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Murljashka [212]
4 years ago
13

A line passes through the point (4,-6) and has a slope of 5/4. Write an equation in slope intercept form for this line.

Mathematics
1 answer:
SIZIF [17.4K]4 years ago
5 0

\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{-6})~\hspace{10em} slope = m\implies \cfrac{5}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-6)=\cfrac{5}{4}(x-4) \\\\\\ y+6=\cfrac{5}{4}x-5\implies y=\cfrac{5}{4}x-11

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CV=0.2 ---- dataset 1

CV = 7.2 --- dataset 2

Step-by-step explanation:

Given

A: 30500, 27500, 31200, 24000, 27100,28600, 39100, 36900, 35000, 21400, 37900, 27900, 18700,33100

B: 4.29, 4.88, 4.34, 4.17, 4.52, 4.80, 3.28, 3.79, 4.84, 4.77, 3.11

Required

The coefficient of variation of each

<u>Dataset A</u>

Calculate the mean

\mu = \frac{\sum x}{n}

\mu = \frac{30500+ 27500+31200+24000+ 27100+28600+ 39100+ 36900+ 35000+ 21400+ 37900+ 27900+ 18700+33100}{14}\mu = \frac{418900}{14}

\mu = 29921.43

Next, calculate the standard deviation using:

\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}}

So, we have:

\sigma= \sqrt{\frac{(30500 - 29921.43)^2 +.................+ (18700- 29921.43)^2 + (33100- 29921.43)^2}{13}}

\sigma= \sqrt{\frac{487723571.42857}{14}}

\sigma= \sqrt{34837397.959184}

\sigma= 5902.32

So, the coefficient of variation is:

CV=\frac{\sigma}{\mu}

CV=\frac{5902.32}{29921.43}

CV=0.2 --- approximated

<u>Dataset B</u>

Calculate the mean

\mu = \frac{\sum x}{n}

\mu = \frac{4.29+ 4.88+ 4.34+ 4.17+ 4.52+ 4.80+ 3.28+ 3.79+ 4.84+ 4.77+ 3.11}{11}

\mu = \frac{46.79}{11}

\mu = 4.25

Next, calculate the standard deviation using:

\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}}

\sigma = \sqrt{\frac{(4.29 - 4.25)^2 + (4.88- 4.25)^2 +.........+ (3.11- 4.25)^2}{11}}

\sigma = \sqrt{\frac{3.859}{11}}

\sigma = \sqrt{0.35081818181}

\sigma = 0.593

So, the coefficient of variation is:

CV=\frac{\sigma}{\mu}

CV = \frac{4.25}{0.5903}

CV = 7.2 -- approximated

3 0
3 years ago
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