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Ksju [112]
3 years ago
5

I need help with this question.?

Mathematics
2 answers:
vladimir2022 [97]3 years ago
4 0

Answer:

A.

Step-by-step explanation:

MissTica3 years ago
4 0

Answer:

D

Step-by-step explanation:

just did it

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. Osteen has 32 marbles in his collection. 18 of his marbles are red, four are blue, and the rest are
nalin [4]

Answer: B

Explanation:  

32 marbles in total.

18 red marbles

4 blue marbles

10 green marbles.

You are asked what part of Osteen's marbles are red?

Probability = number of possible outcomes/total number of out outcomes.

P(Red) = 18/32 = 9/16

Both, 18 and 32 are divisible by 2 which gives 9/16. Since 9/16 cannot be reduced any further, we live it as it is. If you wanted to plug this into a calculator that would give 0.5625 which is 56.25% of Osteen's marbles are red.

P(Blue) = 4/32 = 1/8. 1/8 = 0.125 which 12.5% of his marbles are blue.

P(Green) = 10/32 = 5/16. 5/16= 0.3125 which 31.25 of his marbles are green. When you add 0.5625 + 0.125 + 0.3125 = 1 which is 100%. All probabilities add up to 1.

Hope this elaborate explanation helps.

6 0
2 years ago
-3x - 6+ (-1)<br> help!!!!
guapka [62]

Answer:

-3x = 7 .  

Step-by-step explanation:

-3x - 6 - 1 =

-3x - 7

Hope that helps!

5 0
3 years ago
Read 2 more answers
What is the mode of the following set of data? 34, 23, 12, 11, 11, 25, 25, 26, 25, 27, 30
Olin [163]
1 - B

Mode - The mode the number that appears most often in a data set. In this question, 25 was repeated 3 times and no other ones where repeated 3 times. Thats why 25 is the mode.

2 - D

Median - The middle number of a data set that is arranged from least to greatest.

3 - True

There is no mode because all of the numbers only appear once.

4 - A

5 - C

4 0
3 years ago
WILL GIVE BRAINLIEST!!!
Valentin [98]

Answer:

A

Step-by-step explanation:

LOL

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
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