1. Let:

   • $X$ = number of blue balls drawn
   • $Y$ = number of yellow balls drawn

   Since exactly 3 balls are drawn, $X+Y=3.$

2. We need the means $\bar X$ and $\bar Y$, i.e. $E[X]$ and $E[Y]$.

3. For a random draw without replacement, the expected number of blue balls is: $E[X]=3\cdot \frac{5}{9}=\frac{15}{9}=\frac{5}{3}.$

   Similarly, the expected number of yellow balls is: $E[Y]=3\cdot \frac{4}{9}=\frac{12}{9}=\frac{4}{3}.$

4. Now compute: $7\bar X+4\bar Y=7\cdot \frac{5}{3}+4\cdot \frac{4}{3}$ $=\frac{35}{3}+\frac{16}{3}=\frac{51}{3}=17.$

5. Therefore, the required integer is: $\boxed{17}$

Comparison with stored correct answer:

• Derived answer = $17$
• Stored correct answer = $17$
• They match.