Question

1. based on the 2006 world health statistics, a population model for cambodia was $c(t)=9.038(1.0199)^t$ where $c(t)$ is cambodia’s population in millions, and $t$ is years after 2006. use the given information to complete the statement below. the population of cambodia in 2006 was __________ and had a ________ (growth or decay) factor of ________ such that its population was ________ (increasing or decreasing) at a rate of __________% per year.

Explanation:

Step1: Find initial population

When \(t = 0\) (year 2006), \(C(0)=9.038\times(1.0199)^{0}\). Since any non - zero number to the power of 0 is 1, \(C(0) = 9.038\) million.

Step2: Identify growth/decay factor

The general form of an exponential growth/decay model is \(y = a\times b^{t}\), where \(b\) is the growth/decay factor. Here \(b = 1.0199\). Since \(b>1\), it is a growth factor.

Step3: Calculate growth rate

The growth rate \(r\) can be found from the growth factor \(b\) using the formula \(b=1 + r\). So, \(r=b - 1\). Substituting \(b = 1.0199\), we get \(r=1.0199-1=0.0199\) or \(1.99\%\). And since \(r>0\), the population is increasing.

Answer:

The population of Cambodia in 2006 was 9.038 million and had a growth factor of 1.0199 such that its population was increasing at a rate of 1.99% per year.