# 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 $$A = 64(.5)^\frac{t}{5.7}$$