Answer:
[tex]A(t)=6\cdot2^{-\frac{t}{5730 }[/tex]
Step-by-step explanation:
The general formula for exponential decay given a half-life [tex]t_\frac{1}{2}[/tex] is
[tex]N(t)=N_{0}\cdot2^{\frac{-t}{t_{1/2} }[/tex]
where N(t) is the amount at time t, [tex]N_0[/tex] is the initial amount (at time t=0), and [tex]t_\frac{1}{2}[/tex] is the half life of the substance.
The half life of carbon-14, is approximately 5,730 years.
[tex]t_\frac{1}{2}[/tex] = 5730 years
[tex]N_0[/tex]= 6 milligrams
Therefore:
Amount of carbon-14 remaining in the bone fragment after t years is:
[tex]A(t)=6\cdot2^{-\frac{t}{5730 }[/tex]
