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Novosadov [1.4K]
3 years ago
9

Can I get some help on these please

Mathematics
2 answers:
Gekata [30.6K]3 years ago
8 0

Answer:

1.2/5

2. 5/8

3. Proper Fraction

4. Mixed Fraction

6. 23/5

7. 19/5

8. 11 2/3

9. 50 4/5

Gelneren [198K]3 years ago
4 0

Answer:

1. \frac{2}{5}

2. \frac{5}{8}

3. Proper Fraction

4. Mixed Fraction

6. \frac{23}{5}

7. \frac{19}{5}

8. 11 \frac{2}{3}

9. 50 \frac{4}{5}

10. \frac{1}{8}

Sorry, didn't see the last question

Hope this helped!

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The average size of a new house built in a Hays county in 2010 was 1,872 square feet. A random sample of 40 new homes built in H
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Answer:

The test statistic is z = 2.7.

The pvalue of the test is 0.0035 < 0.05, which means that this sample provides enough evidence to conclude that the average house size has increased in the Hays County since 2010.

Step-by-step explanation:

The average size of a new house built in a Hays county in 2010 was 1,872 square feet. Evidence that it has increased?

This means that the null hypothesis is:

H_0: \mu = 1872

And the alternate hypothesis is:

H_a: \mu > 1872

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

1872 is tested at the null hypothesis:

This means that \mu = 1872

Sample of 40. The average square footage was 2,031 with a standard deviation of 373 square feet.

This means that n = 40, X = 2031, \sigma = 373

Value of the test-statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{2031 - 1872}{\frac{373}{\sqrt{40}}}

z = 2.7

Pvalue and decision:

The pvalue of the test is the probability of finding a sample mean above 2031. This is 1 subtracted by the pvalue of z = 2.7.

Looking at the z-table, z = 2.7 has a pvalue of 0.9965

1 - 0.9965 = 0.0035

The pvalue of the test is 0.0035 < 0.05, which means that this sample provides enough evidence to conclude that the average house size has increased in the Hays County since 2010.

7 0
3 years ago
Find the next two terms 1/8, 2/7,1/2,4/5
Verizon [17]

Answer:  5/4 and 6/3

Step-by-step explanation:

1/2 can be written as 3/6 so the series now becomes,

1/8 , 2/7 , 3/6 , 4/5 ,…….

i think you can now guess the answer, look at the<u> numerators which are increasing by ‘1’ </u>and then look at the <u>denominators which are decreasing by ‘1’.</u>

so the next number in the series is, 1/8 , 2/7 , 3/6 , 4/5 , 5/4 , 6/3

hope this helped

5 0
3 years ago
13=2f+5<br> What does f =
Stells [14]

Answer:

4

Step-by-step explanation:

Step 1:

13 = 2f + 5         Equation

Step 2:

8 = 2f      Subtract 5 on both sides

Step 3:

f = 8 ÷ 2     Divide

Answer:

f = 4

Hope This Helps :)

7 0
4 years ago
Read 2 more answers
A pretzel company calculated that there is a mean of 73.5 broken pretzels in each production run with a standard deviation of 5.
SVEN [57.7K]

Answer:

0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.

Step-by-step explanation:

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.

A pretzel company calculated that there is a mean of 73.5 broken pretzels in each production run with a standard deviation of 5.1.

This means that \mu = 73.5, \sigma = 5.1

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This is the p-value of Z when X = 69.

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

Z = \frac{69 - 73.5}{5.1}

Z = -0.88

Z = -0.88 has a p-value of 0.1894

0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.

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