anonymous
  • anonymous
I am not a student or a minor, I just need help with an algebraic percentage problem, can you help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Owlfred
  • Owlfred
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anonymous
  • anonymous
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
anonymous
  • anonymous
what would the overall average rate of consumption be for this lot of welding wire?

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amistre64
  • amistre64
given 2 data points; the best solution would be a linear function right?
anonymous
  • anonymous
I beleive... its been a long time since I have used algebra (sorry)
amistre64
  • amistre64
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...
anonymous
  • anonymous
I know that the different values have to be weighted per their value against the whole
amistre64
  • amistre64
theres not enough data to get any relevant statistical measurements from tho, weigthed or not... i believe
amistre64
  • amistre64
we could use the point 0 pounds for 0 hours right?
anonymous
  • anonymous
we dont need to figure rate, just an overall average of the rate
amistre64
  • amistre64
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...
anonymous
  • anonymous
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
anonymous
  • anonymous
I need an average pounds per hour for the whole 385,852
amistre64
  • amistre64
When I fit this data to a quadratic I pretty much get this result for x = hours -3466.265785332662 x^2 + 57901.08900977116 x
amistre64
  • amistre64
if i flip my data for x = pounds I can get one to estimate the whole of 385k
amistre64
  • amistre64
-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 :)
amistre64
  • amistre64
but then the rate does increase as the hours drop so it might be an accurate portrayal of the data point
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=-5.750456292871241e-10x^2+%2B+0.00017649192015923025x
amistre64
  • amistre64
its either that or add the data points to each other and /2 to get the midpoint
amistre64
  • amistre64
148,864 @ 13.53 236,988 @ 9.53 ---------------- (385852 @ 23.06)/2 = (192926 @ 11.53)
anonymous
  • anonymous
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
shadowfiend
  • shadowfiend
planechkr, I'll let you click `good answer' for the answer that helped you the most :)
amistre64
  • amistre64
fiddles looks to be a better interpretation of the information given :)

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