The training rate is basically the slope, we can compare the slopes of Patrick and Jennifer to determine whose training rate is faster.
Patrick's slope:
[tex]y=\frac{1}{9}x[/tex]
From the equation, we see that the slope is 1/9 (the coefficient of the variable x is slope)
Jennifer's slope:
We can find the slope for Jennifer by using the given table.
The slope is given by
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Where y is the distance (km) and x is the time (mins)
You can select any two x, y pairs from the given table.
Let us select the first two x,y pairs so
[tex]\begin{gathered} (x_1,y_1)=(40,5) \\ (x_2,y_2)=(64,8) \end{gathered}[/tex]
Substitute these points into the above slope formula.
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{8-5}{64-40}=\frac{3}{24}=\frac{1}{8}[/tex]
So, the slope for Jennifer's training is 1/8
Comparison:
[tex]\begin{gathered} \frac{1}{8}>\frac{1}{9} \\ 0.125>0.111 \end{gathered}[/tex]
Since the slope of Jennifer is greater than the slope of Patrick
Therefore, Jennifer has a faster training rate.
