Question

PLEASE HELP!! The amount of​ carbon-14 present in animal bones after t years is given by ​P(t)equalsUpper P 0 e Superscript negative 0.00012 t. A bone has lost 10​% of its​ carbon-14. How old is the​ bone?

Answer 1

878 years

Explanation:

The amount of carbon-14 present in animal bones after t years is given by:

`P(t)=P_oe^(-0.00012t)`

If the bone has lost 10% of its carbon-14.

Its Initial Amount of C-14,
`P_o`=100%=1

Present Amount, P(t)=(100-10)%=90%=0.9

Substituting these values in the model

`0.9=1*e^(-0.00012t)\\$Taking natural logarithm of both sides$\\ln(0.9)=-0.00012t\\t=(ln(0.9))/(-0.00012) \\t\approx 878\: years`

The bone is 878 years old.