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BartSMP [9]
3 years ago
12

Chahana acquired and placed in service $1,185,000 of equipment on August 1, 2019 for use in her sole proprietorship. The equipme

nt is 5-year recovery property. No other acquisitions are made during the year. Chahana elects to expense the maximum amount under Sec. 179, and bonus depreciation is not applied. Chahana's total deductions for 2019 (including Sec. 179 and depreciation) are:___________.
A) $1,020,000.
B) $237,000.
C) $1.185,000.
D) $1,053,000
Business
1 answer:
algol [13]3 years ago
7 0

Answer:

D) $1,053,000

Explanation:

Calculation to determine what Chahana's total deductions for 2019 (including Sec. 179 and depreciation) are

Sec 179 immediate expensing $1,020,000

MACRS depreciation:

Add Basis for depreciation $33,000

[($1,185,000 - $1,020,000 Sec. 179) × .20]

Total depreciation $1,053,000

($1,020,000+$33,000)

Therefore Chahana's total deductions for 2019 (including Sec. 179 and depreciation) are:$1,053,000

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In a unionized firm, the _____ clause of the collective bargaining agreement typically retains for management the authority to i
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The answer is:  "management rights" .
_________________________________________________________
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Click this link to view O*NET’s Work Context section for Human Resources Managers. Note that common contexts are listed toward t
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Read 2 more answers
A portfolio manager sells Treasury bonds and buys corporate bonds because the spread between corporate- and Treasury-bond yields
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Answer: The correct answer is "an intermarket spread".

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The cob Douglas production function is given by Q(K,L)=AK^1.4*L^1.6
Alexeev081 [22]

Part a) The Cob Douglas production function is given as:

Q(K,L)=AK^{1.4} L^ {1.6 } .

To show that this function is homogeneous with degree 3, we introduce be a parameter, t.

Q(tK,tL)=A(tK)^{1.4} (tL)^ {1.6 } .

Using properties of exponents, we on tinder:

Q(tK,tL)=At^{1.4}K^{1.4} t^ {1.6 }L^ {1.6 } .

This implies that:

Q(tK,tL)=t^{1.4} \times t^ {1.6 }(AK^{1.4} L^ {1.6 } )

Q(tK,tL)=t^{1.4 + 1.6}(AK^{1.4} L^ {1.6 } )

Simplify the exponent of t to get;

Q(tK,tL)=t^{3}(AK^{1.4} L^ {1.6 } )

Hence the function is homogeneous with degree, 3

Part b) To verify Euler's Theorem, we must show that:

K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L}=3AK^{1.4}L^{1.6}

Verifying from the left:

K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =K(1.4AK^{0.4} L^{1.6}) + L(1.6AK^{1.4} L^{0.6})

K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =1.4(AK^{1.4} L^{1.6}) + 1.6(AK^{1.4} L^{1.6})

K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =(1.4 +  1.6)(AK^{1.4} L^{1.6})

K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =3(AK^{1.4} L^{1.6})

Q•E•D

8 0
3 years ago
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