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rusak2 [61]
3 years ago
10

You have the following expenses: clothes—$98.95, utilities—$234.98, rent—$875.34, and groceries—$103.98. Which expenses are disc

retionary?
Mathematics
2 answers:
Furkat [3]3 years ago
7 0
Utilities and groceries
belka [17]3 years ago
5 0

Answer:

B

B

A

D

Extra:

1. To calculate discretionary money, you

B) <em>Calculate the difference between your net income and fixed expenses</em>

2. Deductions from your pay such as FICA, federal withholding, and state withholding are _____

B)<em> Involuntary and fixed</em>

3. You have the following expenses: clothes—$98.95, utilities—$234.98, rent—$875.34, and groceries—$103.98. Which expenses are discretionary?

A) <em>Clothes</em>

4. Suppose that you work at the Peacock Blue store for 40 hours over five days at a rate of $9.25/hour. You then quit your job. Deductions are FICA (7.65%), federal withholding (12%), and state withholding (8%). Your expenses are transportation at $5.25/day, lunch at $3.85/day, and black slacks required for work at $19.95. How much is your discretionary income for the week?

D) <em>$202.24</em>

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A basketball team wants to paint half of a free-throw circle grey. If the circumference of the free-throw circle is 30.77 feet,
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Answer:

37.7 ft²

Step-by-step explanation:

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<em>Divide both sides by 2</em>

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5 0
3 years ago
A random sample of 500 of these people is drawn, of whom 194 turn out to be currently enrolled in college. Estimate the percenta
Rzqust [24]

Answer:

0.388 - 2.00\sqrt{\frac{0.388(1-0.388)}{500}}=0.344

0.388 + 2.00\sqrt{\frac{0.388(1-0.388)}{500}}=0.432

The 95.5% confidence interval would be given by (0.344;0.432)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

The estimated proportion on this case is given by:

p = \frac{X}{n}= \frac{194}{500}=0.388

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95.5% of confidence, our significance level would be given by \alpha=1-0.955=0.045 and \alpha/2 =0.0225. And the critical value would be given by:

z_{\alpha/2}=-2.00, z_{1-\alpha/2}=2.00

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.388 - 2.00\sqrt{\frac{0.388(1-0.388)}{500}}=0.344

0.388 + 2.00\sqrt{\frac{0.388(1-0.388)}{500}}=0.432

The 95.5% confidence interval would be given by (0.344;0.432)

4 0
3 years ago
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