John ran a 450m race (2sf) in a time of 62 seconds (nearest second). Calculate the difference between his maximum and minimum average speed. (3sf)

S=D/T Max= 455(upper bound)/61.5(lower bound)= 910/123 m/s Min= 445(lower bound)/62.5 (upper bound) = 178/25 m/s Max - Min= 856/3075 m/s

CR

Related Maths GCSE answers

All answers ▸

When I multiple two negative numbers together is my answer positive or negative?


How do you factorise a quadratic with a co-efficient in front of the x^2 - e.g: 3x^2 + 14x + 8


Solve simultaneous equations: 3x + y = 12 and 5x + 5y = 30


Expand and simplify (b-4)(b+5)