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iren [92.7K]
3 years ago
5

Teagan Company uses Departmental Overhead allocation to allocate its manufacturing overhead costs. It has identified two​ depart

ments: Machining​ & Assembly. The Machining department allocates overhead using machine​ hours, and the Assembly department allocates overhead using direct labor hours. It has accumulated the following data for​ 2018: Machining Department estimated manufacturing​ overhead: ​ $670,000 Assembly Department esimated manufacturing​ overhead: ​$450,000 Total estimated overhead costs ​ $1,120,000 Estimated machine hours in the machining​ department: ​ 10,000 hours Estimated direct labor hours in the assembly​ department: ​ 15,000 hours The following shows the amount of machine hours and direct labor hours incurred by Job​ #601 within each​ department: Machining​ Department: 9 Machine Hours 4 DL hours Assembly​ Department: 15 DL hours What is the amount of total allocated overhead for Job?
Business
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

Machining:

Allocated MOH= $603

Assembly:

Allocated MOH= $450

Explanation:

Giving the following information:

Machining:

Allocates overhead using machine-hours

Estimated manufacturing​ overhead: ​ $670,000

Estimated machine-hours= 10,000

Assembly:

Allocates overhead using direct labor hours.

Estimated manufacturing​ overhead: ​$450,000

Estimated direct labor hours= 15,000 hours

First, we need to calculate the estimated manufacturing overhead rate for each department:

To calculate the estimated manufacturing overhead rate we need to use the following formula:

Estimated manufacturing overhead rate= total estimated overhead costs for the period/ total amount of allocation base

Machining:

Estimated manufacturing overhead rate= 670,000/10,000= $67 per machine hour

Assembly:

Estimated manufacturing overhead rate= 450,000/15,000= $30 per direct labor hour.

Job​ 601:

Machining​ Department: 9 Machine Hours

Assembly​ Department: 15 DL hours

To allocate overhead we use the following formula:

Allocated MOH= Estimated manufacturing overhead rate* Actual amount of allocation base

Machining:

Allocated MOH= 67*9= $603

Assembly:

Allocated MOH= 30*15= $450

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3. Debit    Salaries expense            $32,000

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4. Debit    Prepaid Rent                   $21,000

Credit               Cash                                     $21,000

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Credit                  Accounts payable            $31,000

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7 0
3 years ago
Elizabeth Kennedy sells beauty supplies. Her annual demand for a particular skin sparkle is 17,000 units. The cost of placing an
elena-s [515]

Answer:

The minimum cost will be "$214085".

Explanation:

D = 1700 units \\\\S =  \$ 50 \\\\H=  20%\\

i) When quantity = 1-1500,  price = $ 12.50 , and holding price is $12.50 * 20 %= $2.50.

ii) When quantity = 1501 -10,000,  price = $ 12.45 , and holding price is $12.45 * 20 %= $2.49.

iii) When quantity = 10,0001- and more,  price = $ 12.40 , and holding price is $12.40 * 20 %= $2.48.

EOQ= \sqrt{\frac{2DS}{H}} \\\\EOQ1= \sqrt{\frac{2\times 17000\times 50}{2.50}} \\\\EOQ1=824.62 \ \ \ or \ \ \ 825\\

EOQ2= \sqrt{\frac{2\times 17000\times 50}{2.49}} \\\\EOQ1=826.2T \ \ \ or \ \ \ 826\\

EOQ3= \sqrt{\frac{2\times 17000\times 50}{2.48}} \\\\EOQ3=827.93 \ \ \ or \ \ \ 828\\

know we should calculate the total cost of EOQ1 and break ever points (1501 to 10,000)units

total \ cost = odering \ cost + holding \ cost + \ Annual \ product \ cost\\\\total_c  = \frac{D}{Q} \times S +  \frac{Q}{2} \times H + (p \times D) \\\\T_c  = \frac{17000}{825} \times 50+  \frac{825}{2} \times 2.50 + (12.50 \times 17000)\\\\T_c = 1030 .30 +1031.25+212500\\\\T_c =$ 214561.55\\\\

T_c  = \frac{17000}{1501} \times 50+  \frac{1501}{2} \times 2.49 + (12.45 \times 17000)\\\\T_c = 566.28 +1868.74+211650\\\\T_c =$ 214085.02 \ \ \ or \ \ \  $ 214085\\\\

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The total cost is less then 15001. So, optimal order quantity is 1501, that's why cost is = $214085.

5 0
3 years ago
Pls help
Arturiano [62]

Answer:

True

Explanation:

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4 0
3 years ago
A researcher wants to test the order of integration of some time series data. He decides to use the DF test. He estimates a regr
pav-90 [236]

Answer:

a) H0: u = presence of a unit root

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b) t stat = -0.064

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d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise

Explanation:

<u>a) stating the null and alternative hypothesis</u>

H0: u = presence of a unit root

HA: u ≠ presence of a unit root  ( i.e. stationary series )

<u>b) performing the test </u>

critical value = -2.88

T stat = coefficient / std error

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c) From the test, the value of T stat > critical value we will reject the Null hypothesis hence the next step will be to accept the alternative hypothesis

d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise

   

5 0
3 years ago
...<br><br><br><br><br>Great <br><br>-----------------
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Answer:

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