Solve the problem.
-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 2004 ? Use $\mathrm { y } = \mathrm { y } _ { 0 } \mathrm { e } ^ { 0.013 \mathrm { t } }$ and round to the nearest ten-thousand.
A) $8,150,000$
B) $8,970,000$
C) $7,990,000$
D) $9,780,000$

Question

Quizplus · Accepted Answer

Using the formula $y = y_0 e^{0.013t}$, where $y_0 = 7,638,000$, $t = 2004 - 1999 = 5$ years, and the growth rate is $1.3\%$ or $0.013$ in decimal form, we calculate the population in 2004. Substituting the values, we get $y = 7,638,000 \times e^{0.013 \times 5}$. Solving this gives a value close to $8,150,000$, rounding to the nearest ten-thousand.

Solve the Problem 1.3%1.3 \%1.3% Per Year If 7,638,0007,638,0007,638,000 People Lived There in 1999 , How Many Will There

Solve the Problem $1.3 \%$ Per Year If $7,638,000$ People Lived There in 1999 , How Many Will There