# The table gives estimates of the world population, In million, from 1750 to 2000. (Round your...

## Question:

The table gives estimates of the world population, In million, from 1750 to 2000. (Round your answers to the nearest millions.)

Year Population
1750 750
1800 980
1850 1260
1900 1650
1950 2560
2000 6080

(a) Use the experimental model and the population figures for 1750 and 1800 to predict the world population in 1900 and 1950

(b) Use the experimental model and the population figures for 1800 and 1850 to predict the world population in 1950.

(c) Use the experimental model and the population figures for 1900 and 1950 to predict the world population in 2000.

## Exponential Growth

We will be solving the given question using an exponential growth model. The model will be:

$$P(t)=P(0)e^{rt}$$

For each of the sub-parts, the parameters of the formula change as shown below.

To solve each of the questions, we are going to use an exponential model. The model will be:

$$P(t)=P(0)e^{rt}$$

P(t) is the population after t years when the initial population at time t=0 was P(0). r is the rate of growth of the population which needs to be calculated from the data.

a) Here, P(0) is 750 and P(50) is 980. The rate of growth is calculated as follows.

\begin{align} &750e^{50r}=980\\ &50r=\ln \left ( \frac{980}{750} \right )\\ &r\approx 0.005350\\ \therefore &P(t)=750e^{0.00530t} \end{align}

The predicted populations in 1900 (t=150) and 1950 (t=200) will be:

\begin{align} &P(150)=750e^{0.00530*150}=1660.831\\ &P(200)=750e^{0.00530*200}=2164.778\\ \end{align}

b) Here, P(0) is 980 and P(50) is 1260. The rate of growth is calculated as follows.

\begin{align} &980e^{50r}=1260\\ &50r=\ln \left ( \frac{1260}{980} \right )\\ &r\approx 0.005026\\ \therefore &P(t)=980e^{0.005026t} \end{align}

The predicted populations in 1950 (t=150) will be:

\begin{align} &P(150)=980e^{0.005026*150}=2082.767\\ \end{align}

c) Here, P(0) is 1650 and P(50) is 2560. The rate of growth is calculated as follows.

\begin{align} &1650e^{50r}=2560\\ &50r=\ln \left ( \frac{2560}{1650} \right )\\ &r\approx 0.008785\\ \therefore &P(t)=1650e^{0.008785t} \end{align}

The predicted populations in 2000 (t=100) will be:

\begin{align} P(100)=1650e^{0.008785*100}=3972.022 \end{align}