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lesya692 [45]
2 years ago
8

Bogle Company purchased machinery for $320,000 on January 1, 2014. Straight-line depreciation has been recorded based on a $20,0

00 salvage value and a 5-year useful life. The machinery was sold on May 1, 2018 at a gain of $6,000. How much cash did Bogle receive from the sale of the machinery?
Mathematics
1 answer:
horrorfan [7]2 years ago
7 0

Answer:

$66,000

Step-by-step explanation:

Data provided in the question:

Purchasing cost of the machine = $320,000

Salvage value = $20,000

Useful life of the machine = 5 years

The gain on selling of the machine = $6,000

Now,

The annual depreciation of the machine

= [Cost - salvage value] ÷ ( Useful life )

= [ $320,000 - $20,000 ] ÷ 5

= $60,000

Accumulated depreciation from January 1, 2014 to May 1, 2018

= Annual depreciation ×  (Duration from January 1, 2014 to May 1, 2018 )

= $60,000 × ( 4 years 4 months )

= $60,000 × \frac{52}{12} years      

[48 months + 4 months = 52 months]

= $260,000

Therefore,

Book value on May 1, 2018 = Cost - Accumulated depreciation

= $320,000 - $260,000

= $60,000

Now,

Gain = Selling cost - Book value

$6,000 = Selling cost - $60,000

or

Selling cost = $66,000

Hence,

cash Bogle received from the sale of the machinery was $66,000

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Suppose you were exploring the hypothesis that there is a relationship between parents’ and children’s party identification. Wou
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Answer:

No

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For example in a population of smokers some may be in the habit of taking cocaine. But a sample of cocaine users does not mean the whole population uses it.

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2 years ago
Consider the function represented by the graph. What is the domain of this function
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A school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. one week the total sales for these items were
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solution:

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Answer:

Step-by-step explanation:

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For the null hypothesis,

µ = 17

For the alternative hypothesis,

µ < 17

This is a left tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80,

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ)/(s/√n)

Where

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The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.

4 0
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