A new method was developed for the analysis of opioids in urine samples. The following data

was obtained via the calibration curve method for opioid A$.$

$)$

Signal intensity $+-$ standard deviation $($a$.$ u$.)*$

$0$

$20+-6\mathrm{.}6$

$5$

$30+-4\mathrm{.}5$

$10$

$40+-2$

$20$

$65+-5$

$40$

$105+-10\mathrm{.}5$

$60$

$150+-22\mathrm{.}5$

$80$

Calculate the limit of detection $($LOD$)$ and the limit of quantitation $($LOQ$)$ of the method.

a$)$ LOD$=9\mathrm{.}1ppm$;LOQ$=30\mathrm{.}4ppm$

b$)$ LOD$=10\mathrm{.}6ppb$;LOQ$=35\mathrm{.}2ppb$

c$)$ LOD$=9\mathrm{.}1ppb$;LOQ$=30\mathrm{.}4ppb$

d$)$ LOD$=10\mathrm{.}6ppm$;LOQ$=35\mathrm{.}2ppm$

e$)$ None of the above.

What is the centroid of the calibration curve?

a$)(20,65)$

b$)(40,105)$

c$)(30\mathrm{.}7,81\mathrm{.}4)$

d$)(22\mathrm{.}5,68\mathrm{.}3)$

e$)$ None of the above.

The number of blank determinations was $16.$ All the other signals are the average of $3$

determinations each.analytical Chemistry.