SOLUTION: Question: A bone was found to contain 40% of the Carbon-14 that it contained when it was part ofa living animal. If the decay of Carbon-14 is exponential with an annual rate of dec
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-> SOLUTION: Question: A bone was found to contain 40% of the Carbon-14 that it contained when it was part ofa living animal. If the decay of Carbon-14 is exponential with an annual rate of dec
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Question 1164123: Question: A bone was found to contain 40% of the Carbon-14 that it contained when it was part ofa living animal. If the decay of Carbon-14 is exponential with an annual rate of decay of 0.0124%, how long ago did the animal die?
How far I got:
The amount lost was 60% or 0.6
Rate is -0.0124% or -0.000124
Equation created: 0.6 = 1e^-0.000124t | t is the number of years
I get an answer: -14694.5065 meaning the animal died 14695 year ago
Apparantly I am wrong by one digit, the answer being 1471 years.
Where did I insert the extra 0?
Thank you
Does NOT yield the amount lost, it yields the amount remaining, therefore
Note that 1471 years is about 25% of a half-life of 5730 years, after 1471 years you would have about 84% remaining. 40% remaining after 7600 years makes sense because 40% is between 50% and 25% which occur at 5730 and 11460 years respectively.
John
My calculator said it, I believe it, that settles it
