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2 years ago

15

If a residence is rented for 15 days or more and is used for personal purposes for not more than 14 days or 10 percent of the da

ys rented, whichever is greater, no allocation of expenses is required and the taxpayer may claim a deduction for the full amount of the expenses.

1 answer:

11Alexandr11 [23.1K]2 years ago

7 0

Answer:

False.

Step-by-step explanation:

For a residence whose primary usage is for rental purpose, its expenses must still be allocated between personal and rental days as agreed between the client and the owner

All terms of agreement remains valid and there's no other allocation of expenses that will be required. The taxpayer cannot claim any deduction for the full amount of the expenses.

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Step-by-step explanation:

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A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a yea

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The 95% confidence interval is

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Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 36$

The first sample mean is $\= x_1 = 3.5$

The first standard deviation is $\sigma_1 = 5.9 \ pounds$

The second sample size is $n_2 = 36$

The second sample mean is $\= x_2 = 0.6$

The second standard deviation is $\sigma = 4.4$

Generally the degree of freedom is mathematically represented as

$df = \frac{ [ \frac{s_1^2 }{n_1 } + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2 }$

=> $df = \frac{ [ \frac{5.9^2 }{34 } + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2 }$

=> $df =63$

Generally the standard error is mathematically represented as

$SE = \sqrt{ \frac{s_1 ^2 }{n_1} + \frac{s_2^2 }{ n_2 } }$

=> $SE = \sqrt{ \frac{ 5.9 ^2 }{ 36 } + \frac{ 4.4^2 }{36} }$

=> $SE = 1.227$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2}, 63 } = 1.998$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2}, 63 } * SE$

=> $E = 1.998 * 1.227$

=> $E = 2.45$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 - \x_2) -E < \mu$

=> $(3.5 - 0.6) - 2.45 < \mu_1 - \mu_2 < ( 3.5 - 0.6) + 2.45$

=> $0.45 < \mu_1 - \mu_2 < 5.35$

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