anonymous
  • anonymous
Suppose you find a piece of human bone that contains 48% of the amount of radioactive carbon-14 normally found in the bone of a living person. How old is the bone? (Half-life of carbon-14 is 5730 years)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dumbcow
  • dumbcow
use decay function \[\large P_n = P_0 e^{-kt}\] given the half life, you can find "k" \[\large 0.5 = e^{-5730k}\] \[k = \frac{\ln 0.5}{-5730} \] now solve for "t" \[\large 0.48 = e^{\frac{\ln 0.5}{5730} t}\] \[t = \frac{5730 \ln 0.48}{\ln 0.5}\]
anonymous
  • anonymous
you can also do this without any \(e\)
anonymous
  • anonymous
\[\left(\frac{1}{2}\right)^{\frac{t}{5730}}=.48\] solve for \(t\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
change of base \[\frac{t}{5730}=\frac{\ln(.48)}{\ln(.5)}\]\[t=\frac{5730\ln(.48)}{\ln(.5)}\]
anonymous
  • anonymous
thanks to both of you soo much!! i actually understand it!

Looking for something else?

Not the answer you are looking for? Search for more explanations.