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

Answered by Christopher R. Maths tutor

3226 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

A sequence increases by 5 each time and the first term is x. The sum of the first four terms is 54. Set up and solve an equation to work out the value of x.


There are 3 red beads and 1 blue bead in a jar. A bead is taken at random from the jar. what is the probability that the bead is blue?


Solve the simultaneous equations. 2x + y =10 and x + y = 4


Question: What proportion of the clock is the area covered when the time is 12:10? (Here the question should indicate the time stated and shade in the proportion of the clock to be computed.)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences