Question

the table shows the relationship between time spent running and distance traveled.
which type of model best describes the relationship?
linear, because the r value for the linear model is closest to 1
exponential, because the r value for the exponential model is closest to 0
linear, because the rate of change between each pair of points is exactly 520
exponential, because the rate of change between each pair of points is 1.98

Explanation:

Step1: Calculate the rate of change (slope)

The formula for the rate of change (slope) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \((x_1 = 1,y_1=530)\) and \((x_2 = 2,y_2 = 1050)\), \(m_1=\frac{1050 - 530}{2 - 1}=\frac{520}{1}=520\).
For \((x_2 = 2,y_2 = 1050)\) and \((x_3 = 3,y_3 = 1600)\), \(m_2=\frac{1600 - 1050}{3 - 2}=\frac{550}{1}=550
eq520\). So the "linear, because the rate of change between each pair of points is exactly 520" is wrong.
For an exponential model, we check the ratio of \(y\) - values. \(\frac{y_2}{y_1}=\frac{1050}{530}\approx1.98\), \(\frac{y_3}{y_2}=\frac{1600}{1050}\approx1.52
eq1.98\). So the "exponential, because the rate of change between each pair of points is 1.98" is wrong.
In general, when choosing between linear and exponential models, the \(r\) - value (correlation coefficient) is used. The closer the \(r\) - value is to \(1\) (for a linear model) or \(0\) (for an exponential model in the context of comparing with a linear model), the better the fit. A linear model is a better fit when \(|r|\) is closer to \(1\) compared to the \(r\) - value of an exponential model.

Answer:

linear, because the \(r\) value for the linear model is closest to \(1\)