# Math Help - Help needed on averages please :)

1. ## Help needed on averages please :)

Hi,
Im a bit stuck on a question I have to do.
I need to find the annual average decrease between 1982-7.45 and 2003-6.15. Is it 6.15+7.45/12=1.13 per year Or do I just divide by 2 or the number of years inbetween? Also if I want to say 1.13 million is it written as 1.135 (to the powers of 5) and does anyone know how to get the little 5 above the bigger numbers?

Any advice would be absolutely fantastic.
Thank you

2. ## Re: Help needed on averages please :)

I take it to mean average decrease per year. That is 'total decrease divided by number of years'

3. ## Re: Help needed on averages please :)

$\frac{6.15-7.45}{2003-1982}$

next time, post your question in the appropriate forum.

4. ## Re: Help needed on averages please :)

next time, post your question in the appropriate forum.[/QUOTE]

Its nice that theres FRIENDLY and helpful people on here!

5. ## Re: Help needed on averages please :)

Thank you biffboy