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svetlana [45]
3 years ago
10

Sooner Machinery Company purchased a delivery truck at a cost of $56,000 on March 10, 2018. The truck has a useful life of six y

ears with an estimated salvage value of $5,000. Compute the annual depreciation for the first two calendar years using
a. The straight-line method.
b. The 150% declining-balance method.
Business
1 answer:
Fofino [41]3 years ago
4 0

Answer:

Results are below.

Explanation:

Giving the following information:

Purchase price= $56,000

Useful life= 6 yearsd

Salvage value= $5,000

<u>a. To calculate the annual depreciation, we need to use the following formula:</u>

Annual depreciation= (original cost - salvage value)/estimated life (years)

Annual depreciation= (56,000 - 5,000) / 6= $8,500

<u>Year 1</u>:

Annual depreciation= (8,500/12)*10= $7,083.33

<u>Year 2:</u>

Annual depreciation= $8,500

<u>b. To calculate the annual depreciation, we need to use the following formula:</u>

Annual depreciation= 1.5*[(book value)/estimated life (years)]

<u>Year 1:</u>

Annual depreciation= [(1.5*8,500)/12]*10= $10,625

<u>Year 2:</u>

Annual depreciation= [(51,000 - 10,625)/6]*1.5

Annual depreciation= $10,093.75

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a project that will last for 8 years is expected to have equal annual cash flows of $97,900. If the required return is 7.6 perce
iVinArrow [24]

Question:

MC algo 5-28 Calculating NPV A project that will last for 8 years is expected to have equal annual cash flows of $97,900. If the required return is 7.6 percent, what maximum initial investment would make the project acceptable?

Multiple Choice $516,751.56 $571,237.51 $1,026,395.85 $482,301.46 $550,008.71

Answer:

PV of cash inflow = $571,237.5  

Explanation:

T<em>he maximum initial investment amount to be paid is the present value of the series of the annual cash inflow discounted at the opportunity cost rate of 7.6% per annum. </em>

<em>In other words,the maximum to be paid for the investment should be equal to the value today of the series of eight equal annual cash flow of $97,900 discounted at 7.6%</em>

This is given in the relationship below:

PV of cash inflow = A ×( 1- (1+r)^(-n))/r )  

A- equal annual cash - 97,900. r-rate of return - 7.6%, n-number of years- 8

PV = 97,900 × ( 1 - (1+0.076)^(-8)/0.76)=  571,237.5  

PV of cash inflow = $571,237.5  

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3 years ago
Which of the following is an example of an ethical standard you may find in
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B) bribing government officials
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3 true statements about recording journal entries in Quickbooks
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Answer:

Explanation:

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3 years ago
Clark Oil and Gas incurred costs of $15.3 million for the rights to extract resources from a natural gas deposit. The company ex
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Answer and Explanation:

The computation is shown below

Depletion expense per cubic feet  would be

= Total cost ÷ Total cubic feet

= $15.3 million ÷ 8 million

= $1.9125 per cubic feet

Now

Depletion for year 1

= Number of cubic feet extracted × Depletion per cubic feet

= 800,000 × $1.9125

= $1,530,000

And, for the depletion for year 2

= 1,600,000 × $1.9125

= $3,060,000

7 0
3 years ago
EBike, an electronic bicycle manufacturer, has identified two customer segments, one is willing to pay a higher price for a cust
kramer

Solution:

1)

Profit function of customised bicycle customers, P1 = d1*(p1-c)

= (11000-25p1)*(p1-160)

= 11000p1-1760000-25p1^2+4000p1

= -25p1^2+15000p1-1760000

In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP1/dp1 = d(-25p1^2+15000p1-1760000)/dp1 = 0

=> -50p1+15000 = 0

=> p1 = 300

Profit function of price sensitive customers, P2 = d2*(p2-c)

= (11000-45p2)*(p2-160)

= 11000p2-1760000-45p2^2+7200p2

= -45p2^2+18200p2-1760000

In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP2/dp2 = d(-45p2^2+18200p2-1760000)/dp2 = 0

=> -90p2+18200 = 0

=> p2 = 202.22

Price to be charged for customised segment = $ 300

Price to be charged for price sensitive segment = $ 202.22

------------------------------------------------------------

2)

Considering single price, p

Total profit from both segments, P = (d1+d2)*(p-160)

= (11000-25p+11000-45p)*(p-160)

= (22000-70p)*(p-160)

= 22000p-3520000-70p^2+11200p

= -70p^2+33200p-3520000

In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP/dp = d(-70p^2+33200p-3520000)/dp = 0

=> -140p+33200 = 0

=> p2 = 237.14

3)

This is solved by Solver as follows:

[ Find the attachments ]

8 0
3 years ago
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