Answer :
We start by assuming that the population grows continuously according to the model
[tex]$$
P(t) = P_0 \, e^{rt},
$$[/tex]
where [tex]$P_0$[/tex] is the initial population, [tex]$r$[/tex] is the continuous growth rate, and [tex]$t$[/tex] is the number of years.
1. Determine the Growth Rate
The population in 1985 is given as [tex]$145$[/tex] million and in 1995 as [tex]$190$[/tex] million. Since the time between 1985 and 1995 is [tex]$10$[/tex] years, we can set up the equation
[tex]$$
190 = 145 \, e^{r \cdot 10}.
$$[/tex]
To solve for [tex]$r$[/tex], divide both sides by [tex]$145$[/tex]:
[tex]$$
\frac{190}{145} = e^{10r}.
$$[/tex]
Taking the natural logarithm of both sides gives
[tex]$$
\ln\left(\frac{190}{145}\right) = 10r.
$$[/tex]
Therefore, the growth rate [tex]$r$[/tex] is
[tex]$$
r = \frac{1}{10} \ln\left(\frac{190}{145}\right).
$$[/tex]
Calculating the value yields approximately
[tex]$$
r \approx 0.027 \quad \text{per year}.
$$[/tex]
2. Calculate the Population in 2005
The year 2005 is [tex]$10$[/tex] years after 1995. Using the population from 1995 ([tex]$190$[/tex] million) as the starting point, we calculate the population in 2005 by
[tex]$$
P(2005) = 190 \, e^{r \cdot 10}.
$$[/tex]
Substituting the approximate value for [tex]$r$[/tex], we get
[tex]$$
P(2005) = 190 \, e^{0.027 \cdot 10}.
$$[/tex]
Evaluating this expression gives an approximate population of about [tex]$248.97$[/tex] million.
3. Matching with the Options
Among the choices provided:
- Option A: [tex]$P=145 e^{(0.027)(10)}$[/tex]
- Option B: [tex]$P=145 e^{(0.027)(20)}$[/tex]
- Option C: [tex]$P=190 e^{(0.027)(10)}$[/tex]
- Option D: [tex]$P=190 e^{(0.027)(20)}$[/tex]
The correct expression that we derived is
[tex]$$
\boxed{190 \, e^{(0.027)(10)}},
$$[/tex]
which corresponds to Option C.
Thus, the correct answer is Option C.
[tex]$$
P(t) = P_0 \, e^{rt},
$$[/tex]
where [tex]$P_0$[/tex] is the initial population, [tex]$r$[/tex] is the continuous growth rate, and [tex]$t$[/tex] is the number of years.
1. Determine the Growth Rate
The population in 1985 is given as [tex]$145$[/tex] million and in 1995 as [tex]$190$[/tex] million. Since the time between 1985 and 1995 is [tex]$10$[/tex] years, we can set up the equation
[tex]$$
190 = 145 \, e^{r \cdot 10}.
$$[/tex]
To solve for [tex]$r$[/tex], divide both sides by [tex]$145$[/tex]:
[tex]$$
\frac{190}{145} = e^{10r}.
$$[/tex]
Taking the natural logarithm of both sides gives
[tex]$$
\ln\left(\frac{190}{145}\right) = 10r.
$$[/tex]
Therefore, the growth rate [tex]$r$[/tex] is
[tex]$$
r = \frac{1}{10} \ln\left(\frac{190}{145}\right).
$$[/tex]
Calculating the value yields approximately
[tex]$$
r \approx 0.027 \quad \text{per year}.
$$[/tex]
2. Calculate the Population in 2005
The year 2005 is [tex]$10$[/tex] years after 1995. Using the population from 1995 ([tex]$190$[/tex] million) as the starting point, we calculate the population in 2005 by
[tex]$$
P(2005) = 190 \, e^{r \cdot 10}.
$$[/tex]
Substituting the approximate value for [tex]$r$[/tex], we get
[tex]$$
P(2005) = 190 \, e^{0.027 \cdot 10}.
$$[/tex]
Evaluating this expression gives an approximate population of about [tex]$248.97$[/tex] million.
3. Matching with the Options
Among the choices provided:
- Option A: [tex]$P=145 e^{(0.027)(10)}$[/tex]
- Option B: [tex]$P=145 e^{(0.027)(20)}$[/tex]
- Option C: [tex]$P=190 e^{(0.027)(10)}$[/tex]
- Option D: [tex]$P=190 e^{(0.027)(20)}$[/tex]
The correct expression that we derived is
[tex]$$
\boxed{190 \, e^{(0.027)(10)}},
$$[/tex]
which corresponds to Option C.
Thus, the correct answer is Option C.
