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jeyben [28]
3 years ago
8

Which of these phrases contain a variable? Check all that apply. A three cups of salsa B the weight of the puppy C the cost the

of dinner D twenty feet of wire
Mathematics
1 answer:
Lyrx [107]3 years ago
4 0

Answer:

phrases B and C are variables.

Step-by-step explanation:

Any thing whose value can be changed is called variable.

Whereas, if the value is fixed then it is called a constant.

We can use these definitions to find the phrases that contains a variable.

First option is: A three cups of salsa - It is constant because, the number of cups of salsa is fixed which is 3.

Second option is: the weight of the puppy - It is variable because, the weight of the puppy can be changed and not always fixed.

Third option is : the cost the of dinner- It is variable because, the cost of the dinner can be changed and depends on the quantity of the food ordered.

Last option: twenty feet of wire: It is a constant because the length of the wire is fixed which is 20 feet.

Therefore, the phrases B and C are variables.

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A manufacturer of nails claims that only 4% of its nails are defective. A random sample of 20 nails is selected, and it is found
irakobra [83]

Answer:

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

Step-by-step explanation:

A manufacturer of nails claims that only 4% of its nails are defective.

At the null hypothesis, we test if the proportion is of 4%, that is:

H_0: p = 0.04

At the alternative hypothesis, we test if the proportion is more than 4%, that is:

H_a: p > 0.04

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.

4% is tested at the null hypothesis

This means that \mu = 0.04, \sigma = \sqrt{0.04*0.96}

A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.

This means that n = 20, X = 0.1

Value of the test statistic:

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

z = \frac{0.1 - 0.04}{\frac{\sqrt{0.04*0.96}}{\sqrt{20}}}

z = 1.37

P-value of the test and decision:

Considering an standard significance level of 0.05.

The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.

Looking at the z-table, z = 1.37 has a p-value of 0.9147

1 - 0.9147 = 0.0853

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

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3 years ago
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SLOVE THIS QUESTION
eduard
The answer is 1561 yuans and 27 jiao.
6 0
3 years ago
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