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2 years ago

Trinkle Co., Inc. made several purchases of long-term assets in Year 1. The details of each purchase are presented here.

Business

1 answer:

Orlov [11]2 years ago

5 0

Answer:

New Office Equipment $42,863

Basket Purchase Of Copier, Computer, Scanner $61,500

Land For New Warehouse $310,050

Explanation:

Calculation to determine the amount of cost to be capitalized in the asset accounts

NEW OFFICE EQUIPMENT

Amount of cost to be capitalised in the asset accounts = $41,900*0.98+$860+$510+$431

Amount of cost to be capitalised in the asset accounts =$41,062+$860+$510+$431

Amount of cost to be capitalised in the asset accounts =$42,863

BASKET PURCHASE OF COPIER, COMPUTER AND SCANNER

Amount of cost to be capitalised in the asset accounts = $22,755 + $6,765 + $31,980

Amount of cost to be capitalised in the asset accounts= $61,500

LAND FOR NEW WAREHOUSE with an old building torn down

Amount of cost to be capitalised in the asset accounts = $82,400 + $4,750 - $1,800 + $7,700 + $217,000

Amount of cost to be capitalised in the asset accounts = $310,050

Therefore The Amount of cost to be capitalised in the asset accounts are:

New Office Equipment $42,863

Basket Purchase Of Copier, Computer, Scanner $61,500

Land For New Warehouse $310,050

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3 years ago

This morning, DJ’s invested $238,000 to help fund a company expansion project planned for three years from now. How much additio

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Answer:

$3,842.78

Explanation:

We must determine the future value of the money invested and then calculate the difference between both return rates. We can use the future value formula: FV = present value x (1 + return rate)ⁿ

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3 years ago

Automobile repair costs continue to rise with the average cost now at $367 per repair.† Assume that the cost for an automobile r

vovikov84 [41]

Answer:

a) 0.1728

b) 0.09183

c) 0.7354

d) $ 222.25

Explanation:

Given

mean = $\mu$ = $367

Standard deviation = $\sigma$ =$88

Cost of automobile repair is normally distributed.

a) We have to find P( x > 450 )

P( x > 450 ) = 1 - P( x <= 450 )

Using excel function, P( x <= x ) = NORMDIST (x, $\mu$ , $\sigma$ , 1 )

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P( x <450 ) = NORMDIST( 450 , 367, 88, 1 ) = 0.8272

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We have to find a.

Using Excel, = NORMINV ( Probability, $\mu$ , $\sigma$ )

a = NORMINV ( 0.05 , 367, 88 ) = 222.2529

Cost = $ 222.25

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