Answer :
We are given an initial population of
[tex]$$P_0 = 141300$$[/tex]
and an annual growth rate of
[tex]$$5\% = 0.05.$$[/tex]
When a quantity grows by a fixed percentage every year, the formula to calculate the population after [tex]$t$[/tex] years is given by
[tex]$$P = P_0 (1 + r)^t,$$[/tex]
where [tex]$r$[/tex] is the annual growth rate.
Here, we substitute the values:
- [tex]$P_0 = 141300$[/tex]
- [tex]$r = 0.05$[/tex]
- [tex]$t = 6$[/tex]
This gives
[tex]$$P = 141300 \times (1 + 0.05)^6 = 141300 \times (1.05)^6.$$[/tex]
Calculating the exponentiation results in
[tex]$$P \approx 141300 \times (1.05)^6 \approx 189355.51402031255.$$[/tex]
Thus, the equation that represents the town's population after 6 years is
[tex]$$\boxed{P = 141300(1.05)^6}.$$[/tex]
[tex]$$P_0 = 141300$$[/tex]
and an annual growth rate of
[tex]$$5\% = 0.05.$$[/tex]
When a quantity grows by a fixed percentage every year, the formula to calculate the population after [tex]$t$[/tex] years is given by
[tex]$$P = P_0 (1 + r)^t,$$[/tex]
where [tex]$r$[/tex] is the annual growth rate.
Here, we substitute the values:
- [tex]$P_0 = 141300$[/tex]
- [tex]$r = 0.05$[/tex]
- [tex]$t = 6$[/tex]
This gives
[tex]$$P = 141300 \times (1 + 0.05)^6 = 141300 \times (1.05)^6.$$[/tex]
Calculating the exponentiation results in
[tex]$$P \approx 141300 \times (1.05)^6 \approx 189355.51402031255.$$[/tex]
Thus, the equation that represents the town's population after 6 years is
[tex]$$\boxed{P = 141300(1.05)^6}.$$[/tex]
