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Elza [17]
3 years ago
15

Solve this question—————-

Mathematics
1 answer:
Ber [7]3 years ago
5 0

Answer: (2, -5) (light blue answer choice)

Solving by substitution

−7x+y=−19;−2x+3y=−19

−7x+y+7x=−19+7x (Add 7x to both sides)

y=7x−19

Step: Substitute7x−19 for y in −2x+3y=−19:

−2x+3y=−19

−2x+3(7x−19)=−19

19x−57=−19 (Simplify both sides of the equation)

19x−57+57=−19+57 (Add 57 to both sides)

19x=38

19x/19 = 38/19 (Divide both sides by 19)

x=2

Step: Substitute 2 for x in y=7x−19:

y=7x−19

y=(7)(2)−19

y= −5 (Simplify both sides of the equation)

Therefore: x = 2 and y = −5

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Answer:

1/24

Step-by-step explanation:

5/12 - 3/8=?

The first step is to find a denominator (bottom number) that is the same for both fractions. In this case, 24 works.

10/24 - 9/24 = ?

1/24

8 0
2 years ago
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solniwko [45]

Answer:

-1

Step-by-step explanation:

-2(x-7)=16

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and get -1 as your answer.

Hope my answer has helped you!

5 0
2 years ago
There are twice as many gray kittens than white
Marianna [84]

Answer:

8 gray kittens

Step-by-step explanation:

There are double the amount of grey kittens compared to white kittens so, you would do 4 × 2 which is 8. There are 8 grey kittens.

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7 0
2 years ago
228 + 6 =<br> R<br> What is the quotient
cricket20 [7]

228 + 6 = 224 + 6 + 4 = 230 + 4

= 234

Answer: 234

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4 0
2 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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