Solve.
-The population in a particular country is growing at the rate of $1.3 \%$ per year. If $7,638,000$ people lived there in 1999 , how many will there be in the year 2007? Use $\mathrm { y } = \mathrm { y } _ { 0 } \mathrm { e } ^ { 0.013 \mathrm { t } }$ and round to the nearest ten-thousand.
A) $8,310,000$
B) $9,320,000$
C) $8,480,000$
D) $10,170,000$

Question

Quizplus · Accepted Answer

The formula for population growth is given as $y = y_0 e^{0.013t}$, where $y_0$ is the initial population, $t$ is the time in years, and $y$ is the population after $t$ years. Given $y_0 = 7,638,000$ and $t = 2007 - 1999 = 8$ years, we substitute these values into the formula to find the population in 2007. $y = 7,638,000 \cdot e^{0.013 \cdot 8}$. Calculating this gives a value close to $8,480,000$, rounding to the nearest ten-thousand.

Solve -The Population in a Particular Country Is Growing at the Rate

Solve
-The Population in a Particular Country Is Growing at the Rate