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sladkih [1.3K]
2 years ago
10

The following formula is used in economics to find a factory's unit labor cost U, where O is the

Mathematics
1 answer:
Sauron [17]2 years ago
4 0

The formula for W (The hourly compensation per worker) is W=O/U

Given that The following formula is used in economics to find a factory's unit labour cost U, where O is the hourly output per worker and W is the hourly compensation per worker

U=\frac{O}{W},

Asked to Rearrange the formula to highlight the hourly compensation per worker

From the given question,

U(factory's unit labour cost)=O( hourly output per worker)/W(hourly compensation per worker)

Rearranging the formula to highlight the hourly compensation per worker,

W(hourly compensation per worker) × U(factory's unit labour cost)=O( hourly output per worker)

W(hourly compensation per worker)=\frac{O}{U}

Therefore W=\frac{O}{U} the formula for W (The hourly compensation per worker)

Hence, The formula for W (The hourly compensation per worker) is W=\frac{O}{U}

Complete question: The following formula is used in economics to find a factory's unit labour cost U, where O is the hourly output per worker and W is the hourly compensation per worker

U=\frac{O}{W},

Asked to Rearrange the formula to highlight the hourly compensation per worker

Learn more about compensation  here:

brainly.com/question/27129728

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A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand
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We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

<em>We have to test the claim that the population mean is 10.5.</em>

Let, NULL HYPOTHESIS, H_0 : \mu = 10.5  {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, H_a : \mu \neq 10.5  {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

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<em>Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.</em>

Therefore, we conclude that the population mean is different from 10.5.

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