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3 years ago

Peter attempted to use the divide-center method to find the line of best fit on a scatterplot.

Mathematics

1 answer:

SpyIntel [72]3 years ago

8 0

Answer:

He had a different number of points above the line of best fit than below the line of best fit.

Step-by-step explanation:

This is the best likely answer to the question about Peter's attempt to use divide-center method to find the line of best fit on a scatterplot.

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Can someone help (8th grade math)

Nadusha1986 [10]

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3 years ago

If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

svetoff [14.1K]

$sin2\alpha = \frac{260}{269}$

<u>Step-by-step explanation:</u>

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$ ,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒ $sin2\alpha =$ $\frac{10}{\sqrt{269} }$ × $\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$ ×2

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )$

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

⇒ $sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )$

⇒ $sin2\alpha = \frac{260}{269}$

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Step-by-step explanation:

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