Use the half-life decay model as follows:
"A = A _{0}(\\frac{1}{2})\\\\^\\frac{-t}{h} \\\\= A _{0}2^\\frac{-t}{h}"
Using "A _{0} = 100" and "h = 46.5", we obtain:
"100(2^\\frac{-t}{46.5})"
After 24 hours:
"100(2\\\\^\\frac{-24}{46.5})\\\\= 69.9245 \\ milligrams"
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