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ratelena [41]
2 years ago
12

Tenet Engineering, Inc. operates two user divisions as separate cost objects. To determine the costs of each division, the compa

ny allocates common costs to the divisions. During the past month, the following common costs were incurred: Computer services (85% fixed) $260,000 Building occupancy 600,000 Personnel costs 110,000 Total common costs $970,000 The following information is available concerning various activity measures and service usages by each of the divisions: Division A Division B Area occupied (square feet) 20,000 40,000 Payroll $380,000 $180,000 Computer time (hours) 200 220 Computer storage (megabytes) 4,050 -0- Equipment value $200,000 $250,000 Operating profit (pre-allocations) $555,000 $495,000 If common computer service costs are allocated using computer time as the allocation basis, what is the computer cost allocated to Division B
Business
1 answer:
skad [1K]2 years ago
5 0

Answer:

$136,190

Explanation:

The computation of computer cost allocated to Division B is shown below:-

Computer cost allocated to Division B = Computer service cost × Computer time of Division B ÷ Total computer time.

= $260,000 × 220 ÷ (200 + 220)

= $260,000 × 220 ÷ 420

= $260,000 × 0.5238

= $136,190

Therefore for computing the computer cost allocated to Division B we simply applied the above formula.

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Helpful team members strive to resolve differences and encourage a warm, supportive climate by praising and agreeing with others
Salsk061 [2.6K]

Answer:

The statement is: True.

Explanation:

<em>Helpful team members</em> are those who take advantage of their expertise to help the team achieve its goals faster or at least at the expected time. More than troublemakers they are problem solvers. Helpful team members act proactively, tend do be charismatic and enthusiastic, and motivate others to continue working towards a common goal.

7 0
2 years ago
A graphics reproduction firm has four units of equipment that are automatic but occasionally become inoperative because of the n
elena55 [62]

Answer:

(a) Average number of unit in line  = 0.256

(b) Average number of unit in operation= 3.209

(c) Average number of unit being service in operation = 0.535

Explanation:

Given Data:

Number of machine N = 4

Number of attendant (S) = 1

Service time (T)= 5 mins

Time required by the machine before servicing = 30 mins

Calculating the service factor (X) using the formula;

X = T/(T+U)

    = 5/(5+30)

    = 5/35

     = 0.1429

(a) Calculating the average number of unit in line (L) using the formula;

L = N* (1-F)

where, N = Number of unit

F = efficiency factor

L = average number of unit in line

Using the finite queuing table at X = 0.1429 and S = 1,

Efficiency factor = 0.936

Substituting, we have;

L = 4*(1-0.936)

   = 4* 0.064

   = 0.256

(c) Calculating the average number of unit being service in operation (H) using the formula;

H = N*F*X

   = 4 *0.936*0.1429

   = 0.535

(b) Calculating the average number of unit in operation using the formula;

Average number of unit in operation= Number of unit-down unit

But down unit = L+H

The formula becomes;

Average number of unit in operation= Number of unit-(L+H)

                                                             = 4 - (0.256+0.535)

                                                             = 4-0.791

                                                             = 3.209

6 0
2 years ago
The 6.3 percent, semi-annual coupon bonds of PE Engineers mature in 13 years and have a price of $992. These bonds have a curren
ludmilkaskok [199]

Answer:

6.35, 6.39 and 6.49

Explanation:

6.3% = 0.063

yield = 0.063 ×$1,000/ 0.992 yield = 0.063 ×$1,000)/ 0.992 ×$1,000)

Current yield = 0.0635, or 6.35 percent PV = $992 = 0.063× $1,000 / 2) ×{(1 - {1 / [1 + (r / 2)]26}) / (r/ 2)} + $1,000 / [1 + (r / 2)]26 r = .0639, or 6.39 percent EAR = [1 + .0639 / 2)]2 - 1 EAR = .0649, or 6.49

7 0
2 years ago
Read 2 more answers
A manufacturer shipped units of a certain product to two locations. The equation above shows the total shipping cost TTT, in dol
andriy [413]

Answer: 2,200 units.

Explanation:

The complete exercise is:

T = 5c + 12 f

A manufacturer shipped units of a certain product to two locations. The equation above shows the total shipping cost T, in dollars, for shipping c units to the closer location and shipping f units to the farther location. If the total shipping cost was $47,000 and 3,000 units were shipped to the farther location, how many units were shipped to the closer location?

Given the following equation:

T = 5c + 12 f

You know that "T" is the total shipping cost (in dollars), "c" is the number of units shipped to the closer location and "f" is the number of units shipped to the farther location.

Based on the information given in the exercise, you can identify that, in this case:

T=47,000\\\\f=3,000

Then, knowing those values, you need to substitute them into the given equation:

47,000 = 5c + 12(3,000)

And finally, you must solve for "c" in order to calculate the number of units that  were shipped to the closer location.

You get that this is:

47,000 = 5c + 12(3,000)\\\\47,000-36,000 = 5c\\\\11,000=5c\\\\\frac{11,000}{5}\\\\c=2,200

3 0
2 years ago
Read 2 more answers
Alejandro's supervisor notices that sales were down in february. he asks alejandro to look up past sales figures in the company'
Nana76 [90]
Monthly sales over five years

5 0
2 years ago
Read 2 more answers
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