since the half-life is 5730 years, the amount left after t years is
a = (1/2)^(t/5730)
If 81% is lost, that leaves a = 0.19, then just solve for t.
Since 1/8 < .19 < 1/4, the answer should be between 2 and 3 half-lives.
Skeletal remains had lost 81% of the C-14 they originally contained. Determine the approximate age of the bones. (Assume the half life of carbon-14 is 5730 years. Round your answer to the nearest whole number.)
yr
1 answer