The half life of phosphorus -32 is 14.3 days. What percentage of an original samples radioactivity remains after 57.2 days?

Answer 1

k = 0.693/t_{1/2/sub>
ln(No/N) = kt
No = 100 (for convenience)
N = ?
t = 57.5 days
k from above.
Then (N/No)*100 = %remaining.}

Answer 2

Poooty butt

Answer 3

potty butt