Find the value of x. Round your answer to the nearest tenth.<br> Help please!

Write an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the line described by y=1/2

Marat540 [252]

Given:

The point, (4, -3)

The line,

$y=\frac{1}{2}x+5$

To find an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the given line:

The slope of the line is,

$m=\frac{1}{2}$

Since the given line is parallel to the new line, so the slope will be same for the both.

Using the point-slope formula,

$y-y_1=m(x-x_1)$

Substitute the point and slope we get,

$\begin{gathered} y-(-3)=\frac{1}{2}(x-4) \\ y+3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2-3 \\ y=\frac{1}{2}x-5 \end{gathered}$

Hence, the equation in slope-intercept form for the line is,

$y=\frac{1}{2}x-5$

8 0

1 year ago

The school is taking 80 students to the state fair for a field trip. There are students going from three different classes. Mr.

Dmitry_Shevchenko [17]

I very much believe it is C but not quite sure

6 0

2 years ago

Read 2 more answers

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

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