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son4ous [18]
3 years ago
9

Suppose the population of a town is 100,000 in 1999. The population increases at a rate of 4.5% every year. What will be the pop

ulation of the town in 2005? Round your answer to the nearest whole number.
Mathematics
2 answers:
Bumek [7]3 years ago
7 0

Answer:

130,226.

Step-by-step explanation:

We have been given that the population of a town is 100,000 in 1999. The population increases at a rate of 4.5% every year.

We will use exponential growth function to solve our given problem. We know that exponential function is in form f(x)=a\cdot b^x, where,

f(x) = Final value,

a = Initial value,

b = For growth b is in form 1+r, where r represents rate in decimal form.  

Let us convert 4.5% into decimal form.

4.5\%=\frac{4.5}{100}=0.04

Upon substituting our given values in exponential growth formula, we will get:

f(x)=100,000\cdot (1.045)^x, where x represents number of years after 1999.

To find population of town in 2005, we will substitute x=6 (2005-1999=6) in our function.

f(6)=100,000\cdot (1.045)^6

f(6)=100,000\cdot 1.3022601248475156

f(6)=130,226.01248475156

f(6)\approx 130,226

Therefore, the population of the town in 2005 will be 130,226.

uysha [10]3 years ago
3 0
100,000 * (1.045)^years
2005 Population = 100,000 * (1.045)^6
<span><span>= 100,000 * 1.3022601248 </span>
= </span><span><span><span>130,226</span></span></span>



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