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Andru [333]
3 years ago
6

Which group represents equivalent ratios?

Mathematics
1 answer:
charle [14.2K]3 years ago
8 0

The group 70 \%, \frac{7}{10}, 0.7 represents equivalent ratios.

Answer: Option B

<u>Step-by-step explanation:</u>

As we see carefully, the equation B represents the equivalent ratio. Because if you divide the 7 by 10 then you will get 0.7 and this 0.7 can also be written as 70% in percentage form. So, the correct answer is Option B.

Case A: 75 \%, \frac{2}{4}, 0.8

It can be written as that 0.75, 0.5, 0.8.this group does not represent equivalent ratios

Case B: 70 \%, \frac{7}{10}, 0.7

If you divide the 7 by 10 then you will get 0.7 and this 0.7 can also be written as 70% in percentage form. This group represents equivalent ratios

Case C: 70 \%, \frac{7}{10}, 0.25

It can writes as that 0.7,0.7.0.25 so this group does not represent equivalent ratios

Case D: 70 \%, \frac{1}{7}, 0.7

This gives 0.7, 0.14, 0.7 so this group does not represent equivalent ratios.

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Answer:

Pvalue of 0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

Step-by-step explanation:

The average weight of a package of rolled oats is supposed to be at most 18 ounces.

This means that the null hypothesis is:

H_{0}: \mu \leq 18

Is the true mean larger than the specification?

Due to the question asked, the alternate hypothesis is:

H_{a}: \mu > 18

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.

18 is tested at the null hypothesis:

This means that \mu = 18

A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces.

This means that n = 18, X = 18.2, \sigma = 0.5

Value of the test statistic:

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

z = \frac{18.2 - 18}{\frac{0.5}{\sqrt{18}}}

z = 1.7

Pvalue of the test:

Probability of finding a sample mean above 18.2, which is 1 subtracted by the pvalue of z = 1.7.

Looking at the z-table, z = 1.7 has a pvalue of 0.9554.

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Hope this helps :)
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