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%

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