A specific radioactive substance follows a continuous exponential decay model, It has a half-life of 18 minutes. At the start of the experiment, $29.9 \mathrm{~g}$ is present. (a) Let $t$ be the time (in minutes) 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. \[ y=29.9 e^{\left(\frac{\ln 0.5}{18}\right) t} \] (b) How much will be present in 11 minutes? Do not round any intermediate computations, and round your answer to the nearest tenth. $\square \mathrm{g}$ Explanation Check

Step 1 ：The formula relating y to t is given by: \(y = 29.9 \times e^{\left(\frac{\ln 0.5}{18}\right) \times t}\)

Step 2 ：This formula is derived from the exponential decay model, where the amount of substance y at time t is equal to the initial amount (29.9g) times e raised to the power of the decay rate times time. The decay rate is given by \(\frac{\ln(0.5)}{18}\), which is the natural logarithm of the half-life divided by the half-life in minutes.

Step 3 ：To find out how much will be present in 11 minutes, we substitute t = 11 into the formula: \(y = 29.9 \times e^{\left(\frac{\ln 0.5}{18}\right) \times 11}\)

Step 4 ：Solving the above equation gives: \(y = 29.9 \times e^{-0.3857}\)

Step 5 ：Further simplifying gives: \(y = 29.9 \times 0.6801\)

Step 6 ：\(\boxed{y = 20.3}\)g. So, approximately 20.3g of the substance will be present after 11 minutes.