Respuesta :

Using exponential decay formula:

[tex]N=N_o(\frac{1}{2})^{\frac{t}{\tau}}[/tex]

Where:

[tex]\begin{gathered} N_o=3mg \\ t=222.5_{\text{ }}days \\ \tau=44.5_{\text{ }}days \end{gathered}[/tex]

Therefore:

[tex]\begin{gathered} N=3(\frac{1}{2})^{\frac{222.5}{44.5}} \\ N\approx0.09mg \end{gathered}[/tex]

Answer:

About 0.09 miligrams

Otras preguntas