A specific radioactive substance follows a continuous exponential decay model. It has a half-life of 23 days. At the start of the experiment, $576.3 \mathrm{~g}$ is present. (a) Let $t$ be the time (in days) 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 \[ y= \] \[ y= \] (b) How much will be present in 5 days? Do not round any intermediate computations, and round your answer to the nearest tenth.

Step 1 ：The formula for exponential decay is given by \(y = y_0 * e^{-kt}\), where \(y_0\) is the initial amount, \(k\) is the decay constant, and \(t\) is the time. In this case, we know that the half-life is 23 days, so we can use the formula for the decay constant in terms of the half-life: \(k = \ln(2) / T_{1/2}\), where \(T_{1/2}\) is the half-life. We can substitute these values into the decay formula to get the formula relating \(y\) to \(t\).

Step 2 ：Given that \(y_0 = 576.3\), \(T_{1/2} = 23\), we can calculate \(k = 0.030136833937388925\).

Step 3 ：Substituting these values into the decay formula, we get the formula relating \(y\) to \(t\) as \(y = 576.3 * e^{-0.030136833937388925t}\).

Step 4 ：To answer part (b) of the question, we can substitute \(t = 5\) into the formula to find out how much of the substance will be present in 5 days.

Step 5 ：Given that \(y_0 = 576.3\), \(T_{1/2} = 23\), \(k = 0.030136833937388925\), we can calculate \(y_5 = 495.68675692144217\).

Step 6 ：The amount of the substance that will be present in 5 days is approximately \(\boxed{495.7 \mathrm{~g}}\).