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anonymous

  • 5 years ago

I am not a student or a minor, I just need help with an algebraic percentage problem, can you help?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    I am a certified weld inspector, and this is a real world problem.... I have a sum of 385852 pounds of welding wire used last year at different rates of consumption. I need an average consumption of this gross. 148,864 were used at 13.53 pounds per hour, and 236,988 were used at 9.53 pounds per hour

  3. anonymous
    • 5 years ago
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    what would the overall average rate of consumption be for this lot of welding wire?

  4. amistre64
    • 5 years ago
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    given 2 data points; the best solution would be a linear function right?

  5. anonymous
    • 5 years ago
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    I beleive... its been a long time since I have used algebra (sorry)

  6. amistre64
    • 5 years ago
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    if we knew the total hours for last year we might be able to fit it to a quadratic function; or use statistical measures to get a best fit line...

  7. anonymous
    • 5 years ago
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    I know that the different values have to be weighted per their value against the whole

  8. amistre64
    • 5 years ago
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    theres not enough data to get any relevant statistical measurements from tho, weigthed or not... i believe

  9. amistre64
    • 5 years ago
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    we could use the point 0 pounds for 0 hours right?

  10. anonymous
    • 5 years ago
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    we dont need to figure rate, just an overall average of the rate

  11. amistre64
    • 5 years ago
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    the rate of change depends on the type of function that best fits the data... if that function is linear, then we have a line; if its curved then we gotta bend the line to fit...

  12. anonymous
    • 5 years ago
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    Sum = 385852 pounds used at the following rates (to make 100%) 148,864 @ 13.53 pounds per hour, and 236,988 @ 9.53 pounds per hour

  13. anonymous
    • 5 years ago
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    I need an average pounds per hour for the whole 385,852

  14. amistre64
    • 5 years ago
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    When I fit this data to a quadratic I pretty much get this result for x = hours -3466.265785332662 x^2 + 57901.08900977116 x

  15. amistre64
    • 5 years ago
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    if i flip my data for x = pounds I can get one to estimate the whole of 385k

  16. amistre64
    • 5 years ago
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    -5.750456292871241e-10x^2 + 0.00017649192015923025x at an hour output of: -17.514048386364664; which leads me to believe that what im thinking of is quite possibly wrong :)

  17. amistre64
    • 5 years ago
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    but then the rate does increase as the hours drop so it might be an accurate portrayal of the data point

  18. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=-5.750456292871241e-10x^2+%2B+0.00017649192015923025x

  19. amistre64
    • 5 years ago
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    its either that or add the data points to each other and /2 to get the midpoint

  20. amistre64
    • 5 years ago
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    148,864 @ 13.53 236,988 @ 9.53 ---------------- (385852 @ 23.06)/2 = (192926 @ 11.53)

  21. anonymous
    • 5 years ago
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    148864 pounds --------------------- = 11002.51 hours 13.53 pounds / hour 236988 pounds --------------------- = 24867.57 hours 9.53 pounds / hour total hours = 11002.51 + 24867.57 = 35870.08 385852 pounds -------------- = 10.756 pounds / hour 35870.08 hours

  22. shadowfiend
    • 5 years ago
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    planechkr, I'll let you click `good answer' for the answer that helped you the most :)

  23. amistre64
    • 5 years ago
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    fiddles looks to be a better interpretation of the information given :)

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