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koban [17]
3 years ago
9

Farmer Company purchased equipment on January 1, Year 1 for $82,000. The equipment is estimated to have a 5-year life and a salv

age value of $4,000. The company uses the straight-line depreciation method. At the beginning of Year 4, Farmer revised the expected life to eight years.
The annual amount of depreciation expense for each of the remaining years would be:

a) $6,240.
b) $4,400.
c) $7,040.
d) $3,900.
Mathematics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

option (a) $6,240

Step-by-step explanation:

Given:

Purchasing cost of the equipment = $82,000

Estimated life = 5 years

Salvage value = $4,000

Revised expected life = 8 years

Now,

Depreciation per year = \frac{\textup{82,000-4,000}}{\textup{5}}

therefore,

The accumulated Depreciation at the beginning of year 4

= Annual depreciation × years passed

= 15,600 × 3

= $46,800

Thus,

The book value at the beginning of year 4

= Purchasing cost - Depreciation

= $82,000 - $46,800

= $35,200

Now,

The remaining life = Revised estimated life - Years passed

= 8 - 3

= 5 years

therefore,

Depreciation expense =\frac{\textup{book value at the beginning of year 4 - salvage value}}{\textup{Revised estimated life}}

= \frac{\textup{35,200  - 4000}}{\textup{5}}

= $6,240

Hence,

The correct answer is option (a) $6,240

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mariarad [96]

Answer:

a. \frac{35}{51}

b. \frac{51}{100}

c. \frac{1}{5}

Step-by-step explanation:

Suppose cities represented by C', suburbs represented by S and rural represented by R,

Let x be the total number of bonds issued throughout the US,

According to the question,

n(A) = 70% of x = 0.7x,

n(B) = 10% of x = 0.1x,

n(C) = 20% of x = 0.2x,

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n(A∩S) = 20% of n(A) = 0.2 × 0.7x = 0.14x,

n(A∩R) = 30% of n(A) = 0.3 × 0.7x = 0.21x,

n(B∩C') = 40% of n(B) = 0.4 × 0.1x = 0.04x,

n(B∩S) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(B∩R) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(C∩C') = 60% of n(C) = 0.6 × 0.2x = 0.12x,

n(C∩S) = 15% of n(C) = 0.15 × 0.2x = 0.03x,

n(C∩R) = 25% of n(C) = 0.25 × 0.2x = 0.05x,

n(C') = n(A∩C')  + n(B∩C')  + n(C∩C')  = 0.35x + 0.04x + 0.12x = 0.51x

n(S) = n(A∩S) + n(B∩S) + n(C∩S) = 0.14x + 0.03x + 0.03x = 0.20x

a. The probability that it will receive an A rating, if a new municipal bond is to be issued by a city,

P(\frac{A}{C'})=\frac{P(A\cap C')}{P(C')}=\frac{0.35x/x}{0.51x/x}=\frac{0.35}{0.51}=\frac{35}{51}

b. The proportion of municipal bonds are issued by cities = \frac{n(C')}{x}

=\frac{0.51x}{x}

=\frac{51}{100}

c. The proportion of municipal bonds are issued by suburbs = \frac{n(S)}{x}

=\frac{0.20x}{x}

=\frac{20}{100}

=\frac{1}{5}

3 0
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