Answer:
An old bone contains 80% of its original carbon-14 in 1844.6479 years
Step-by-step explanation:
We know that
half life time of C-14 is 5730 years
so, h=5730
now, we can use formula
[tex]P(t)=P_0(\frac{1}{2})^{\frac{t}{h} }[/tex]
we can plug back h
and we get
[tex]P(t)=P_0(\frac{1}{2})^{\frac{t}{5730} }[/tex]
An old bone contains 80% of its original carbon-14
so,
P(t)=0.80Po
we can plug it and then we solve for t
[tex]0.80P_0=P_0(\frac{1}{2})^{\frac{t}{5730} }[/tex]
[tex]0.80=(\frac{1}{2})^{\frac{t}{5730} }[/tex]
[tex]\ln \left(0.8\right)=\ln \left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)[/tex]
[tex]t=-\frac{5730\ln \left(0.8\right)}{\ln \left(2\right)}[/tex]
[tex]t=1844.6479[/tex]
So,
An old bone contains 80% of its original carbon-14 in 1844.6479 years
Answer:
Part 1 = C. (the LAST one) /// Part 2 = 0.8 ////// Part 3 = B. (about 1,845 years)
Step-by-step explanation:
theyre all correct.
