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Tcecarenko [31]
3 years ago
9

Which best describes bias, if any, from the sample surveyed? survey: favorite sports sample: randomly selected fans at a basebal

l game A. Since the sample is randomly selected, there is unlikely to be any bias. B. Since baseball fans are also likely to be fans of other sports, there is unlikely to be any bias. C. Since baseball is played in the summer, there is likely to be bias against sports played in the winter. D. Since the sample is taken at a baseball game, it is likely to be biased in favor of baseball.
Mathematics
1 answer:
Ugo [173]3 years ago
7 0

Answer:

c

Step-by-step explanation:

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Find the lowest common denominator for the set of fractions.
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3 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
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