Suppose that a company introduces a new computer game in a city using television advertisements. Surveys show that P\% of the target audience buy the game after $\mathrm{x}$ ads are broadcast, satisfying the equation below. Complete parts (a) through (d). \[ P(x)=\frac{100}{1+47 e^{-0.16 x}} \] a) What percentage buy the game without seeing a TV ad $(x=0)$ ? $2.1 \%$ (Type an integer or a decimal rounded to the nearest tenth as needed.) b) What percentage buy the game after the ad is run 35 times? $\%$ (Type an integer or a decimal rounded to the nearest tenth as needed.)

Step 1 ：The problem provides us with the function \(P(x)=\frac{100}{1+47 e^{-0.16 x}}\), which represents the percentage of the target audience that buys the game after \(x\) ads are broadcast.

Step 2 ：For part a), we need to find \(P(0)\), which represents the percentage of the target audience that buys the game without seeing any ads. Substituting \(x = 0\) into the function, we get \(P(0) = 2.0833333333333335\).

Step 3 ：Rounding to the nearest tenth, we get approximately \(2.1\%\). So, the percentage of the target audience that buys the game without seeing any ads is approximately \(\boxed{2.1\%}\).

Step 4 ：For part b), we need to find \(P(35)\), which represents the percentage of the target audience that buys the game after seeing 35 ads. Substituting \(x = 35\) into the function, we get \(P(35) = 85.19341841118424\).

Step 5 ：Rounding to the nearest tenth, we get approximately \(85.2\%\). So, the percentage of the target audience that buys the game after seeing 35 ads is approximately \(\boxed{85.2\%}\).