Average Rate of Change: CCSS.Math.Content.HSF-IF.B.6 - Common Core: High School - Functions

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Question

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.

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

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

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

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

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

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

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

Since,

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

therefore Diane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

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

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

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

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

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

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

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

Since,

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

therefore Hector had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 34 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 34 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{34\textup{ min.}}{5.2\textup{ miles}}=\frac{6.5\textup{min}}{1\textup{mile}}$

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

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

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

Since,

$\6.5<7.9<10 \P_D<P_J<P_H$

therefore Diane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 57 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 57 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{57\textup{ min.}}{5.2\textup{ miles}}=\frac{10.96\textup{ min}}{1\textup{mile}}$

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

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

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

Since,

$\7.9<10<10.96 \P_J<P_H<P_D$

therefore Diane had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\10.8<12<13 \P_J<P_H<P_D$

therefore Jane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\10.8<12<13 \P_J<P_H<P_D$

therefore Diane had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&45 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&45 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{45\textup{ min.}}{5\textup{ miles}}=\frac{9\textup{min}}{1\textup{mile}}$

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

Since,

$\9<10.8<12 \P_H<P_J<P_D$

therefore Hector had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&30 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&30 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{30\textup{ min.}}{2.5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\12=12<13 \P_J=P_H<P_D$

therefore Hector and Jane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{52\textup{ min.}}{4\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\12<13=13 \P_D=P_J>P_H$

therefore Diane and Jane had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{52\textup{ min.}}{4\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\12<13=13 \P_D=P_J>P_H$

therefore Hector had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&65 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 4&52 min.\ Hector &5&65 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{52\textup{ min.}}{4\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{65\textup{ min.}}{5\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

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

Since,

$\13=13=13 \P_D=P_J=P_H$

therefore everyone's pace is the same.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 38 min. \ Jane & 4&35 min.\ Hector &5&72 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 38 min. \ Jane & 4&35 min.\ Hector &5&72 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{38\textup{ min.}}{3\textup{ miles}}=\frac{12.6\overline{6}\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{35\textup{ min.}}{4\textup{ miles}}=\frac{8.75\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{72\textup{ min.}}{5\textup{ miles}}=\frac{14.4\textup{min}}{1\textup{mile}}$

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

Since,

$\8.75<12.6\overline{6}<14.4 \P_J<P_D<P_H$

therefore Jane is the fastest on the relay.

Compare your answer with the correct one above

Question

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.

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

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

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

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

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

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

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

Since,

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

therefore Diane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

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

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

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

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

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

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

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

Since,

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

therefore Hector had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 34 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 34 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{34\textup{ min.}}{5.2\textup{ miles}}=\frac{6.5\textup{min}}{1\textup{mile}}$

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

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

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

Since,

$\6.5<7.9<10 \P_D<P_J<P_H$

therefore Diane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 57 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 5.2& 57 min. \ Jane & 3.4&27 min.\ Hector &7&70 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{57\textup{ min.}}{5.2\textup{ miles}}=\frac{10.96\textup{ min}}{1\textup{mile}}$

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

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

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

Since,

$\7.9<10<10.96 \P_J<P_H<P_D$

therefore Diane had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\10.8<12<13 \P_J<P_H<P_D$

therefore Jane had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the slowest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\10.8<12<13 \P_J<P_H<P_D$

therefore Diane had the slowest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&45 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&27 min.\ Hector &5&45 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{27\textup{ min.}}{2.5\textup{ miles}}=\frac{10.8\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{45\textup{ min.}}{5\textup{ miles}}=\frac{9\textup{min}}{1\textup{mile}}$

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

Since,

$\9<10.8<12 \P_H<P_J<P_D$

therefore Hector had the fastest pace for the relay.

Compare your answer with the correct one above

Question

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.

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&30 min.\ Hector &5&60 min.\ \hline \end{tabular}$

Who had the fastest pace?

Answer

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.

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

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.

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

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

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

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

$\begin{tabular}{ |c|c|c|c| } \hline & Miles& Time \ \hline Diane & 3& 39 min. \ Jane & 2.5&30 min.\ Hector &5&60 min.\ \hline \end{tabular}$

$\textup{Diane's Pace: }P_D=\frac{39\textup{ min.}}{3\textup{ miles}}=\frac{13\textup{min}}{1\textup{mile}}$

$\textup{Jane's Pace: }P_J=\frac{30\textup{ min.}}{2.5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

$\textup{Hector's Pace: }P_H=\frac{60\textup{ min.}}{5\textup{ miles}}=\frac{12\textup{min}}{1\textup{mile}}$

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

Since,

$\12=12<13 \P_J=P_H<P_D$

therefore Hector and Jane had the fastest pace for the relay.

Compare your answer with the correct one above

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