Suppose that a company introduces a new computer game in a city using television advertisements. Surveys show that $\mathrm{P} \%$ of the target audience buy the game after $x$ ads are broadcast, satisfying the equation below. Complete parts (a) through (d). \[ P(x)=\frac{100}{1+51 e^{-0.1 x}} \] a) What percentage buy the game without seeing a TV ad $(\mathrm{x}=0)$ ? $\%$ (Type an integer or a decimal rounded to the nearest tenth as needed.)

Step 1 ：The question is asking for the percentage of the target audience that buys the game without seeing a TV ad. This corresponds to the value of P when x=0. We can substitute x=0 into the equation to find this value.

Step 2 ：Substitute x=0 into the equation: \(P(x)=\frac{100}{1+51 e^{-0.1 x}}\)

Step 3 ：After substituting, we get: \(P = 1.9230769230769231\)

Step 4 ：Final Answer: The percentage of the target audience that buys the game without seeing a TV ad is approximately \(\boxed{1.9\%}\).