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

Answer:

878 years

Step-by-step 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

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

The bone is 878 years old.