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

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

Mathematics
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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
what would the overall average rate of consumption be for this lot of welding wire?

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given 2 data points; the best solution would be a linear function right?
I beleive... its been a long time since I have used algebra (sorry)
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...
I know that the different values have to be weighted per their value against the whole
theres not enough data to get any relevant statistical measurements from tho, weigthed or not... i believe
we could use the point 0 pounds for 0 hours right?
we dont need to figure rate, just an overall average of the rate
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...
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
I need an average pounds per hour for the whole 385,852
When I fit this data to a quadratic I pretty much get this result for x = hours -3466.265785332662 x^2 + 57901.08900977116 x
if i flip my data for x = pounds I can get one to estimate the whole of 385k
-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 :)
but then the rate does increase as the hours drop so it might be an accurate portrayal of the data point
http://www.wolframalpha.com/input/?i=-5.750456292871241e-10x^2+%2B+0.00017649192015923025x
its either that or add the data points to each other and /2 to get the midpoint
148,864 @ 13.53 236,988 @ 9.53 ---------------- (385852 @ 23.06)/2 = (192926 @ 11.53)
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
planechkr, I'll let you click `good answer' for the answer that helped you the most :)
fiddles looks to be a better interpretation of the information given :)

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