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Flashcards: Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6

Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \\ \hline Diane & 5.2& 39 min. \\ Jane & 3.4&27 min.\\ Hector &7&70 min.\\ \hline \end{tabular}$ $\displaystyle \begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \\ \hline Diane & 5.2& 39 min. \\ Jane & 3.4&27 min.\\ Hector &7&70 min.\\ \hline \end{tabular}$

Who had the fastest pace?

$\textup{Not Enough Information}$ $\displaystyle \textup{Not Enough Information}$

$\textup{Diane}$ $\displaystyle \textup{Diane}$

$\textup{Jane}$ $\displaystyle \textup{Jane}$

$\textup{Diane and Hector have the same pace.}$ $\displaystyle \textup{Diane and Hector have the same pace.}$

$\textup{Hector}$ $\displaystyle \textup{Hector}$

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Flashcards: Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6

Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.

Who had the fastest pace?

$\textup{Not Enough Information}$ $\displaystyle \textup{Not Enough Information}$

$\textup{Diane}$ $\displaystyle \textup{Diane}$

$\textup{Jane}$ $\displaystyle \textup{Jane}$

$\textup{Diane and Hector have the same pace.}$ $\displaystyle \textup{Diane and Hector have the same pace.}$

$\textup{Hector}$ $\displaystyle \textup{Hector}$

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Flashcards: Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6

Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.

Who had the fastest pace?

Your answer:

$\textup{Not Enough Information}$ $\displaystyle \textup{Not Enough Information}$

Your answer:

$\textup{Jane}$ $\displaystyle \textup{Jane}$

Your answer:

$\textup{Diane and Hector have the same pace.}$ $\displaystyle \textup{Diane and Hector have the same pace.}$

Your answer:

$\textup{Hector}$ $\displaystyle \textup{Hector}$

Correct answer:

$\textup{Diane}$ $\displaystyle \textup{Diane}$

Explanation:

This question is testing one's ability interpret a table and identify the rate of change.

For the purpose of Common Core Standards, "calculate and interpret the average rate of change of a function over a specific interval" falls within the Cluster B of "interpret functions that arise in applications in terms of the context" concept (CCSS.MATH.CONTENT.HSF-IF.B.5).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the average rate of change formula for calculations.

$\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$ $\displaystyle \frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$

In the particular case, to find the fastest pace person, there will need to an average rate of change calculation done for each person to find their mile pace.

Step 2: Calculate the mile pace for Diane, Jane, and Hector.

$\textup{Diane's Pace: }P_D=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$ $\displaystyle \textup{Diane's Pace: }P_D=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$

$\textup{Jane's Pace: }P_J=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$ $\displaystyle \textup{Jane's Pace: }P_J=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$

$\textup{Hector's Pace: }P_H=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$ $\displaystyle \textup{Hector's Pace: }P_H=\frac{\Delta \textup{Time}}{\Delta \textup{Miles}}$

Using the information provided by the table, calculate the pace of each individual.

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{5.2\textup{ miles}}=\frac{7.5\textup{min}}{1\textup{mile}}$ $\displaystyle \textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{5.2\textup{ miles}}=\frac{7.5\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{3.4\textup{ miles}}=\frac{7.9\textup{min}}{1\textup{mile}}$ $\displaystyle \textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{3.4\textup{ miles}}=\frac{7.9\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{70\textup{ min.}}{7\textup{ miles}}=\frac{10\textup{min}}{1\textup{mile}}$ $\displaystyle \textup{Hector's Pace: }P_H=\frac{70\textup{ min.}}{7\textup{ miles}}=\frac{10\textup{min}}{1\textup{mile}}$

Step 3: Compare the paces to arrive at a solution.

Since,

$\\7.5<7.9<10 \\P_D<P_J<P_H$ $\displaystyle \\7.5< 7.9< 10 \\P_D< P_J< P_H$

therefore Diane had the fastest pace for the relay.

Flashcards: Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6

Diane, Jane, and Hector participated in a relay race. The following table shows the distance they each ran and the amount of time it took them.

Who had the slowest pace?

$\text{Diane}$ $\displaystyle \text{Diane}$

$\text{Hector}$ $\displaystyle \text{Hector}$

$\text{Jane and Diane have the same pace.}$ $\displaystyle \text{Jane and Diane have the same pace.}$

$\text{Jane}$ $\displaystyle \text{Jane}$

$\text{More information is needed}$ $\displaystyle \text{More information is needed}$