Chapter 3.3, Problem 2CYP

To determine

The 2-year period that has the least rate of change.

Answer to Problem 2CYP

  $2002-2004$

Explanation of Solution

Given:

The graph:

  

Concept Used:

Equation of the line passing through the points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given by $y-y_1=m\left(x-x_1\right)$ , where $\text{slope, }m=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

The rate of change of a line between two points is given by the slope of the line.

From the given graph it is clear that the slope of the line segment between 2002 to 2004 is smallest. So, during this period of 2 years, the rate of change is least.

From the graph it is clear that for first two year period, the attendance is 317 and 324 respectively. So, the rate of change during 2000 to 2002 is

  $\begin{array}{c}\text{Rate of change}=\frac{324-317}{2002-2000}\\ =\frac{7}{2}\\ =3.5\end{array}$

Similarly, the rate of change during year 2002 to 2004 is

  $\begin{array}{c}\text{Rate of change}=\frac{325-324}{2004-2002}\\ =\frac{1}{2}\\ =0.5\end{array}$

And, the rate of change during year 2004 to 2006 is

  $\begin{array}{c}\text{Rate of change}=\frac{335-325}{2006-2004}\\ =\frac{10}{2}\\ =5\end{array}$

Thus, the least rate of change occur during the time period 2002 to 2004.