Yefim S. answered • 02/02/21

Math Tutor with Experience

A(t) = 0.26A_{0}; so, 0.26A_{0}e^{-0.00012t}; ln(0.26) = - 0.00012; t = ln(0.26)/(-0.00012) = 11225.6 years

Jessica R.

asked • 02/02/21The amount of carbon-14 present in animal bones after t years is given by A(t)=A_{0}e^{-0.00012t}. A sample of fossil had 26% of the carbon 14 of a contemporary living sample. Estimate the age of the sample. Round to the nearest year as needed.

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Yefim S. answered • 02/02/21

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Math Tutor with Experience

_{0}; so, 0.26A_{0}e^{-0.00012t}; ln(0.26) = - 0.00012; t = ln(0.26)/(-0.00012) = 11225.6 years

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