Solve the problem.
-The population in a particular country is growing at the rate of $1.7 \%$ per year. If $7,638,000$ people lived there in 1999 , how many will there be in the year 2004 ? Use $f ( x ) = y _ { 0 } e ^ { 0.017 t }$ and round to the nearest ten-thousand.
A) $8,320,000$
B) $9,150,000$
C) $8,150,000$
D) $9,980,000$

Question

Quizplus · Accepted Answer

To find the population in 2004, we use the given formula $f(x) = y_0 e^{0.017t}$, where $y_0 = 7,638,000$ is the initial population in 1999, and $t$ is the number of years from 1999 to 2004, which is $5$ years. Plugging in the values, we get $f(x) = 7,638,000 \times e^{0.017 \times 5}$. Calculating this gives a value close to $8,320,000$, rounding to the nearest ten-thousand.

Solve the Problem 1.7%1.7 \%1.7% Per Year If 7,638,0007,638,0007,638,000 People Lived There in 1999 , How Many Will There

Solve the Problem $1.7 \%$ Per Year If $7,638,000$ People Lived There in 1999 , How Many Will There