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anonymous
 5 years ago
The amount of a drug in a person's bloodstream can be modeled with exponential decay. Suppose that, in 3 hours, 18% of the drug is removed from the bloodstream. What is the halflife of the drug?
A) 8.1 hours
B) 8.7 hours
C) 9.3 hours
D) 9.9 hours
E) 10.5 hours
anonymous
 5 years ago
The amount of a drug in a person's bloodstream can be modeled with exponential decay. Suppose that, in 3 hours, 18% of the drug is removed from the bloodstream. What is the halflife of the drug? A) 8.1 hours B) 8.7 hours C) 9.3 hours D) 9.9 hours E) 10.5 hours

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0E. Use the exponential decay formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you show me how to set it up?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first write the formula for exponential decay: I'm going to use my own variables for convience F=Ae^(kt)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F=amount you have A=amount you start with k=constant t=time elapsed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what you have to do first is find k, so using 3 hours and 18% we find k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0assume you start with 100% so A=1 F=1.18=.82 < this is cause you remove .18 from the blood, leaving you with .82

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you have .82=1e^(3k)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then plug k into the equation. this time F=.5 since you are left with half the amount and t is what you are trying to find.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that give me 10.5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do I solve for K? im sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0.82=1e^(3k) Ln(.82)=3k k=Ln(.82)/3
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