,570.80

Explanation:

The computation of the withheld during the year is shown below:

= Social security tax + medicare tax

where,

Social security tax is

= $128,400 × 6.2%

= $7,960.80

And, the medicare tax is

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So, the total withheld during the year is

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We simply added the social security and medicare taxes so that the withheld could come

5 0
4 years ago
The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume that Fore and Aft decides to build
Nataly [62]

Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.

<u>Explanation:</u>

The problem is that of Multiple-server Queuing Model.

Number of servers, M = 2.

Arrival rate, \lambda= 6 boats per hour.

Service rate, \mu= 10 boats per hour.

Probability of zero boats in the system,\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\} = 0.5385

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Lq =[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0

= [\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385 = 0.0593 boats.

The average time a boat will spend waiting for service, Wq  =  0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.

The average time a boat will spend at the dock, W =  0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.

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Ivanshal [37]

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ikadub [295]

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