Answer
$1.97\times 10^4y$
Work Step by Step
We can find the age of the basket as follows:
$\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{5730 y}=1.210\times 10^{-4}y^{-1}$
We know that
$0.0925N_{\circ}=N_{\circ}e^{-\pi t}$
$\implies 0.0925=e^{-\pi t}$
Taking the natural log of both sides and solving for t, we obtain:
$t=\frac{\ln 0.0925}{\pi}$
$t=-\frac{\ln 10^4}{1.210\times 10^{-4}y^{-1}}=1.97\times 10^4y$
