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denis23 [38]
2 years ago
10

Cainas Cookies purchased a commercial oven on 1/1/14 for a total cost of 35,000. Estimated useful life is 6 years, with a salvag

e value of 5,000 at the end of that time. Cainas estimates that the equipment will be used for 12,000 baking hours. For the first year of operations, Cainas had 2,500 backing hours. For the second year Cainas had 1,700 hours. Compute the depreciation for YEAR 2. Group of answer choices
Business
1 answer:
Anit [1.1K]2 years ago
3 0

Answer:

Units of production = $4250

Straight line depreciation expense = $5,000

Double declining method = $7.777

Explanation:

The depreciation method to he used wasn't stated, so I calculated the depreciation expense using 3 depreciation methods

Straight line depreciation expense = (Cost of asset - Salvage value) / useful life

(35,000 - 5,000) / 6 = $5,000

The depreciation expense each year would be $5000

Depreciation expense using the double declining method = Depreciation factor x cost of the asset

Depreciation factor = 2 x (1/useful life)

2 / 6 = 0.3333

Deprecation expense in year 1 = 0.3333 x $35,000 = $11,666.67

Book value = $35,000 - $11,666.67 = $23,333.33

Depreciation expense in year 2 = $23,333.33 × 0.3333 = $7.777

Depreciation expense using units of production = ( hours used in year / total estimated hours of the machine) x (Cost of asset - Salvage value)

(1,700 / 12,000) x (35,000 - 5,000) = $4250

I hope my answer helps you

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An appliance dealer must decide how many (if any) new microwave ovens to order for next month. The ovens cost $220 and sell for
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Answer:

Explanation:

Order 0: we have unsold items for which the return is -25

return is -25*(.4*1+.2*2+.1*3) = -25*1.1 = $-27.50

Order 1: we have to sell at a discount if no orders, otherwise sell 1, and unsold items if demand 2 or 3

return is .3*(1/2*300-220) + (1-.3)*(300-220) + -.25*(.2*1+.1*2) = .3*-70+.7*80+-25*(.4) =

-21 + 56 - 10 = $25

Order 2: we have to sell at a discount if 0 or 1 orders, sell 1 or 2, and unsold items if demand 3

return is (.3*2+.4*1)*(1/2*300-220)+(.4*1+(.2+.1)*2)*(300-220)+-25*.1 =1*-70+1*80-25*.1 =

-70 + 80 - 2.5 = $7.50

Order 3:

return is (.3*3+.4*2+.2*1)*(1/2*300-220)+(.4*1+.2*2+.1*3)*(300-220) = 1.9*-70 + 1.1*80 =

-133 + 88 = -$45

Order 1, with a return of $25, as this is the highest return.

b) If we had a perfect information, we would never pay a penalty for underordering or suffer a discounted return from over-ordering

(.4*1+.2*2+.1*3)*(300-220) = 1.1*80 = $88

Then, the value of perfect information is $88 - $25 = $63

c) P(D=0|F) = P(F|D=0)*P(D=0)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.1*.3/(.1*.3+.2*.4+.3*.2+.9*.1)=.03/.26 = 3/26

P(D=1|F) = P(F|D=1)*P(D=1)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.2.4/(.1*.3+.2*.4+.3*.2+.9*.1)=.08/.26 = 4/13

P(D=2|F) = P(F|D=2)*P(D=2)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.3*.2/(.1*.3+.2*.4+.3*.2+.9*.1)=.06/.26 = 3/13

P(D=3|F) = P(F|D=3)*P(D=3)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.9*.1/(.1*.3+.2*.4+.3*.2+.9*.1)=.09/.26 = 9/26

P(D=0|U) = P(U|D=0)*P(0)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.8*.3/(.8*.3+.3*.4+.1*.2+.1*.1)=.24/.39 = 8/13

P(D=1|U) = P(U|D=1)*P(1)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.3*.4/(.8*.3+.3*.4+.1*.2+.1*.1)=.12/.39 = 4/13

P(D=2|U) = P(U|D=`)*P(`)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.1*.2/(.8*.3+.3*.4+.1*.2+.1*.1)=.02/.39 = 2/39

P(D=3|U) = P(U|D=3)*P(3)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.1*.1/(.8*.3+.3*.4+.1*.2+.1*.1)=.01/.39 = 1/39

P(N|D=0 = 1-.1-.8 = .1

P(N|D=1) = 1 - .2 - .3 = .5

P(N|D=2) = 1 - .3 - .1 = .6

P(N|D=3) = 1 - .9 - .1 = 0

P(D=0|N) = P(N|D=0)*P(D=0)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))=.1*.3/(.1*.3+.5*.4+.6*.2+.0*.1)= .03/.35 = 3/35

P(D=1|N) = P(N|D=1)*P(D=0)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))= .5*.4/(.1*.3+.5*.4+.6*.2+.0*.1)= .20/.35 = 4/7

P(D=2|N) = P(N|D=2)*P(D=2)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))= .6*.2/(.1*.3+.5*.4+.6*.2+.0*.1)= .12/.35 = 12/35

P(D=3|N) = 0

If the result of the survey is an F, we have

P(D=0|F) = 3/26

P(D=1|F) = 4/13

P(D=2|F) = 3/13

P(D=3|F) = 9/26

If the order is 0, the return is -25*(1*4/13+2*3/13+3*9/26) = -25*47/26 = -1175/26 = -$45.19

If the order is 1, the return is 3/26*-70+(1-3/26)*80+-25*(1*3/13+2*9/26) = 515/13 = $39.62

If the order is 2, the return is (3/26*2+4/13)*-70+(1*4/13+2*(3/13+9/26))*80 + -25*9/26 =

1835/26 = $70.58

If the order is 3, the return is (3/26*3+4/13*2+3/13)*-70+(1*4/13+2*3/13+3*9/26)*80 =

795/13 = $61.15

We should order 2.

P(D=0|U) = 8/13

P(D=1|U) = 4/13

P(D=2|U) = 2/39

P(D=3|U) = 1/39

If we order 0, the return is (4/13*1+2/39*2+1/39*3)*-25 = -475/39 = -$12.18

If the order is 1, the return is 8/13*-70+(1-8/13)*80+-25*(1*2/39+2*1/39) =-580/39= -14.87

If the order is 2, the return is (8/13*2+4/13)*-70+(1*4/13+2*(2/39+1/39))*80 + -25*1/39 =

-2785/39= -$71.41

If the order is 3, the return is (8/13*3+4/13*2+2/39*1)*-70+(1*4/13+2*2/39+3*1/39)*80 =

-1780/13 = -$136.92

Order 0

P(D=0|N) = 3/35

P(D=1|N) = 4/7

P(D=2|N) = 12/35

P(D=3|N) = 0

If we order 0, the return is (4/7*1+12/35*2)*-25 = -220/7 = -$31.43

If the order is 1, the return is 3/35*-70+(1-3/35)*80+-25*(1*12/35) = 410/7 = $58.57

If the order is 2, the return is (3/35*2+4/7)*-70+(1*4/7+2*12/35)*80 = 340/7 = $48.57

We don't order 3, as the probability of 3 is 0

we order 1

We order 2 if there is an F, 0 if there is an N, and 1 if there is a U.

d) P(F) = .26

P(N) = .39

P(U) = .35

Then, the expected return is .26*1835/26 +-475/39*.39 + 410/7*.35 = $34.10

Since we make $25 if we just take 1, we should pay up to $34.10-$25 = $9.10 for the survey.

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C. Tenancy in common.

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