Question 1005726
A = Ao * (1/2)^t/h, where Ao is initial amount, h is decay rate in hours, t is time
.28*Ao = Ao * (1/2)^(t/h)
.28 = (1/2)^(t/0.008333333)
note h = 30/3600 = 0.008333333
use definition of logarithm
t/0.008333333 = log base (1/2) (0.28)
t =  1.8365 * 0.008333333 * 3600
t =  55.09 seconds