Trapezoid RSTU with vertices R(-5, 15), S(0, 10), T(-5, 5) and U(-15, 5): scale factor = 1/5, centered at the origin

A welder works 8 hours a day at a rate of $25.00 per hour. The overtime rate is doubled the normal rate. On a particularday, the

Alborosie

Answer:

C.

Step-by-step explanation:

Reeeeeeeeeeeeeeeeeeee

5 0

3 years ago

Josh can paint a whole room in m hours while Kevin can paint a whole room in n hours. How many parts of the room can they paint

Leona [35]

<h3>Answer: (m+n)/(mn)</h3>

Work Shown:

Josh:

1 room = m hours

1/m room = 1 hour

Kevin:

1 room = n hours

1/n room = 1 hour

We see that the rates for Josh and Kevin are 1/m and 1/n respectively. This is the amount of a room they paint in one hour. Combine the fractions

1/m + 1/n = n/(mn) + m/(mn) = (n+m)/(mn) = (m+n)/(mn)

The expression (m+n)/(mn) represents how much of the room they get painted in 1 hour. This is if they work together and it assumes neither worker slows the other one down.

7 0

3 years ago

the figure below is made of rectangles what is the perimeter of the figure below? note that not all the side lengths are labeled

Anit [1.1K]

The perimeter is 30cm

5 0

3 years ago

Select all that apply

katovenus [111]

The range is Y>2. :)

5 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

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

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

LCL = 7.431 hours to UCL = 10.569 hours

8 0

3 years ago