Approximate the number using a calculator. Round your answer to three decimal places.
-The population in a particular country is growing at the rate of $2.2 \%$ per year. If $6,365,000$ people lived there in 1999 , how many will there be in the year 2005 ? Use $\mathrm { f } ( \mathrm { x } ) = \mathrm { y } _ { 0 } \mathrm { e } ^ { 0.022 \mathrm { t } }$ and round to the nearest ten-thousand.
A) $7,260,000$
B) $7,990,000$
C) $7,120,000$
D) $8,720,000$

Question

Quizplus · Accepted Answer

The formula for population growth is given as $ f(x) = y_0 e^{0.022t} $, where $y_0$ is the initial population, $t$ is the time in years, and $e$ is the base of the natural logarithm. Given $y_0 = 6,365,000$ and $t = 2005 - 1999 = 6$ years, we calculate $ f(x) = 6,365,000 \cdot e^{0.022 \cdot 6} $. This calculation yields approximately $7,260,000$ when rounded to the nearest ten-thousand.

Approximate the Number Using a Calculator 2.2%2.2 \%2.2% Per Year If 6,365,0006,365,0006,365,000

Approximate the Number Using a Calculator $2.2 \%$ Per Year If $6,365,000$