Question 873540
 A 20-gram sample of uranium is decaying at a constant rate.
 After 5 days there are 19.6 grams of the uranium remaining.
:
We can find the half-life of the substance with this information using the formula
Ao*2^(-t/h) = A, where:
Ao = initial amt
t = time
h = half-life of substance
A = resulting amt after t time
:
20*2(-5/h) = 19.6
2^(-5/h) = {{{19.6/20}}}
2^(-5/h) = .98
{{{-5/h}}} = {{{ln(.98)/ln(2)}}}
{{{-5/h}}} = -.029146
h = {{{(-5)/(-.029146)}}}
h = 171.55 days is the half life of the substance
:
 After 10 days there are 19.2 grams remaining.
 How much of the sample remains after 30 days?
t = 30
A = 20*2(-30/171.55)
A = 17.7 grams remain after 30 days