The number of bacteria in a certain sample increases according to the following function, where $y_{0}$ is the initial number present, and $y$ is the number present at time $t$ (in hours).
\[
y=y_{0} e^{0.059 t}
\]
How many hours does it take for the size of the sample to double? Do not round any intermediate computations, and round your answer to the nearest tenth. hours
Answer
Final Answer: It takes approximately \(\boxed{11.7}\) hours for the size of the sample to double.
Step 1 ：We are given the function \(y=y_{0} e^{0.059 t}\) and we want to find the time \(t\) it takes for the size of the sample to double. This means we want to find \(t\) such that \(y = 2y_{0}\).
Step 2 ：We can set up the equation \(2y_{0} = y_{0} e^{0.059 t}\) and solve for \(t\).
Step 3 ：Let's assume \(y_{0} = 1\) for simplicity. Then the equation becomes \(2 = e^{0.059 t}\).
Step 4 ：Solving this equation gives \(t = 11.748257297626193\).
Step 5 ：Rounding to the nearest tenth, we get \(t = 11.7\).
Step 6 ：Final Answer: It takes approximately \(\boxed{11.7}\) hours for the size of the sample to double.
