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

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.