#### Answer

about 1056 years old

#### Work Step by Step

In the model for decay, substitute A with $\displaystyle \frac{88}{100}A_{o}$
(88$\%$ of $A_{o})$
and solve for t:
$\displaystyle \frac{88}{100}A_{o}=A_{0}e^{-0.000121t}\qquad .../\times\frac{1}{A_{o}}$
$\displaystyle \frac{88}{100}=e^{-0.000121t}\qquad .../$ take ln( ) of both sides
$\ln 0.88=-0.000121t\qquad .../\div(-0.000121)$
$ t=\displaystyle \frac{\ln 0.88}{-0.000121}\approx$1056.47414471
In 1989, the skeletons were
about 1056 years old.