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.
$$y=29.9e^{\left(\frac{\ln 0.5}{18}\right)t}$$
(b) How much will be present in 11 minutes?
Do not round any intermediate computations, and round your answer to the nearest tenth.
$◻\mathrm{g}$
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 ：$\boxed{{\displaystyle y=20.3}}$g. So, approximately 20.3g of the substance will be present after 11 minutes.