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katrin2010 [14]
3 years ago
13

The taxpayer had a qualifying child who is not disabled and who turned 13 on may 31, 2017. They paid a child care provider to ca

re for the child for the entire year while they worked. How many months of child care expenses can they deduct on their tax return?
Mathematics
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

5 months

Step-by-step explanation:

The taxpayer can only deduct the first five months of the qualifying child. May is the fifth month. On June 1st, the qualifying child is going to be 14 years old. Therefore he cannot be included in work-related expenses as specified in the Publication 503 - Main Content from the IRS. Please see the Working-Related Expenses, Indent (b) which is as follows:

(b) The parent of your qualifying person if your qualifying person is your child and under age 13.

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An electrical rm manufactures light bulbs that have a life span that is approximately normally distributed. The population stand
stira [4]

Answer:

The 't' test statistic = 1.46 < 1.69

The test of hypothesis is H 0 :μ = 800 is accepted

A sample of 30 bulbs are found came from average µ= 800

Step-by-step explanation:

Step 1:-

Given population of mean μ = 800

given size of small sample n =30

sample standard deviation 'S' = 45

Mean value of the sample χ = 788

Null hypothesis H_{0} =µ  =800

alternative hypothesis  H_{1} = µ ≠ 800

<u>Step 2</u>:-

The 't' test statistic t =  \frac{x-μ}{\frac{sig}{\sqrt{n} } }

                             t = \frac{788-800}{\frac{45}{\sqrt{30} } }

                       t = \frac{12}{8.215} = 1.4607

<u>Step 3</u>:-

The degrees of freedom γ = n-1 = 30-1 =29

From "t" value from table at 0.05 level of significance ( t = 1.69)

The calculated value t = 1.4607 < 1.69

Therefore The null hypothesis H_{0} ' is accepted.

<u>conclusion</u>:-

A sample of 30 bulbs are came found from average µ= 800

7 0
4 years ago
Below is a probability distribution for the number of failures in an elementary statistics course. X 0 1 2 3 4 P(X=x) 0.41 0.18
faust18 [17]

Answer:

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Step-by-step explanation:

a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29

b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59

c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88

d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12

e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24

f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59

3 0
4 years ago
Read 2 more answers
Finish the chart for<br> y= 1/2x
11Alexandr11 [23.1K]

Answer:

So we know that y=1/2x. So we apply this equation on each x.

x=-2, so y=1/2*-2 which is -1. so y = -1

x=0, so y = 1/2 * 0 which is 0. so y=0

x=2, so y = 1/2*2 which is 1. so y=1

Hope this helps :))

4 0
3 years ago
Read 2 more answers
The electric potential in a circuit is given by
shtirl [24]
We are given with the equationL
V(t) = 320 e^(-3.1t)
At V = 200
200 = 320 e^(-3.1t)
t = 0.15 s

100 = 320 e^(-3.1t)
t = 0.36 s

It takes 0.36 - 0.15 = 0.21 s for the voltage to drop from 200 to 100 volts.
3 0
3 years ago
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I need help on this pleaseeeeee!!!
GuDViN [60]

Answer:

Answer B.

Step-by-step explanation:

x+2y=10

2x+8y=16

x=-2y+10

2x+8y=16

2(-2y+10)+8y=16

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4y=-4

y=-1

x=-2y+10

x=-2(-1)+10

x=2+10

x=12

one solution

6 0
3 years ago
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