Question

High concentrations of carbon monoxide CO can cause coma and possible death. The time required for a person to reach a COHb level capable of causing a coma can be approximated by the quadratic modelT=00022 - 316x + 127 9, where is the exposure time in hours necessary to reach this level and 500 5x5 800 is the amount of carbon monoxide present in the air in parts per million (ppm)a) What is the exposure time when x=550?b) Estimate the concentration of CO necessary to produce a coma in 6 hWhat is the exposure time when x=550?Round to the nearest tenth as needed)

Answer 1

The function T(x) gives us the time required to reach a coma and x is the amount of carbon monoxide.

$T(x)=0.0002x^2-0.316x+127.9$

a) Set x=550 and solve for T as follows:

$\begin{gathered} T(550)=0.0002(550)^2-0.316(550)+127.9 \\ \Rightarrow T(550)=60.5-173.8+127.9 \\ \Rightarrow T(550)=14.6 \end{gathered}$

And remember that T is given in hours, then the answer to part a) is 14.6 hr.

b) Set T(x)=6 and solve for x as follows:

$\begin{gathered} T(x)=6 \\ \Rightarrow0.0002x^2-0.316x+127.9=6 \\ \Rightarrow0.0002x^2-0.316x+121.9=0 \\ \Rightarrow x\approx669.2\text{ or }x\approx910.8 \end{gathered}$

You can solve the quadratic equation by any method you want.

Remember that T(x) requires that 500<=x<=800, we cannot know the effect that 910.8 ppm can cause in a person. Then, the answer to part b) is x=669.2 ppm