Mr. Wang took a taxi to the Capital International Airport. The initial fare is 10 yuan (for the first $3 \mathrm{~km}$), and for every additional $0.6 \mathrm{~km}$ beyond $3 \mathrm{~km}$, the fare increases by 1 yuan. When Mr. Wang arrived at the airport, the meter showed a fare of 34 yuan. Let the distance Mr. Wang traveled be $s \mathrm{~km}$. Then the range of values for $s$ is $\qquad$
19.17.4 $\leqslant s<18$

According to the problem, we can get the equation $10+\left[\frac{s-3}{0.6}\right] \times 1=34$ (where $[x]$ represents the greatest integer not exceeding $x$). Solving this, we get
$$
24 \leqslant \frac{s-3}{0.6}<25 \text {. }
$$

Therefore, $17.4 \leqslant s<18$.