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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 formula model T=.0002xsq - .316x +127.9, where T is the exposure time in hours necessary to reach this level and 500≤x≤800 is the amount of CO present in the air in parts per million (ppm).

What is the exposure time when x = 610?

Answer 9.6 hr I got this easy

Estimate the concentration of CO necessary to produce a coma in 6 hr.
This is the part I can’t find any help on.

Answer :

nobillionaireNobley
T = 0.0002x^2 - 0.316x + 127.9

When x = 610,
T = 0.0002(610)^2 - 0.316(610) + 127.9 = 0.0002(372100) - 192.76 + 127.9 = 74.42 - 64.86 = 9.56.
The exposure time is 9.56 hours.

T = 0.0002x^2 - 0.316x + 127.9 = 6
0.0002x^2 - 0.316x + 121.9 = 0
x^2 - 1580x + 609500 = 0
x^2 - 1580x + 624100 = -609500 + 624100 = 14600
(x - 790)^2 = 14600
x - 790 = + or -120.83
x = 121 + 790 or -121 + 790
x = 911 or 669

but 500 ≤ x ≤ 800, therefore the required amount of CO is 669 ppm.

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