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3 years ago

Simplify: (8x4)(4x3)/7x2 A. 32x5/7 B. 1/28x C. 1/3x D. 7/9x4

1 answer:

8 0

The Answer is A:

Simplify the following:
(8×4 x^4 x^3)/(7 x^2)

Combine powers. (4 x^3×8 x^4)/(7 x^2) = (4 x^(4 + 3 - 2) 8)/7:
(4×8 x^(4 + 3 - 2))/7

4 + 3 - 2 = 5:
(4×8 x^5)/7

4×8 = 32:
Answer: (32 x^5)/7

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In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time coll

ICE Princess25 [194]

Answer:

a) n = 1037.

b) n = 1026.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use $\pi = 0.5$ , which is when we are going to need the largest sample size.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 2.575*0.5$

$(\sqrt{n}) = \frac{2.575*0.5}{0.04}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}$

$n = 1036.03$

Rounding up

n = 1037.

(b) Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.

$\pi = 0.55$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}$

$0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}$

$(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}$

$(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}$

$n = 1025.7$

Rounding up

n = 1026.

6 0

3 years ago

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

skad [1K]

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.

6 0

2 years ago

When using fraction strips ,how do you know that 2 fractions are equivalent

Zigmanuir [339]

When the longer strip is on top and the litter strips fit perfectly under the longer one. For example, the fraction 1/3 and 3/6. One longer strip will be on the top representing 6. Then 3 litter 3s will be under that perfectly

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3 years ago

Factor with steps please!!!

myrzilka [38]

So you want to find two numbers that add up to -13 and multiply to 40. You can find the factors of 40 and find the pair that adds up to -13. -8 and -5 multiply to 40 and add up to -13.

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You can use foil to test it

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Step-by-step explanation:

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