Respuesta :
The amount of C-14 at time t is
[tex]N(t) = N_{0} e^{-Kt} [/tex]
where
N₀ = inital amount of C-14
t = years
K = 0.0001 = 10⁻⁴
When the skull was discovered, the amount of C-14 contained 43% of N₀.
Therefore
[tex]e^{-10^{-4}t} = 0.43 \\\\ -10^{-4} t = ln(0.43) \\\\ t = - \frac{ln(0.43)}{10^{-4}} =8439.7 \, years[/tex]
Answer:
The age of the skull s 8440 years (nearest integer)
[tex]N(t) = N_{0} e^{-Kt} [/tex]
where
N₀ = inital amount of C-14
t = years
K = 0.0001 = 10⁻⁴
When the skull was discovered, the amount of C-14 contained 43% of N₀.
Therefore
[tex]e^{-10^{-4}t} = 0.43 \\\\ -10^{-4} t = ln(0.43) \\\\ t = - \frac{ln(0.43)}{10^{-4}} =8439.7 \, years[/tex]
Answer:
The age of the skull s 8440 years (nearest integer)
