91Ó°ÊÓ

It is given that:

$\stackrel{Áåœ}{y}=11.0358−0.01466x$

The figure is

The formula used:$Residual=y−\stackrel{Áåœ}{y}$

Let's start by determining the regression line's anticipated weight, which can be done by evaluating the regression line's equation, which is:

$$\begin{gathered} \stackrel{Áåœ}{y}=11.0358−0.01466x \\ =11.0358−0.01466\left(20\right) \\ =10.7426 \end{gathered}$$

In the residual plot, we can see that the residual corresponding to $20$kilometers per hour is around $2.5$

Thus, the residual will be:

$Residual=2.5$

The residual, as we know, is the actual value reduced by the anticipated value, so

$$\begin{gathered} Residual=y−\stackrel{Áåœ}{y} \\ ⇒y=Residual+\stackrel{Áåœ}{y} \\ =2.5+10.7426 \\ =13.2426 \end{gathered}$$

As a result, the fuel consumption units are in litres per $100$ kilometres driven, and the actual fuel consumption at $20$ kilometres per hour is roughly $13.2426$ litres per $100$ kilometres driven.