Explanation:
If the dug dosage is 600 mg and 5.3% is eliminated per hour, we have:
[tex]\begin{gathered} D(t)=600\cdot(1-\frac{5.3}{100})^t \\ D(t)=600\cdot(\frac{94.7}{100})^t \\ D(t)=600\cdot(0.947)^t \end{gathered}[/tex]Because for each hour the drug is eliminated at a rate of 5.3%, so the 94.7% of the previous amount remains.
Answer:
[tex]D(t)=600\cdot(0.947)^t[/tex]How much of the drug is left after 4 hours?
[tex]\begin{gathered} D(4)=600\cdot(0.947)^4 \\ D(4)\approx4.82.5598\text{ mg} \\ \end{gathered}[/tex]