Substance A decomposes at a rate proportional to the amount of A present.
a) Write an equation that gives the amount $A$ left of an initial amount $A_{0}$ after time $t$.
b) It is found that $14 \mathrm{lb}$ of $A$ will reduce to $7 \mathrm{lb}$ in $3.8 \mathrm{hr}$. After how long will there be only $1 \mathrm{lb}$ left?
a) Choose the equation that gives $A$ in terms of $A_{0}, t$, and $k$, where $k> 0$.
A. $A(t)=A_{0} e^{k t}$
B. $A(t)=A_{0}^{k t}$
C. $A(t)=A_{0}^{-k t}$
D. $A(t)=A_{0} e^{-k t}$
b) There will be $1 \mathrm{lb}$ left after $\square \mathrm{hr}$.
(Do not round until the final answer. Then round to the nearest whole number as needed.)

Step 1 ：The problem describes a situation of exponential decay, which is typically modeled by the equation \(A(t) = A_0 e^{-kt}\), where \(A(t)\) is the amount of substance A at time \(t\), \(A_0\) is the initial amount of substance A, and \(k\) is the decay constant. So, the answer to the first part of the question is D. \(A(t)=A_{0} e^{-k t}\).

Step 2 ：For the second part of the question, we need to find the value of \(k\) first using the given information that \(14 \mathrm{lb}\) of \(A\) will reduce to \(7 \mathrm{lb}\) in \(3.8 \mathrm{hr}\).

Step 3 ：Let's denote the initial amount of substance A as \(A_0 = 14\), the amount of substance A after \(3.8\) hours as \(A = 7\), and the time as \(t = 3.8\).

Step 4 ：Substitute these values into the equation \(A = A_0 e^{-kt}\), we get \(7 = 14 e^{-3.8k}\). Solve this equation for \(k\), we get \(k = 0.182407152778933\).

Step 5 ：Then we can use this value of \(k\) to find the time it will take for the amount of substance A to reduce to \(1 \mathrm{lb}\). Substitute \(A = 1\), \(A_0 = 14\), and \(k = 0.182407152778933\) into the equation \(A = A_0 e^{-kt}\), we get \(1 = 14 e^{-0.182407152778933t}\). Solve this equation for \(t\), we get \(t = 14.467948703818893\).

Step 6 ：Round \(t\) to the nearest whole number, we get \(t = 14\).

Step 7 ：Final Answer: There will be \(1 \mathrm{lb}\) left after \(\boxed{14}\) hours.