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mafiozo [28]
3 years ago
8

David bought three DVDs in for books for $40 at a yard sale and I bought one DVD and six books for $18 how much did each DVD and

book cost
Mathematics
1 answer:
grandymaker [24]3 years ago
4 0

Answer: One DVD is $12 and one book is $1

Step-by-step explanation:

Let the DVD be d and the book be b.

Case 1:

3d + 4b = $40.......(1)

Case 2:

1d + 6b = $18........... (2)

In this equations above, I intend to eliminate variable d, so I'm going to multiply the coefficient of d in (1) to multiply (2)

3 × (1d + 6b = 18)

3d + 18b = 54........ .(3)

Then, subtract (1) from (3)

3d + 18b = 54

- ( 3d + 4b = 40)

3d - 3d + 18b - 4b = 54 - 40

0 + 14b = 14

b = 14/14

b = 1.

Then substitute 1 for b in (2)

1d + 6(1) = 18

d + 6 = 18

d = 18 - 6

d = 12.

(d,b) = (12,1)

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The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t
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Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that \mu = 273, \sigma = 100

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 30, s = \frac{100}{\sqrt{30}}

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}

Z = 0.88

Z = 0.88 has a p-value of 0.8106

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}

Z = -0.88

Z = -0.88 has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 50, s = \frac{100}{\sqrt{50}}

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}

Z = 1.13

Z = 1.13 has a p-value of 0.8708

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}

Z = -1.13

Z = -1.13 has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that n = 100, s = \frac{100}{\sqrt{100}}

X = 289

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

By the Central Limit Theorem

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

Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}

Z = 1.6

Z = 1.6 has a p-value of 0.9452

X = 257

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

Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}

Z = -1.6

Z = -1.6 has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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