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3 years ago

The slope of the line containing the points (6, 4) and (-5, 3) is: A. 1/11 B. 1 C. -1

Mathematics

2 answers:

Triss [41]3 years ago

6 0

Answer:

1/11

Step-by-step explanation:

svetoff [14.1K]3 years ago

5 0

$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-4}{-5-6}\implies \cfrac{-1}{-11}\implies \cfrac{1}{11}$

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vladimir1956 [14]

Answer:

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oksano4ka [1.4K]

Answer:

$Lower = 231.134- 3.098=228.036$

$Upper = 231.134+ 3.098=234.232$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case we know that the 95% confidence interval is given by:

$\bar X=\frac{233.002 +229.266}{2}= 231.134$

And the margin of error is given by:

$ME = \frac{233.002 -229.266}{2}= 1.868$

And the margin of error is given by:

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And the critical value for 95% of confidence is $t_{\alpha/2}= 2.776$

So then we can find the deviation like this:

$s = \frac{ME \sqrt{n}}{t_{\alpha/2}}$

$s = \frac{1.868* \sqrt{5}}{2.776}= 1.506$

And for the 99% confidence the critical value is: $t_{\alpha/2}= 4.604$

And the margin of error would be:

$ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098$

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$Upper = 231.134+ 3.098=234.232$

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Answer:

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