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2 years ago

A rectangle has a perimeter of 42 m and a width of 9 m. What is the length in meters.

Mathematics

1 answer:

Vlad [161]2 years ago

5 0

Answer:

12 meters

Step-by-step explanation:

2x9=18 (get both widths)

42-18=24 (take away widths of 9m)

24/2=12 (divide by 2 to get one length)

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A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular store that have an a

storchak [24]

Answer: 267.

Step-by-step explanation:

When there is no prior information for the population proportion, then the formula we use to find the sample size to estimate the confidence interval :

$n=0.25(\dfrac{z^*}{E})^2$ , where z* = Critical z-value and E + amrgin of error.

Let p = proportion of packages of ground beef sold at a particular store that have an actual fat content exceeding the fat content stated on the label.

Since , we have no prior information about p. so we use above formula

with E = 0.06 and critical value for 95% confidence =z* =1.96 [By z-table ] , we get

$n=0.25(\dfrac{1.96}{0.06})^2=0.25(32.6666)^2\\\\=(0.25)(1067.111111)\\\\=266.777777778\approx267$

Hence, the required sample size is 267.

7 0

3 years ago

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho

vitfil [10]

Answer:

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

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