suppose the population of a town increases by 3% each year. the population of the town today is 25,000. use compound interest table to find its population ten years from now. (must give the answer to the nearest thousand)

Answer 1

the calculation is

25000(1.03)^10
which on a calculator is a piece of cake

If you have to use "compound interest tables ??" you probably have a column for 4%, go down that table to the row n=10 to find 1.34392
multiply that by 25000
you should get $33597.91

Answer 2

the above contains a typo, I obviously meant 3% in the third last line

Answer 3

Compound interest table for 3% p.a., compounded each year, obtained by multiplying the factor of the previous year by 1.03. The value is also equal to 1.03ⁿ where n is the number of years compounded.
Year growth factor
0 1.0000
1 1.0300
2 1.0609
3 1.0927
4 1.1255
5 1.1593
6 1.1941
7 1.2299
8 1.2668
9 1.3048
10 1.3439

After 10 years, the population is
25000*1.3439
=33598
=34000 (to the nearest thousand).

Answer 4

Oops, it does not seem to like pasting from Excel!

Here is the compound interest table:
year growth factor
0 1.0000
1 1.0300
2 1.0609
3 1.0927
4 1.1255
5 1.1593
6 1.1941
7 1.2299
8 1.2668
9 1.3048
10 1.3439
After the tenth year, the population is
25000*1.3439
=33598
=34000 (to the nearest thousand)

Answer 5

OK, I give up pasting!
The growth factor for 10 years is 1.3439.
Multiply this with 25000 gives 33598, and rounding to the nearest thousand gives 34000.