Question

spent walking. they kept track of how many dogs they walked and how long they were out each day.

question 3

does this data suggest a linear or nonlinear relationship? why?

Explanation:

In a linear relationship, the rate of change (slope) is constant. Let's check the differences in the number of dogs (which we can consider as the \(x\) - values, \(x_1 = 1,x_2 = 2,\cdots,x_{10}=10\)) and the differences in the time spent walking (which we can consider as the \(y\) - values, \(y_1 = 25,y_2 = 36,\cdots,y_{10}=92\)).

The difference in \(x\) - values: \(\Delta x=x_{i + 1}-x_i=1\) (constant) for \(i=1,\cdots,9\)

The differences in \(y\) - values:
\(\Delta y_1=y_2 - y_1=36 - 25 = 11\)
\(\Delta y_2=y_3 - y_2=40 - 36=4\)
\(\Delta y_3=y_4 - y_3=52 - 40 = 12\)
\(\Delta y_4=y_5 - y_4=60 - 52 = 8\)
\(\Delta y_5=y_6 - y_5=68 - 60 = 8\)
\(\Delta y_6=y_7 - y_6=74 - 68 = 6\)
\(\Delta y_7=y_8 - y_7=110 - 74 = 36\)
\(\Delta y_8=y_9 - y_8=89 - 110=-21\)
\(\Delta y_9=y_{10}-y_9=92 - 89 = 3\)

Since the differences in the \(y\) - values (\(\Delta y\)) are not constant, the rate of change is not constant.

Answer:

The data suggests a nonlinear relationship. Because the differences in the time - spent walking (the \(y\) - values) for a constant difference in the number of dogs (the \(x\) - values, \(\Delta x = 1\)) are not constant.