Bromine- 88 is radioactive and has a half life of 16.3 seconds. What percentage of a sample would be left after 15.2 seconds?
Round your answer to 2 significant digits.

Step 1 ：We are given that Bromine-88 is radioactive and has a half life of 16.3 seconds. We are asked to find what percentage of a sample would be left after 15.2 seconds.

Step 2 ：The half-life of a radioactive substance is the time it takes for half of the substance to decay. We can use the formula for exponential decay to solve this problem. The formula is: \(N = N0 \times (1/2)^{t/T}\)

Step 3 ：In this formula: \n- \(N\) is the final amount of the substance, \n- \(N0\) is the initial amount of the substance, \n- \(t\) is the time that has passed, \n- \(T\) is the half-life of the substance.

Step 4 ：In this case, we want to find \(N/N0\), the percentage of the substance that remains. We know that \(t = 15.2\) seconds and \(T = 16.3\) seconds. We can substitute these values into the formula to find the answer.

Step 5 ：Substituting the given values into the formula, we get: \(N/N0 = (1/2)^{15.2/16.3}\)

Step 6 ：Calculating the above expression, we get \(N/N0 = 0.5239440490606813\)

Step 7 ：Converting this to percentage, we get \(52.394404906068125\%\)

Step 8 ：Rounding this to two significant digits, we get \(52.39\%\)

Step 9 ：Final Answer: The percentage of the sample that would be left after 15.2 seconds is approximately \(\boxed{52.39\%}\)