# Help needed on averages please :)

• Apr 28th 2012, 02:54 AM
dontwannagrowup
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
• Apr 28th 2012, 03:05 AM
biffboy
Re: Help needed on averages please :)
I take it to mean average decrease per year. That is 'total decrease divided by number of years'
• Apr 28th 2012, 04:32 AM
skeeter
Re: Help needed on averages please :)
$\displaystyle \frac{6.15-7.45}{2003-1982}$

next time, post your question in the appropriate forum.
• Apr 28th 2012, 09:01 AM
dontwannagrowup
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!
• Apr 28th 2012, 09:02 AM
dontwannagrowup
Re: Help needed on averages please :)
Thank you biffboy