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cricket20 [7]
2 years ago
15

Deduct 6.2% of the first 113,000 of an employee's annual earnings for social security taxes and 1.45% of all annual earnings for

medicare taxes.
Randy bumgarner earns a monthly gross salary of 3,600.00. To date, he has earned a total of 14,00.00. what is this month's Social Security tax withholding?________ What is this months Medicare tax with holding?____________
Mathematics
1 answer:
seropon [69]2 years ago
5 0

So there is a deduction of 6.2% of the first $113,000.00 and 1.45% of all annual earnings for medicare taxes.

Randy earns a monthly gross salary of $3,600.00. So Randy makes:

3600 \times 12=43,200 dollars annually.

Now,

Randy's this month's Social Security tax withholding will be given by 6.2% of his gross monthly salary that is $3,200.00.

\frac{6.2}{100} \times 3600=223.20

So, month's Social Security tax withholding is $223.20.

Now this month's Medicare tax withholding is given by 1.45% of his monthly gross salary.

That is:

\frac{1.45}{100} \times 3600=52.20

So, month's Medicare tax withholding is $52.20.

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Allison is 5 ft tall. One afternoon her shadow measured 2.5 ft. At the same time, the shadow of her favorite tree was 14.5 ft. W
luda_lava [24]
At the same time, the sun casts a shadow on the girl and tree creating two similar right triangles. You can use a ratio of girl's height to her shadow will be the same as the tree's height to it's shadow.

5 / 2.5 = T / 14.5
2 = T / 14.5
2 * 14.5 = T
29 ft = Tree


8 0
2 years ago
Mike took 12 shots with a basketball and made 9. This can be expressed as a decimal as 0.75. What other person’s successful shot
dimulka [17.4K]
I baseball player swung at the ball 100 times and only made conract 75 times this could be exspressed as 0.75 
hope this helps
5 0
2 years ago
A psychologist is collecting data on the time it takes to learn a certain task. For 50 randomly selected adult subjects, the sam
tatuchka [14]

Answer: (15.263,\ 17.537)

Step-by-step explanation:

According to the given information, we have

Sample size : n= 50

\overline{x}=16.40

s=4.00

Since population standard deviation is unknown, so we use t-test.

Critical value for  95 percent confidence interval  :

t_{n-1,\alpha/2}=t_{49, 0.025}= 2.009575\approx2.010

Confidence interval : \overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}

16.40\pm (2.010)\dfrac{4}{\sqrt{50}}\\\\=16.40\pm1.13702770415\\\\=16.40\pm1.1370\\\\=(16.40-1.1370,\ 16.40+1.1370)\\\\=(15.263,\ 17.537)

Required 95% confidence interval :  (15.263,\ 17.537)

8 0
2 years ago
A charity organization had to sell 18 tickets to their fundraiser just to cover necessary production costs.
meriva

Answer: slope and x-intercept

4 0
2 years ago
Read 2 more answers
Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma
Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that \mu = 14.4, \sigma = 1.1.

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{1.1}

Z = 1.64

Z = 1.64 has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

s = \frac{\sigma}{\sqrt{110}} = 0.1.

Now

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{0.1}

Z = 18

Z = 18 has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

7 0
2 years ago
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