1. **Problem statement:** Calculate an estimate of the mean time taken by 100 students to solve a puzzle given the grouped frequency distribution: | Time (t seconds) | 0 < t \leq 20 | 20 < t \leq 30 | 30 < t \leq 60 | | Frequency | 20 | 30 | 50 | 2. **Formula for mean of grouped data:** $$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is the frequency of the $i$th group and $x_i$ is the midpoint of the $i$th group. 3. **Calculate midpoints:** - For $0 < t \leq 20$, midpoint $x_1 = \frac{0 + 20}{2} = 10$ - For $20 < t \leq 30$, midpoint $x_2 = \frac{20 + 30}{2} = 25$ - For $30 < t \leq 60$, midpoint $x_3 = \frac{30 + 60}{2} = 45$ 4. **Calculate $f_i x_i$ for each group:** - $20 \times 10 = 200$ - $30 \times 25 = 750$ - $50 \times 45 = 2250$ 5. **Sum frequencies and products:** - $\sum f_i = 20 + 30 + 50 = 100$ - $\sum f_i x_i = 200 + 750 + 2250 = 3200$ 6. **Calculate mean:** $$\text{Mean} = \frac{3200}{100} = 32$$ seconds --- 7. **Simplify (a):** Simplify $\left(\frac{8}{a^{12}}\right)^{\frac{1}{3}}$ 8. **Apply power to numerator and denominator:** $$= \frac{8^{\frac{1}{3}}}{(a^{12})^{\frac{1}{3}}}$$ 9. **Simplify powers:** - $8^{\frac{1}{3}} = 2$ because $2^3 = 8$ - $(a^{12})^{\frac{1}{3}} = a^{12 \times \frac{1}{3}} = a^4$ 10. **Final simplified form:** $$= \frac{2}{a^4}$$ --- 11. **Simplify (b):** Simplify $\frac{3x - 6y - ax + 2ay}{x^3 - 2x^2 y}$ 12. **Group like terms in numerator:** $$3x - ax - 6y + 2ay = x(3 - a) + y(-6 + 2a)$$ 13. **Factor denominator:** $$x^3 - 2x^2 y = x^2(x - 2y)$$ 14. **No common factors between numerator and denominator, so the simplified form is:** $$\frac{x(3 - a) + y(-6 + 2a)}{x^2(x - 2y)}$$ --- **Final answers:** - Estimated mean time: $32$ seconds - (a) Simplified: $\frac{2}{a^4}$ - (b) Simplified: $\frac{x(3 - a) + y(-6 + 2a)}{x^2(x - 2y)}$