What percent of the original sample of strontium-90 (half-life = 28.9 years) will remain after the following time periods?after 18 years?

Question

after 47.5 years?

after 1008 years?

Answer 1

k = 0.693/t_1/2
ln(No/N) = kt
Substitute into equation 1 and solve for k, the substitute into equation 2 and solve for N. No will be 18, 47.5, 1008 years.

Answer 2

.....it's actually asking for the % left after so many years

Answer 3

So N = amount left.
% = (N/No)*100 = ?