Is This Cocaine Concentration Calculation Correct?

[Scarpetta]

Can anyone tell me if this concentration calculation is correct please, its my first attempt and a bit unsure.
Two point determination of cocaine sample
Cocaine (mg/ml) Average peak height
0.015 13059
0.20 98496
Unknown (2mg/ml) 61136General equation Y = mx + c
Equation 1 standard 1: 13059 = 0.015m + c
Equation 2 standard 2: 98496 = 0.20m + c
98496 – 13059 = 85437
0.20 – 0.015 = 0.185
therefore 85437 = 0.185m + 0
85437 / 0.185 = m
m = 461821.62
c = 13059 = (0.015 X 461821.62)
13059 – (0.015 X 461821.62) = 6132
So, Y = 461821.62 X + 6132
Unknown therefore:
61136 = 461821.62 X + 0
= 61136 – 0 / 461821.62 = 0.13
0.13 / 2 X 100 = 6.6%

 

[cronxeh]

So, Y = 461821.62 X + 6132
Unknown therefore:
61136 = 461821.62 X + 0
= 61136 – 0 / 461821.62 = 0.13
0.13 / 2 X 100 = 6.6%

Why? You just determed that c is 6132. Following your logic it should be
61136 = 461821.62 X + 6132

0.20 – 0.015 = 0.19 <-- here you have only 2 significant figures.
therefore 85437 = 0.19m
85437 / 0.19 = m = 450000

Hence 61136 = 450000*x + 6132
55004 = 450000*x , x = 0.12
0.12/2 = 0.06 = 6%

 

[Scarpetta]

not bad for a first attempt, close but no cigar, thankx for the pointers