The population of a country between 1921 and 2008 with projections to 2010 can be modeled as \( p(t)=2.50\left(1.15^{t}\right) \) million people where \( t \) is number of years since 1900 . What is the average population of this country from 1912 to 1944 ? The units of your answer should be million people. Round your answer to two decimal places.

To find the average population of the country from 1912 to 1944, we first determine the values of \(t\) corresponding to these years. For 1912, \(t = 1912 - 1900 = 12\), and for 1944, \(t = 1944 - 1900 = 44\). Now we need to calculate the average population over that interval. The formula for average value of a continuous function on the interval \([a, b]\) is given by: \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \] In this case, \(a = 12\) and \(b = 44\) with \(f(t) = 2.50(1.15^t)\). Now we need to compute the integral: \[ \int_{12}^{44} 2.50(1.15^t) \, dt \] The integral of \(1.15^t\) is \(\frac{1.15^t}{\ln(1.15)}\). Thus, we compute: \[ \int 2.50(1.15^t) \, dt = 2.50 \cdot \frac{1.15^t}{\ln(1.15)} + C \] Calculating the definite integral from \(12\) to \(44\): \[ = 2.50 \left[ \frac{1.15^{44}}{\ln(1.15)} - \frac{1.15^{12}}{\ln(1.15)} \right] \] Substituting the values: 1. Calculate \(1.15^{44}\) and \(1.15^{12}\): - \(1.15^{44} \approx 35.066\) - \(1.15^{12} \approx 5.487\) Now we substitute these into the equation: \[ = 2.50 \left[ \frac{35.066 - 5.487}{\ln(1.15)} \right] \] Estimating \(\ln(1.15) \approx 0.13976\): \[ = 2.50 \left[ \frac{29.579}{0.13976} \right] \approx 2.50 \cdot 211.54 \approx 528.85 \] To find the average population: \[ \text{Average} = \frac{1}{44-12} \cdot 528.85 = \frac{528.85}{32} \approx 16.59 \text{ million people} \] Thus, the average population of this country from 1912 to 1944 is approximately **16.59 million people**.