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Juli2301 [7.4K]
3 years ago
9

Which best explains why all equilateral triangles are similar?

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
3 0

Answer:

the correct answer is option 1.

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Use the image to answer the question. Acacia and Eduarte will cut a hole in a package of candy and each pour out one piece to ea
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Less likely and more likely

Step-by-step explanation:

got it right on edge

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Graph the system of equations. y = -x + 3 y = -4x -6<br><br><br>NEED HELL ASAP
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Nine students were asked how many miles they live from school. Their answers are listed below.
Zielflug [23.3K]
Answer:  The mode is: 3 .  The range is:  6 .
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Explanation:
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  It would be best to list this values in the data set given, from least to greatest:
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  { 3, 3, 3, 3, 4, 5, 5, 6, 9 } .
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  The mode is the number that occurs most frequently in the data set, which is: "3".
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{The number, "3", occurs FOUR (4) times.  The number, "4", occurs ONE (1) time.  The number, "5", occurs TWO (2) times.  The number, "6", occurs ONE (1) time.  The number, "9", occurs ONE (1) time.}.
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The range is calculated from subtracting the LOWEST value in the data set FROM the HIGHEST value in the data set.
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 The lowest values in the data set is: "3" .
 
 The highest value in the data set is: "9" .
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 To calculate the range:  9 <span>− 3  = 6 .  The range is:  "6".
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  Answer:  The mode is:  3 .  The range is:  6 .
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8 0
3 years ago
Simplify the following expression by combining liker terms: 4x +9 -6x +6
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Answer:

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which is 4x-6x+6+9 which is

-2x+15.

This is my answer I hope it helps you.

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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.

6 0
3 years ago
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