Asked by Jackie
assume the substance has a half-life of 11 years and the initial amount is 126 grams.How long will it be until only 15 % remains?
Answers
Answered by
Damon
1/2 = e^k(11)
ln .5 = 11 k
11 k = -.5978
k = -.05435
so
.15 = e^-.05435 t
-1.897 = - .05435 t
t = 34.9 years
ln .5 = 11 k
11 k = -.5978
k = -.05435
so
.15 = e^-.05435 t
-1.897 = - .05435 t
t = 34.9 years
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