SOLUTION: This is a word problem for my AlgebraII class that I'm stuck on, because I don't know how to set it up. It's a growth and decay problem of the form A=Pe^rt and the problem is In t

Question 1061424: This is a word problem for my AlgebraII class that I'm stuck on, because I don't know how to set it up. It's a growth and decay problem of the form A=Pe^rt and the problem is
In the early 1960's, radioactive strontium-90 was released during atmospheric testing of nuclear weapons, and infiltrated the bones of people alive at the time. If the half life of strontium-90 is 29 years, what fraction of strontium-90 absorbed in 1960 remains in the bones in 2016.
I don't understand the way the book shows these problems, so if you could help that would be amazing.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
It's a growth and decay problem of the form A=Pe^rt and the problem is
In the early 1960's, radioactive strontium-90 was released during atmospheric testing of nuclear weapons, and infiltrated the bones of people alive at the time. If the half life of strontium-90 is 29 years, what fraction of strontium-90 absorbed in 1960 remains in the bones in 2016.
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Not sure how you book wants you to solve these
decay problems; but here's one way.
---------------
A(t) = P*e^(rt)
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Since the half-life is 29 years,
P*e^(r*29) = (1/2)P
[e^29]^r = (1/2)
r = ln(1/2)/ln(e^29) = ln(1/2)/29 = -0.0239
=======
Your problem::
A(6) = P*e^(-0.0239*6)
A(6) = P*0.8664
Ans: 87/100 of the original amount remains after 6 years
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Cheers,
Stan H.
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