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3 years ago

5

How old is a bone that has 12.5 percent of the original amount of radioactive carbon

1 answer:

pav-90 [236]3 years ago

8 0

The solution would be like this for this specific problem:

T = 12.5%

Initial amount = 1

1*2^(-t/5730) = .125

$\frac{-t}{5730} ln(2) = ln(.125)$ <span>

$\frac{-t}{5730} = -3$

$\frac{-t}{5730} = \frac{ln(.125}{ln(2)}$ </span>

<span>t = -3 * -5730
t = 17,190 yrs</span>

<span>So, <span>a bone that has 12.5 percent of the original amount of radioactive carbon is 17,190 yrs old.</span></span>

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