Answer:
The skeleton is approximately 18970 years old.
Step-by-step explanation:
The amount of Carbon-14 remaining after t years is given by the following equation:
[tex]N(t) = N(0)(1-r)^{t}[/tex]
In which N(0) is the initial amount and r is the decay rate, as a decimal.
Carbon-14 has a half-life of about 5700 years.
This means that [tex]N(5700) = 0.5N(0)[/tex]
So
[tex]N(t) = N(0)(1-r)^{t}[/tex]
[tex]0.5N(0) = N(0)(1-r)^{5700}[/tex]
[tex](1-r)^{5700} = 0.5[/tex]
[tex]\sqrt[5700]{(1-r)^{5700}} = \sqrt[5700]{0.5}[/tex]
[tex]1-r = 0.9999[/tex]
[tex]r = 1 - 0.9999[/tex]
[tex]r = 0.0001[/tex]
So
[tex]N(t) = N(0)(1-r)^{t}[/tex]
[tex]N(t) = N(0)(0.9999)^{t}[/tex]
He finds that the fossil contains 15% of the amount of carbon-14 anticipated when compared to a living femur of the same size.
The age is t for which N(t) = 0.15N(0). So
[tex]N(t) = N(0)(0.9999)^{t}[/tex]
[tex]0.15N(0) = N(0)(0.9999)^{t}[/tex]
[tex](0.9999)^{t} = 0.15[/tex]
[tex]\log{(0.9999)^{t}} = \log{0.15}[/tex]
[tex]t\log{0.9999} = \log{0.15}[/tex]
[tex]t = \frac{\log{0.15}}{\log{0.9999}}[/tex]
[tex]t = 18970.25[/tex]
Rounding to the nearest year.
The skeleton is approximately 18970 years old.
