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tamaranim1 [39]
3 years ago
6

During the current year, Martin purchases undeveloped land as an investment. Martin intends to rent the land as pastureland and

hopefully sell it later for a profit. In the current year, Martin receives no rent but he does pay taxes of $2,800, mortgage interest of $900 and liability insurance of $500. How much of these expenses can Martin deduct (before any limitations) on his current tax return?A) $0B) $1,400C) $3,700D) $4,200
Business
1 answer:
Levart [38]3 years ago
6 0

Answer:

D) $4,200

Explanation:

Martin's expenses on land = taxes of $2,800 + mortgage interest of $900 a+ liability insurance of $500 = $4,200

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2 years ago
Now, suppose the government were to decrease net taxes by $20 billion without changing the level of government purchases. This w
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Answer:

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3 years ago
The Rhaegel Corporation’s common stock has a beta of 1.7. If the risk-free rate is 5.7 percent and the expected return on the ma
Nataliya [291]

Answer:

the cost of equity capital as per CAPM is 14.71%

Explanation:

The computation of the cost of equity capital as per CAPM is shown below:

As we know that

= Risk free rate of return + beta × (market rate of return - risk free rate of return)

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The same is relevant

4 0
3 years ago
Find the mean, median and mode for the above set of data
qwelly [4]

Answer:

a. Mean = 35.2 ≈ 35

b. Median = 35.6 ≈ 36

c. Mode = 36.6 ≈ 37

Step-by-step Explanation:

==>Given:

Class of ages in yrs

No. of cases of each class = f

Midpoint of each class = x

Product of midpoint and no. of cases of each class = fx

==>Required:

a. Mean

b. Median

c. Mode

==>SOLUTION:

a. Mean = (Σfx)/Σf

Σf = sum of no. of cases = 5+10+20+22+13+5 = 75

Σfx = 47.5+195+590+869+643.5+297.5 = 2,642.5

Mean = 2,642.5/75 = 35.2 ≈ 35

b. Median = Lm + [((Σf/2) - Cfb)/fm]Cw

Our median is between the 37th and the 38th term, which can be found in the class interval 35-44. This is our median class.

Lm = Lower class boundary of the median class = lower limit of the Medina class + upper limit of the class before the median class ÷ 2 = (35+34)/2 = 34.5

Σf/2 = 75/2 = 37.5

Cfb = Cumulative frequency of class before the median class = 5+10+20 = 35

fm = frequency of the Medina class = 22

Cw = Class width = upper class boundary - lower class boundary = 44.5-34.5 = 10

Median = 34.5 + [(37.5-35)/22] × 10

= 34.5 + [2.5/22] × 10

= 34.5 + [25/22]

= 34.5 + 1.1

= 35.6 ≈ 36

c. Mode = Lm + [∆¹/(∆¹+∆²)]Cw

Modal class = (35-44) [i.e. the class with the highest frequency, which is where our mode falls in]

Lm = lower class boundary of the modal class = lower limit of the modal class + upper limit of the class before the modal class ÷ 2 = (35+34)/2 = 34.5

∆¹ = difference between the frequency of the modal class & the frequency of the class before the modal class = 22 - 20 = 2

∆² = difference between the frequency of the modal class & the frequency of the class after the modal class = 22 - 13 = 9

Cw == Upper class boundary - Lower class boundary = 44.5 - 34.5 = 10

Mode = 34.5 + [2/(2+9)] × 10

= 34.5 + [2/11] × 10

= 34.5 + [20/11]

= 34.5 + 1.8

Mode = 36.6 ≈ 37

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