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Annette [7]
2 years ago
6

Can anyone please help me answer this Math Question ? As soon as possible !!

Mathematics
1 answer:
NNADVOKAT [17]2 years ago
7 0

Igor's taxable income is the difference between the amount

he earns annually and the amount he earns as exemptions.

Response:

  • The amount he pays in annual state income tax is;<u> $1,497</u>

<h3>Which methods are used to calculate income tax?</h3>

Given;

Annual earnings = $57,900

State tax rate = 3%

Amount earned in exemption = $8,000

Required:

The amount Igor pays in annual state income tax.

Solution;

Taxable Income = Annual Income - Exemptions

Therefore;

Igor's taxable income = $57,900 - $8,000 = $49,900

Taxable Income × Tax rate = Amount paid as tax

  • The amount he pays is therefore; $49,900 × 3% = <u>$1,497</u>

Learn more about income tax here:

brainly.com/question/25278778

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B

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If 7x+2y =3 and x-3y=30, what is the value of y?<br><br><br>​
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Which set of numbers is the most reasonable to describe a person’s hat size? A: Integers. B: Whole numbers. C: Irrational number
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Jim was thinking of a number. Jim halves the number and gets an answer of 61.2. Form an equation with
ioda

Answer:

X/2 = 61.2

X=122.4

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3 0
2 years ago
A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
pshichka [43]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

Z_{0.95}= 1.648

1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

5 0
3 years ago
Read 2 more answers
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