Respuesta :
SOLUTION
A.
To solve this question, we will use the compound interest formula.
Which is:
[tex]\begin{gathered} A=P(1-\frac{r}{100})^{nt} \\ Since\text{ we are dealing with a yearly statistics, n = 1} \end{gathered}[/tex][tex]\begin{gathered} \text{From 1989 to 2007, there is a year difference of 18 years} \\ t=18 \\ A=109,185 \\ P=125,500 \\ We\text{ are looking for the continuous growth rate (r)} \\ \text{Now, we will substitute all these given parameters into the formula } \\ \text{above.} \end{gathered}[/tex][tex]\begin{gathered} 109,185=\text{ 125,500(1-}\frac{r}{100})^{18} \\ \frac{195185}{125500}=\frac{125500}{125500}(1-\frac{r}{100})^{18} \\ 0.87=(1-\frac{r}{100})^{18} \\ \text{take the natural logarithm of both sides:} \\ \ln 0.87=18\ln (1-\frac{r}{100}) \\ -0.1393=18\ln (1-\frac{r}{100}) \\ \frac{-0.1393}{18}=\ln (1-\frac{r}{100})_{}_{}_{}_{}_{} \\ -0.007737=\ln (1-\frac{r}{100}) \\ \end{gathered}[/tex][tex]\begin{gathered} e^{-0.007737}=(1-\frac{r}{100}) \\ 0.9922=1-\frac{r}{100} \\ \frac{r}{100}=1-0.9922 \\ \frac{r}{100}=0.007707 \\ r=100\times0.007707 \\ r=0.771\text{ \%} \end{gathered}[/tex]The continuous decay rate is 0.771%
B.
Using the same formula:
[tex]\begin{gathered} A=P(1-\frac{r}{100})^{nt} \\ t=2021-2007=14 \\ P=109,185 \\ n=1 \\ A=\text{?} \\ r=0.771 \\ \text{Substitute all the parameters into the formula above:} \end{gathered}[/tex][tex]\begin{gathered} A=P(1-\frac{r}{100})^{nt} \\ A=109,185(1-\frac{0.771}{100})^{1\times14} \\ A=109,185\times0.89730607 \\ A=97,972.36 \\ A=97,972\text{ (to the nearest person)} \end{gathered}[/tex]The population of the city in the year 2021 is 97,972.
C.
We will use the same formula:
[tex]\begin{gathered} A=P(1-\frac{r}{100})^{nt} \\ A=97,890 \\ P=125,500 \\ r=0.771 \\ t=\text{?} \\ \text{Substitute all these parameters into the formula above:} \\ \end{gathered}[/tex][tex]\begin{gathered} 97890=125,500(1-\frac{0.771}{100})^t^{} \\ \frac{97890}{125500}=\frac{125500}{125500}(0.99229)^t \\ 0.78=0.99229^t \\ \ln 0.78=t\ln 0.99229 \\ -\frac{0.2485}{\ln 0.99229}=t \\ t=32.101 \\ SO\text{ the year that the population will reach 97,890 will be:} \\ 1989+32.101=2021.101 \\ \text{Which is approximately year 2021.} \end{gathered}[/tex]