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=.0002x^2−.316x+127.9, where T is the exposure time in hours necessary to reach this level and 500≤ x ≤800 is the amount of carbon monoxide present in the air in parts per million? (ppm).
A) What is the exposure time when x= 600? hr
B) Estimate the concentration of CO necessary to produce a coma in 9 hr.

A) 10.38hours

B) 617.95 is the concerntration of CO

Step-by-step explanation:

A) when x=600

T= 0.0002x^2 - 0.316x + 127.9

T= 0.0002(600)^2 -0.316(600) + 127.9

T= 10.38secs

B) to find the concerntration, use Almighty formular

X=-b+-sqrt(b^2-4ac)/2a

0.0002x^2-0.316x-127.9=9

0.0002x^2 - 0.316x + 118.9=0

X= 0.316 +- sqrt(0.316^2) -4(0.0002×118.9)/(2×0.0002)

X= 0.316+(0.68817/0.0004)=962.05 or

X=-0.316 -0.68817/0.0004 =617.95

Therefore X is within the domain of 617.95 is the concerntration

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

answer:

a=9

step-by-step explanation: lmk if you need a explanation