Chapter 3, Problem 24MCQ

(a)

To determine

To calculate: For which time period is the rate of change is the greatest.

Answer to Problem 24MCQ

Therefore, $rateofchange$ is greatest for time period between 2011-12.

Explanation of Solution

Given information: The population growth is shown for Heckertville since 2008,

  $rateofchange=\frac{\Delta y}{\Delta x}$

The rate of change for each time period is given by,

  $\begin{array}{l}rateofchange=\frac{\Delta y}{\Delta x}\\ \text{For 2008-09, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{0.5}{1}=0.5\\ \text{For 2009-10, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{0.5}{1}=0.5\\ \text{For 2010-11, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{0.25}{1}=0.25\\ \text{For 2011-12, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{1}{1}=1\\ \text{For 2012-13, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{0.75}{1}=0.75\\ \text{For 2013-14, }rateofchange=\frac{\Delta y}{\Delta x}=\frac{0.75}{1}=0.75\end{array}$

Therefore, $rateofchange$ is greatest for time period between 2011-12.

(b)

To determine

The meaning of the slope from 2008 to 2014.

Answer to Problem 24MCQ

Therefore, the slope means Annual Growth in population.

Explanation of Solution

Given information: The population growth is shown for Heckertville since 2008,

The slope for the graph shown for time period between 2008 to 2014 represents the increase in population per year.

Therefore, the slope means Annual Growth in population.

(c)

To determine

The mathematical practice used for solving the problem.

Answer to Problem 24MCQ

Therefore $rateofchange=\frac{\Delta y}{\Delta x}=\frac{(y_2-y_1)}{(x_2-x_1)}$ is used for solving this problem.

Explanation of Solution

The mathematical equation used for calculation of growth in population per year is represented as,

  $rateofchange=\frac{\Delta y}{\Delta x}=\frac{(y_2-y_1)}{(x_2-x_1)}$

So, therefore $rateofchange=\frac{\Delta y}{\Delta x}=\frac{(y_2-y_1)}{(x_2-x_1)}$ is used for solving this problem.