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attashe74 [19]
3 years ago
6

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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Step-by-step explanation:

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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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The simplified form of that decimal in fraction form is 3/20. Hope this helps.
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What expression can be used to 80% of 120
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59%

Step-by-step explanation:

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