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Suppose the population of a town is 100,000 in 2014. The population increases at a rate of 4.5% every year. What will be the population of the town in 2020? Round your answer to the nearest whole number.

130,226
929,411
627,000
870,000

Respuesta :

Aurzi
ok. Population = 100,000 *1.045^n (n is number of years since 1999. So, the population in 2005 (where n=6) is 130,226

The population of the town in 2020 is 130,226 , Option A is the correct answer.

What is an Equation ?

An equation is a mathematical statement formed when two algebraic expressions are equated using an equal sign.

It is given that

Initial Population = 100,000

Rate of Increase = 4.5%

Population of the town in 2020 = ?

Let y represent the Population of the town in 2020

and n represent the no. of years elapsed from 2014

The equation that will be formed is

y =    100000 * 1.045 ⁿ

for 2020 , n = 6

y = 100000 * 1.045⁶

y = 130226

Therefore the population of the town in 2020 is 130,226 , Option A is the correct answer.

To know more about Equation

https://brainly.com/question/10413253

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