The town of Springfield, USA has a population of 9900 people. The population is growing at a rate of 2.3% each year. Which expression gives the town's population 5 years from now? Choose 1 answer: (A) 9900*0.023^(5) (B) 9900+0.023^(5) (C) 9900(1+0.023)^(5) (D) 9900+(1+0.023)^(5)

Exponential growth formula: To find the population after a certain number of years when it is growing at a constant percentage rate per year, we use the formula for exponential growth: P = P0(1 + r)^t, where P0 is the initial population, r is the growth rate as a decimal, and t is the time in years.Convert growth rate to decimal: Convert the growth rate from a percentage to a decimal by dividing by 100. The growth rate is 2.3%, so in decimal form, it is 0.023.Substitute values into formula: Substitute the given values into the exponential growth formula: P = 9900(1 + 0.023)^5. This will give us the population of Springfield 5 years from now.Evaluate expression for population: Evaluate the expression to ensure it matches one of the given options. The correct expression based on the exponential growth formula is 9900(1 + 0.023)^5, which corresponds to option (C).

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The town of Springfield, USA has a population of 9900 people. The population is growing at a rate of
2.3% each year.
Which expression gives the town's population 5 years from now?
Choose 1 answer:
(A)
9900*0.023^(5)
(B)
9900+0.023^(5)
(C)
9900(1+0.023)^(5)
(D)
9900+(1+0.023)^(5)

The town of Springfield, USA has a population of $9900$ people. The population is growing at a rate of $2.3 \%$ each year. $\newline$ Which expression gives the town's population $5$ years from now? $\newline$ Choose $1$ answer: $\newline$ (A) $9900 \cdot 0.023^{5}$ $\newline$ (B) $9900+0.023^{5}$ $\newline$ (C) $9900(1+0.023)^{5}$ $\newline$ (D) $9900+(1+0.023)^{5}$

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Q. The town of Springfield, USA has a population of

9900

people. The population is growing at a rate of

2.3 \%

each year.

\newline

Which expression gives the town's population

5

years from now?

\newline

Choose

1

answer:

\newline

(A)

9900 \cdot 0.023^{5}

\newline

(B)

9900+0.023^{5}

\newline

(C)

9900(1+0.023)^{5}

\newline

(D)

9900+(1+0.023)^{5}

Exponential growth formula: To find the population after a certain number of years when it is growing at a constant percentage rate per year, we use the formula for exponential growth: $P = P_0(1 + r)^t$ , where $P_0$ is the initial population, $r$ is the growth rate as a decimal, and $t$ is the time in years.
Convert growth rate to decimal: Convert the growth rate from a percentage to a decimal by dividing by $100$ . The growth rate is $2.3\%$ , so in decimal form, it is $0.023$ .
Substitute values into formula: Substitute the given values into the exponential growth formula: $P = 9900(1 + 0.023)^5$ . This will give us the population of Springfield $5$ years from now.
Evaluate expression for population: Evaluate the expression to ensure it matches one of the given options. The correct expression based on the exponential growth formula is $9900(1 + 0.023)^5$ , which corresponds to option (C).