1. **State the problem:** Calculate the value of the expression: $$\frac{(1.96)^2 \times 0.613 \times (1 - 0.613)}{(0.1)^2 \times (1342 - 1) + (1.96)^2 \times 0.613 \times (1 - 0.613)}$$ 2. **Identify the components:** - $1.96$ is a constant (often a z-score for 95% confidence). - $0.613$ is a proportion $p$. - $1 - 0.613 = 0.387$ is $q$. - $0.1$ is the margin of error $E$. - $1342$ is the population size $N$. 3. **Calculate numerator:** $$ (1.96)^2 = 3.8416 $$ $$ 0.613 \times 0.387 = 0.237031 $$ $$ \text{Numerator} = 3.8416 \times 0.237031 = 0.910 \text{ (approx)} $$ 4. **Calculate denominator:** $$ (0.1)^2 = 0.01 $$ $$ 1342 - 1 = 1341 $$ $$ 0.01 \times 1341 = 13.41 $$ $$ \text{Denominator} = 13.41 + 0.910 = 14.32 \text{ (approx)} $$ 5. **Calculate final value:** $$ \frac{0.910}{14.32} \approx 0.0636 $$ **Final answer:** $$0.0636$$