Posted by **Melia** on Tuesday, October 30, 2012 at 9:20pm.

I know that year 2010, the population is 38.447 million and year 2011, the population is 38.935 million.

The formula given: A=36.1e^(0.0126*t)

Determine the population in 2010 and 2011 to nearest thousand people. Use these values to determine the increase in population to the nearest thousand people over this one year period.

What is this as an annual percentage growth rate with 2 digits after the decimal point?

- Math -
**Melia**, Tuesday, October 30, 2012 at 10:57pm
I subtracted the populations and got 488,000 people.

Then I took that number and divided by 38.935 million to get the decimal answer and after that, I multiplied the decimal by 100 to get the annual percentage growth rate of 1.25%

Would that be the correct answer?

## Answer this Question

## Related Questions

- College alegbra - The population P of a city has been increasing at an annual ...
- math - Date US Population Jul 1, 2012 313.85 million Jul 1, 2011 311.59 million...
- economics - The country of Arkanslavia had money supplies of 183 million ...
- Uninhibited Growth - In 1992, Chad and Denmark each had a population of about 5....
- math - In 2006, the population of the country was 32.3 million. This represented...
- math - can please someone check my answer Q.) 30% of the adult population was ...
- Calculus - The population in certain country is growing at the rate of 7 % a ...
- Algebra word problem - In 2004, the population of Mexico was 106.5 million. If ...
- Trigonometry - In 2010, the population of a country was 70 million and growing ...
- Algebra - On Jan 1, 2010, Chessville has a population of 50,000 people. ...

More Related Questions