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A fossil is found to have a 14^C level of 89.0% co

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Answer #1

solution:

Half life of 14C = 5730 years

(1/2)(x/5730) = 0.89
(x/5730) log (0.5) = log (0.89)
x/5730 = log (0.89) / log (0.5)
x = 5730 log (0.89) / log (0.5)

x = 963.34 years

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