d=20.4 km correct to 1 decimal place
d = 20.4 \text{ km correct to 1 decimal place}
d=20.4 km correct to 1 decimal placet=1.54 h correct to 3 significant figures
t = 1.54 \text{ h correct to 3 significant figures}
t=1.54 h correct to 3 significant figuresv=dt
v = \frac{d}{t}
v=td
vvv = average speed, ddd = distance, ttt = time
By considering bounds, work out the value of v v\,v to a suitable degree of accuracy.
Give a reason for your answer.
