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Anit [1.1K]
4 years ago
11

The following scatter plot represents the relationship between a person's weight and the number of calories the person burns in

one minute of jump roping. Using the line of best fit, about how many calories per minute could you expect a 75-pound person to burn while jump roping?
5
5.5
6
7

Mathematics
2 answers:
Zielflug [23.3K]4 years ago
8 0

Answer:

About 6

Step-by-step explanation:

Look at the 75 pound point, right between 100 and 50. Then go up until that point reaches the red line, and there's your answer!

steposvetlana [31]4 years ago
7 0

Answer:

answer

C.6

Step-by-step explanation:

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Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))
miv72 [106K]
\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{  that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\


\bf \textit{again, using the pythagorean theorem to get the opposite side}
\\\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
-----------------------------\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
\\\\\\

\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
\implies 
tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}


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