Question

The function C(t)=100*e^-0.000121t describes how to find the percentage of carbon-14, C(t), remaining in an object after t years. According to this information, what was the percentage of carbon-14 remaining in an object after 55 years? (Round your answer to the nearest hundredth.)

Answer 1

Answer:

99.34 years

Explanation:

Given function:

$\sf C(t)=100*e^{-0.000121t}$

Insert t = 55:

$\sf C(55)=100*e^{-0.000121(55)}$

Simplify following:

$\sf C(55)=99.3367$

Round to nearest hundredth:

$\sf C(55)=99.34$