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
Answered by Christopher R. Maths tutor

4247 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

In a class of 28 students, the average height of the 12 boys is 1.58 metres. The average height of the class is 1.52 metres. What is the average of the girls?


How to solve a simultaneous equation?


factorise: 3x^2+13x-30


The turning point on a quadratic function


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning