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Mashcka [7]
3 years ago
12

Ares is making 14 jars of honey peanut butter. He wants to use 45 milliliter (ML) of honey in each jar. How much honey (in ML) w

ill ares use in all?
Mathematics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer:

630 milliliters.

Step-by-step explanation:

The statement tells us that the final product is 14 jars of honey peanut butter and that in each jar use 45 mliliters of honey. This means that to know the total honey to be used, the required quantity for each jar must be multiplied by the total number of jars, that is:

14 * 45 = 630

Which means that he would spend a total of 630 milliliters.

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The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho
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Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is  

\sigma = 12.6 \: hours

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

n = 250

The sample mean is  

\bar{x} = 9 \: hours  

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

 \mu = \bar{x} = 9 \: hours

The sample standard deviation would be  

$ s = {\frac{\sigma}{\sqrt{n} }  $

Where   is the population standard deviation and n is the sample size.

$ s = {\frac{12.6}{\sqrt{250} }  $

s = 0.7969 \: hours

Therefore, the sampling distribution of \bar{x} is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

\text {confidence interval} = \mu \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sampling size, s is the sample standard deviation and  is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\

So the required 95% confidence interval is

\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569

The 95% interval estimate of the population mean \mu is

LCL = 7.431 hours to UCL = 10.569 hours

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