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Math Help - Help needed on averages please :)

  1. #1
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    Exclamation 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
    Last edited by dontwannagrowup; April 28th 2012 at 02:58 AM.
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  2. #2
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    Re: Help needed on averages please :)

    I take it to mean average decrease per year. That is 'total decrease divided by number of years'
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  3. #3
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    Re: Help needed on averages please :)

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

    next time, post your question in the appropriate forum.
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    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!
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  5. #5
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    Re: Help needed on averages please :)

    Thank you biffboy
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