Question 476690
the half life of carbon -14 is 5730
how much of 10 mg sample will remain after 4500 years
:
The half-life decay formula: 
A = Ao*2^(-t/h)
Where
A = resulting amt after t
Ao = initial amt (t=0)
t = time of decay
h = half-life of substance
:
A = 10*2^(-4500/5730)
Find 2^(-4500/5730)
A = 10*.58021
A = 5.8021 mg after 4500 yrs