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

Using the formula, we can rewrite it as:

$d = \frac{1}{8} c^2$

Now, we calculate d using the lower and upper bounds of c:

For the lower bound: $d_{lower} = \frac{1}{8} (10.85)^2 = \frac{1}{8} (117.6225) \approx 14.703$

For the upper bound: $d_{upper} = \frac{1}{8} (10.95)^2 = \frac{1}{8} (119.6025) \approx 14.950$

We have:

• Lower bound for d: 14.703
• Upper bound for d: 14.950

Thus, we can state that the bounds for d are approximately 14.703 and 14.95.

Since both bounds for d are approximately calculated, we can round the final result of d. Given the bounds, it is appropriate to report d to two decimal places. Therefore, d can be expressed as:

$d \approx 14.8$

This reflects the degree of accuracy based on the bounds calculated.