Answer:
878 years
Step-by-step explanation:
The amount of carbon-14 present in animal bones after t years is given by:
[tex]P(t)=P_oe^{-0.00012t}[/tex]
If the bone has lost 10% of its carbon-14.
Its Initial Amount of C-14, [tex]P_o[/tex]=100%=1
Present Amount, P(t)=(100-10)%=90%=0.9
Substituting these values in the model
[tex]0.9=1*e^{-0.00012t}\\$Taking natural logarithm of both sides$\\ln(0.9)=-0.00012t\\t=\dfrac{ln(0.9)}{-0.00012} \\t\approx 878\: years[/tex]
The bone is 878 years old.
