Answer:
15502 years
Step-by-step explanation:
From;
0.693/t1/2 = 2.303/t log [Ao]/[A]
Where;
t1/2 = half life of C-14
t = time taken
[Ao] = count rate of living sample
[A] = count rate of the charcoal under study.
0.693/5568 = 2.303/t log (6.68/ 0.97)
1.245 * 10^-4 = 2.303/t log (6.887)
1.245 * 10^-4 = 1.92998/t
t = 1.92998/1.245 * 10^-4
t = 15502 years