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rusak2 [61]
3 years ago
15

How are the rates 6 pounds per day and 6 days per pound difference? When would each be useful

Mathematics
1 answer:
loris [4]3 years ago
4 0
6 lbs per day means that every day 6 pounds are gained. 6 days per lb means that every 6 days 1 pound is gained (or lost)
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Which calcuation and answer show how to convert 5/16 to a decimal?
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25.69 + .75= $26.44

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6 1/2 - 7/8 : 5 11/16
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Step-by-step explanation:

brainly.com/question/6752935

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A researcher conducted an experiment to investigate the effectiveness of a medicated lotion in treating a skin irritation. A gro
Helga [31]

Answer:

It can be concluded that the medicated lotion has an effect on treating skin irritations.

Step-by-step explanation:

In this case we need to determine whether the medicated lotion was effective in treating the skin irritation or not.

The hypothesis can be defined as follows:  

<em>H₀</em>: There is no difference between the two proportions, i.e. <em>p</em>₁ - <em>p</em>₂ = 0.  

<em>Hₐ</em>: There is a significant difference between the two proportions, i.e. <em>p</em>₁ - <em>p</em>₂ ≠ 0.  

The information provided is:

n₁ = 40

n₂ = 40

X₁ = 36

X₂ = 16

Compute the sample proportions and total proportions as follows:

 \hat p_{1}=\frac{36}{40}=0.90\\\\\hat p_{2}=\frac{16}{40}=0.40\\\\\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{36+16}{40+40}=0.65

Compute the test statistic value as follows:

 z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}=\frac{0.90-0.40}{\sqrt{0.65(1-0.65)[\frac{1}{40}+\frac{1}{40}]}}=4.69

The test statistic value is 4.69.

The decision rule is:

The null hypothesis will be rejected if the <em>p</em>-value of the test is less than the significance level.

Compute the <em>p</em>-value as follows:

 p-value=2\times P(Z

The <em>p</em>-value of the test is very small.

The null hypothesis will be rejected at any significance level.

Thus, there is a significant difference between the two proportions.

So, it can be concluded that the medicated lotion has an effect on treating skin irritations.

6 0
3 years ago
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