What percent of a sample of As-81 remains un-decayed after 43.2 seconds

Where Ao = starting amount, At = amount after time t, and k = rate constant

You are given that the half-life is 33 seconds

Calculate the rate constant

ln (2/1) = kx33 sec

0.693 = kx33

k = 0.0210 s^-1

Now calculate the amount left after 43.2 sec

ln (100%/A1%) = 0.0210 x 43.2

ln (100/A1) = 0.9074

100/A1 = 2.478

A1 = 40.4%