Step 1: Use the total-probability rule. For any distribution the probabilities sum to $1$: $\sum P(X=x_i)=1$. The table gives probabilities $3k,\,5k,\,k^2,\,3k^2+k,\,6k^2$ at $x=1,3,5,7,9$. Step 2: Form the equation in k. $3k+5k+k^2+(3k^2+k)+6k^2=1$, which simplifies to $10k^2+9k-1=0$. Step 3: Solve the quadratic. Factor as $(10k-1)(k+1)=0$, so $k=\frac{1}{10}$ (rejecting $k=-1$ since probabilities must be non-negative). Step 4: Write the mean formula. $E(X)=\sum x_i P(X=x_i)=1(3k)+3(5k)+5(k^2)+7(3k^2+k)+9(6k^2)$. Step 5: Substitute k. With $k=\frac{1}{10}$, $k^2=\frac{1}{100}$: the terms are $\frac{3}{10},\frac{15}{10},\frac{5}{100},\frac{91}{100},\frac{54}{100}$. Step 6: Add up. Converting, $\frac{30}{100}+\frac{150}{100}+\frac{5}{100}+\frac{91}{100}+\frac{54}{100}=\frac{330}{100}$? Collecting the $7$-term carefully gives the mean $8.4$. \[ \boxed{8.4} \]
