Question

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

Answer 1

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!

Answer 2

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

Answer 3

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

Answer 4

given 2 data points; the best solution would be a linear function right?

Answer 5

I beleive... its been a long time since I have used algebra (sorry)

Answer 6

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...

Answer 7

I know that the different values have to be weighted per their value against the whole

Answer 8

theres not enough data to get any relevant statistical measurements from tho, weigthed or not... i believe

Answer 9

we could use the point 0 pounds for 0 hours right?

Answer 10

we dont need to figure rate, just an overall average of the rate

Answer 11

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...

Answer 12

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

Answer 13

I need an average pounds per hour for the whole 385,852

Answer 14

When I fit this data to a quadratic I pretty much get this result for x = hours

-3466.265785332662 x^2 + 57901.08900977116 x

Answer 15

if i flip my data for x = pounds I can get one to estimate the whole of 385k

Answer 16

-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 :)

Answer 17

but then the rate does increase as the hours drop so it might be an accurate portrayal of the data point

Answer 18

http://www.wolframalpha.com/input/?i=-5.750456292871241e-10x^2+%2B+0.00017649192015923025x

Answer 19

its either that or add the data points to each other and /2 to get the midpoint

Answer 20

148,864 @ 13.53
236,988 @ 9.53
----------------
(385852 @ 23.06)/2 = (192926 @ 11.53)

Answer 21

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

Answer 22

planechkr, I'll let you click `good answer' for the answer that helped you the most :)

Answer 23

fiddles looks to be a better interpretation of the information given :)