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marta [7]
3 years ago
5

Minor Company installs a machine in its factory at the beginning of the year at a cost of $135,000. The machine's useful life is

estimated to be 5 years, or 300,000 units of product, with a $15,000 salvage value. During its first year, the machine produces 64,500 units of product. Determine the machines' first year depreciation under the straight-line method. Multiple Choice $27,000. $29,025. $23,779. $24,000. $25,800.
Business
1 answer:
VikaD [51]3 years ago
7 0

Answer:

The straight line depreciation for the first year is $24000

Explanation:

The straight line method of depreciation charges/allocates a constant amount of depreciation through out the useful life of the asset. The straight line depreciation expense for the year is calculated as follows,

Straight line depreciation = (Cost - Salvage Value) / Estimated useful life

Straight line depreciation = (135000 - 15000) / 5  = $24000 per year

Thus, the amount of depreciation for first year under straight line method is $24000

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The content areathe point of zero profit is called the: Break-even point.

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A deduction from adjusted gross income for yourself, your spouse, and qualified dependents is:
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Explanation:

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2 years ago
If the supply function for a commodity is p = q2 + 6q + 16 and the demand function is p = −7q2 + 2q + 436, find the equilibrium
adell [148]

Answer:

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equilibrium price = 107

Explanation:

Data provided in the question:

supply function, p = q² + 6q + 16  ........(1)

demand function is p = −7q² + 2q + 436

Now at equilibrium

Demand = Supply

Thus,

q² + 6q + 16 = −7q² + 2q + 436

or

q² + 6q + 16 + 7q² - 2q - 436 = 0

or

8q² + 4q - 420 = 0

or

2q² + q - 105 = 0

on solving for the roots of q

using the Quadratic Formula where

a = 2, b = 1, and c = -105

[ x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }]

a = 2, b = 1, and c = -105

Thus,

[ q = \frac{ -1 \pm \sqrt{1^2 - 4(2)(-105)}}{ 2(2) }]

[ q = \frac{ -1 \pm \sqrt{1 - -840}}{ 4 }]

[ q = \frac{ -1 \pm \sqrt{841}}{ 4 }]

The discriminant ( b² - 4ac > 0)

so, there are two real roots.

Therefore,

q = [\frac{ -1 \pm 29}{ 4 }]

[ q = \frac{ 28 }{ 4 } \; \; \; q = -\frac{ 30 }{ 4 }]

[ q = 7 \; \; \; q = -\frac{ 15}{ 2 }]

since,

Quantity cannot be negative

Thus,

q = 7

therefore, substituting q in (1)

p = 7² + 6(7) + 16

or

p = 49 + 42 + 16

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