Solve for x.
-A city is growing at the rate of $0.3 \%$ annually. If there were $5,226,000$ residents in the city in 2008 , find how many (to the nearest ten-thousand) are living in that city in 2012 . Use $y = 5,226,000 ( 2.7 ) ^ { 0.003 t }$.
A) $5,290,000$
B) $5,320,000$
C) $14,110,000$
D) 170,000

Question

Quizplus · Accepted Answer

The correct formula for exponential growth is $y = P(1 + r)^t$, where $P$ is the initial amount, $r$ is the rate of growth, and $t$ is the time in years. However, the given formula uses a base of $2.7$ raised to $0.003t$, which seems to be a misinterpretation. Assuming a $0.3\%$ annual growth rate, the correct formula should be $y = 5,226,000(1 + 0.003)^t$. Since the growth rate in the question is mistakenly represented, but we proceed with the provided formula for calculation: $y = 5,226,000(2.7)^{0.003t}$. Given $t = 2012 - 2008 = 4$ years, we substitute $t$ in the formula. However, to correct the approach for an annual growth rate of $0.3\%$, we should use $y = 5,226,000(1.003)^4$. Calculating this correctly would not match the given formula, but aiming to match the closest answer based on the provided formula and the growth rate, we approximate the population after 4 years, which would be closest to $5,290,000$ when using a more standard approach to model a $0.3\%$ growth rate over 4 years.

Solve for X 0.3%0.3 \%0.3% Annually If There Were 5,226,0005,226,0005,226,000 Residents in the City in 2008 , Find How Many

Solve for X $0.3 \%$ Annually If There Were $5,226,000$ Residents in the City in 2008 , Find How Many