From 2000-2010 a city had a 2.5% annual decrease in population. If the city had 2,950,000 people in 2000, determine the city's population in 2008.

So you first need to find how much the population drops each year. For that you need to do 2.5%/10years which would be .25 then you would multiply by 8 to find for 2008 which you would get 2.Then you need to get 2% of 2950000. To do this you need to do 2,950,000*.02 which finally gives you 59,000 then you would subtract 2,950,000-59000 which the answer to this is 2,891,000! Hope it helps.

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erinna erinna · Answer 2 · 2022-04-15T07:25:13-04:00

The population of the city in the year 2008 is 2409122.

Important information:

Initial population in 2000 is 2,950,000.
Decreasing rate is 2.5%.

Exponential formula:

According to the exponential decay formula:

\(f(t)=a(1-r)^t\)

Where a is the initial value, r is the decay rate and t is the time period.

The difference between 2008 and 2000 is 8.

Substitute \(a=2950000, r=0.025, t=8\) in the above formula.

\(f(8)=2950000(1-0.025)^8\)

\(f(8)=2950000(0.975)^8\)

\(f(8)=2409122.8208\)

Round the value to the previous integer.

\(f(8)\approx 2409122\)

Therefore, the population of the city in 2008 is 2409122.

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