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Goshia [24]
3 years ago
14

Emily is itemizing deductions on her federal income tax return and had $5200

Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0

Answer:

Emily can deduct $1975 for medical expenses.

Step-by-step explanation:

Emily is itemizing deductions on her federal income tax return and had $5200. Her AGI was $43,000.

We have:

The AGI is =$43,000

Medical expenses amount = $5200

If medical expenses are deductible to the extent that they exceed 7.5% or 0.075  of a taxpayer’s AGI, means  = 43,000 * 0.075 = $3225

Now the deductible amount is  = $5200 - $3225 =  $ 1975

<em>Hence,Emily can deduct $1975 for medical expenses.</em>

ivann1987 [24]3 years ago
6 0

Answer:

Emily can deduct the $1,975 in medical expenses from her adjusted gross income.

Step-by-step explanation:

Before an individual could deduct medical expenses as long as they exceeded 7.5% of their adjusted gross income (currently it's 10%).

If Emily's AGI = $43,000, then she will be able to deduct medical expenses that exceed $3,225 (= $43,000 x 7.5%).

Emily's medical expense deductions = $5,200 - $3,225 = $1,975.

If this situation happened this year, Emily would have only been able to deduct $900  [= $5,200 - ($43,000 x 10%)]

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Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

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Step-by-step explanation:

Let the events be:

W = Wednesday

T = Thursday

F = Friday

S = Saturday

Their corresponding probabilities are

P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1

Since Y = number of days beyond Wednesday that it takes for both magazines to arrive(so possible Y values are 0, 1, 2 or 3)

The possible number of outcomes are therefore 4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)

The values associated for each of the outcomes are as follows:

Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3

The probability mass function of Y is,

P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19

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