FNGR 1 | Lesson 2 | Practice 3 Solution

a.  He will have 32 mg 5.7 hours after 8:00 am or at approximately 1:45.

He will have \(64(.5)^\frac{4}{5.7} ≈ 39.3\);  \(39.3\) mg left at noon.

He will have \(64(.5)^\frac{12}{5.7}≈14.9\);  \(14.9\) mg left at 8 pm.

He will have \(64(.5)^\frac{18}{5.7}≈7.2\); \(7.2\)mg left at 2 am.

 

b.  Equation \(A = 64(.5)^\frac{t}{5.7}\)

 

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