A specific radioactive substance follows a continuous exponential decay model. It has a half-life of 6 hours. At the start of the experiment, $727.7 \mathrm{~g}$ is present. Españ (a) Let $t$ be the time (in hours) since the start of the experiment, and let $y$ be the amount of the substance at time $t$. Write a formula relating $y$ to $t$. Use exact expressions to fill in the missing parts of the formula. Do not use approximations. \[ y=\square e^{(\square)} \] (b) How much will be present in 15 hours? Do not round any intermediate computations, and round your answer to the nearest tenth. \begin{tabular}{|c|c|} \hline$\frac{\text { 믐 }}{}$ & $\square \ln \square$ \\ \hline$x$ & \\ \hline \end{tabular}

Step 1 ：The formula for exponential decay is given by: \(y = y_0 * e^{kt}\) where \(y\) is the final amount of the substance, \(y_0\) is the initial amount of the substance, \(k\) is the decay constant, and \(t\) is the time elapsed.

Step 2 ：Given that the half-life of the substance is 6 hours, we can find the decay constant \(k\) using the formula for half-life: \(t_{1/2} = \frac{ln(2)}{k}\).

Step 3 ：Solving for \(k\), we get: \(k = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{6}\).

Step 4 ：Substituting \(y_0 = 727.7\) and \(k = \frac{ln(2)}{6}\) into the exponential decay formula, we get: \(y = 727.7 * e^{\frac{ln(2)}{6}t}\). So, the missing parts of the formula are 727.7 and \(\frac{ln(2)}{6}\).

Step 5 ：To find out how much of the substance will be present in 15 hours, we substitute \(t = 15\) into the formula: \(y = 727.7 * e^{\frac{ln(2)}{6}*15}\).

Step 6 ：Calculating this gives: \(y \approx 727.7 * e^{1.15} \approx 727.7 * 3.16 \approx 229.9\).

Step 7 ：\(\boxed{229.9}\)g of the substance will be present in 15 hours, rounded to the nearest tenth.