20 d = \frac{1}{8} c^2 \epsilon = 10.9 \text{ correct to 3 significant figures} - Edexcel - GCSE Maths - Question 21 - 2019 - Paper 2

Using the equation $d = \frac{1}{8} c^2$, we need to calculate d for both bounds of \epsilon.

1. For the lower bound:
   $d_{LB} = \frac{1}{8} \times 10.85^2$
   $= \frac{1}{8} \times 117.6225 \approx 14.703\text{ (rounded to 3 decimal places)}$

2. For the upper bound:
   $d_{UB} = \frac{1}{8} \times 10.95^2$
   $= \frac{1}{8} \times 120.9025 \approx 15.113\text{ (rounded to 3 decimal places)}$

Thus, we have the following bounds for d:

• Lower Bound: $d_{LB} \approx 14.703$
• Upper Bound: $d_{UB} \approx 15.113$
  This gives us a range of \text{ d between 14.703 and 15.113}.

Considering the bounds calculated, the suitable degree of accuracy for d should also adhere to the significant figures in \epsilon. Since both bounds are within a range that can be accurately expressed as \text{15.1 (to 3 significant figures)}, it is appropriate to report d as 15.1.