Answer:
It remais 128 mg of the medication at 10:00 a.m.
It takes 1 day for the concentration to drop below 1 mg
Step-by-step explanation:
The formula of the decays of the concentration of the medications is:
[tex]f(t) = d*(0.8)^t[/tex]
So, if she takes the medication ate 6:00 a.m., at 10:00 a.m., t= 2, so:
f(2) = 200*(0.8)²
f(2) = 128 mg
For the concentration drop below 1 mg:
[tex]1 = 200*(0.8)^t[/tex]
[tex](0.8)^t = 0.005[/tex]
Applying ln:
[tex]ln(0.8)^t = ln0.005[/tex]
t*ln0.8 = -5.2983
-0.2231t = -5.2983
t = 23.74 hours
So, it will take 24 hours, or 1 day, for the concentration drop below 1 mg.
