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The table below gives the fish population of a small lake. Use the information to answer the following questions and round each answer to two decimal places.

  1. Use a graphing calculator to fit a regression line to the data and write the linear linear model here.
  2. Use the linear regression line to predict the value of the function at \(\displaystyle x=10\).
  3. Find the correlation coefficient, \(\displaystyle r\) and use it to determine if the line a good fit and explain the solution to part (b).

  1. \(\displaystyle y=787.2 x + 4024.31\)
  2. Evaluating at \(\displaystyle x=10\) gives \(\displaystyle \approx 11896\)
  3. The \(\displaystyle r\) value is \(\displaystyle 0.91\). The line is a good fit.

Download \(\LaTeX\) 

\begin{question}The table below gives the fish population of a small lake. Use the information to answer the following questions and round each answer to two decimal places.\newline
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\hline
Year, $x$ & $2010,0$ & $2011,1$ & $2012,2$ & $2013,3$ & $2014,4$ & $2015,5$ & $2016,6$ & $2017,7$ & $2018,8$ \\ \hline
Fish Population & 4904 & 5472 & 3817 & 6595 & 7748 & 6944 & 7977 & 9809 & 11292 \\ \hline
\end{tabular}\newline

\begin{enumerate}
    \item (5pts) Use a graphing calculator to fit a regression line to the data and write the linear linear model here.
        \soln{4cm}{
            $y=787.2 x + 4024.31$ 
        }
    \item (5pts) Use the linear regression line to predict the value of the function at $x=10$.
        \soln{4cm}{
            Evaluating at $x=10$ gives $\approx 11896$ 
        }
    \item (5pts) Find the correlation coefficient, $r$ and use it to determine if the line a good fit and explain the solution to part (b).
        \soln{4cm}{
            The $r$ value is $0.91$. The line is a good fit. 
        }
\end{enumerate}
\end{question}

Download Question and Solution Environment\(\LaTeX\)
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}
HTML for Canvas 
<p> <p>The table below gives the fish population of a small lake. Use the information to answer the following questions and round each answer to two decimal places.<br>

<ol type="a">
	<li>Use a graphing calculator to fit a regression line to the data and write the linear linear model here.</li>
	<li>Use the linear regression line to predict the value of the function at  <img class="equation_image" title=" \displaystyle x=10 " src="/equation_images/%20%5Cdisplaystyle%20x%3D10%20" alt="LaTeX:  \displaystyle x=10 " data-equation-content=" \displaystyle x=10 " /> .</li>
	<li>Find the correlation coefficient,  <img class="equation_image" title=" \displaystyle r " src="/equation_images/%20%5Cdisplaystyle%20r%20" alt="LaTeX:  \displaystyle r " data-equation-content=" \displaystyle r " />  and use it to determine if the line a good fit and explain the solution to part (b).</li>

</ol>
</p> </p>
HTML for Canvas
<p> <p>
<ol type="a">
	<li> <img class="equation_image" title=" \displaystyle y=787.2 x + 4024.31 " src="/equation_images/%20%5Cdisplaystyle%20y%3D787.2%20x%20%2B%204024.31%20" alt="LaTeX:  \displaystyle y=787.2 x + 4024.31 " data-equation-content=" \displaystyle y=787.2 x + 4024.31 " /> </li>
	<li>Evaluating at  <img class="equation_image" title=" \displaystyle x=10 " src="/equation_images/%20%5Cdisplaystyle%20x%3D10%20" alt="LaTeX:  \displaystyle x=10 " data-equation-content=" \displaystyle x=10 " />  gives  <img class="equation_image" title=" \displaystyle \approx 11896 " src="/equation_images/%20%5Cdisplaystyle%20%5Capprox%2011896%20" alt="LaTeX:  \displaystyle \approx 11896 " data-equation-content=" \displaystyle \approx 11896 " /> </li>
	<li>The  <img class="equation_image" title=" \displaystyle r " src="/equation_images/%20%5Cdisplaystyle%20r%20" alt="LaTeX:  \displaystyle r " data-equation-content=" \displaystyle r " />  value is  <img class="equation_image" title=" \displaystyle 0.91 " src="/equation_images/%20%5Cdisplaystyle%200.91%20" alt="LaTeX:  \displaystyle 0.91 " data-equation-content=" \displaystyle 0.91 " /> . The line is a good fit.</li>
</ol></p> </p>