Solve the problem.
-The population in a particular country is growing at the rate of $2.8 \%$ per year. If $7,638,000$ people lived there in 1999 , how many will there be in the year 2005 ? Use $\mathrm { f } ( \mathrm { x } ) = \mathrm { y } _ { 0 } \mathrm { e } ^ { 0.028 \mathrm { t } }$ and round to the nearest ten-thousand.
A) $8,850,000$
B) $10,840,000$
C) $9,940,000$
D) $9,040,000$

Question

Quizplus · Accepted Answer

Using the formula $ f(x) = y_0 e^{0.028t} $, where $ y_0 = 7,638,000 $ and $ t = 6 $ (from 1999 to 2005), we calculate $ f(x) \approx 9,040,000 $.

Solve the Problem 2.8%2.8 \%2.8% Per Year If 7,638,0007,638,0007,638,000 People Lived There in 1999 , How Many Will There

Solve the Problem $2.8 \%$ Per Year If $7,638,000$ People Lived There in 1999 , How Many Will There