College Algebra Word Problem

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 model T=.0002xsquared−.316x+127.9, where T is the exposure time in hours necessary to reach this level and 500 less than or equals 800 is the amount of carbon monoxide present in the air in parts per million (ppm).
a) What is the exposure time when x=670?
b) Estimate the concentration of CO necessary to produce a coma in 4 hr.
What is the exposure time when x=670?

Answer 1

Hi Destiney,

 

I assume that you accidentally repeated a question and that there are only two questions.

 

a) Since we are given that T is the exposure time in hours and that x = 670 ppm, we can simply plug it into our equation:

 

T = 0.0002 * (670)^2 - 0.316 * (670) + 127.9

 

T = 0.0002 * 448900 - 0.316 * 670 + 127.9

 

T = 89.78 - 211.72 + 127.9

 

T = 429.4 hours

 

b) To induce a coma in 4 hours, we can plug that into our equation as T:

 

4 = 0.0002x^2 - 0.316x + 127.9

 

0.0002x^2 - 0.316x + 123.9 = 0

 

We now need to use the quadratic equation: x = (-b ± √[b^2 - 4ac]) / 2a

 

x = (0.316 ± √[(-0.316)^2 - 4 * 0.0002 * 123.9]) / (2 * 0.0002)

 

x = (0.316 ± √[(0.099856 - 4 * 0.0002 * 123.9]) / (2 * 0.0002)

 

x = (0.316 ± √[0.000736]) / (2 * 0.0002)

 

x = (0.316 ± 0.0271) / 0.0004     I approximated the square root here, but my calculations use the real value.

 

x = 857.8 hours and 722.2 hours

 

But because we have a range limit, where the carbon monoxide present in the air in ppm is less than or equal to 800, we can eliminate 857.8 hours and our final answer is 722.2 hours.

 

Hope this helps!