The amount of carbon-14 present in animal bones after $t$ years is given by $P (t) = P_{0} e^{- 0.00012 t}$ . A bone has lost $36 %$ of its carbon-14. How old is the bone?
The bone is about years old.
(Round to the nearest integer as needed.)

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Final Answer: The bone is approximately $3719$ years old.

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Step 1 ：The problem is asking for the time $t$ when the amount of carbon-14 in the bone is $64 %$ of the original amount (since it has lost $36 %$ ). This can be represented by the equation $0.64 P_{0} = P_{0} e^{- 0.00012 t}$ .

Step 2 ：We can solve this equation for $t$ by setting $P_{0} = 1$ and $P = 0.64$ .

Step 3 ：Solving the equation gives $t = 3719.0591885701624$ .

Step 4 ：Rounding to the nearest integer gives $t = 3719$ .

Step 5 ：Final Answer: The bone is approximately $3719$ years old.

The amount of carbon-14 present in animal bones after t years is given by P(t)=P0e−0.00012t. A bone has lost 36% of its carbon-14. How old is the bone?The bone is about years old.(Round to the nearest integer as needed.)

Answer

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The amount of carbon-14 present in animal bones after $t$ years is given by $P (t) = P_{0} e^{- 0.00012 t}$ . A bone has lost $36 %$ of its carbon-14. How old is the bone?
The bone is about years old.
(Round to the nearest integer as needed.)