91Ó°ÊÓ

The given data is

Use the given data to draw the scatter plot diagram.

As, we can see that the relation between weight and calories is looking linear. In the obtained diagram, the slants downward from left to right.

Take two points from the given table:

$\left(39.52,210\right),\mathrm{and}\left(66.45,280\right)$

Now, the equation of the line passing through two points is,

$\begin{array}{rcl}\left(y-210\right)& =& \frac{280-210}{66.54-39.52}\left(x-39.52\right)\\ y-210& =& \frac{70}{66.54-39.52}\left(x-39.52\right)\\ y& =& 2.59\left(x-39.52\right)+210\\ & =& 2.59x+107.6432\end{array}$

As we have taken randomly two points from the table. Thus, the maybe changed while the points will change.

The graph of the line on the scatter plot is,

Substitute 62.3 for x in the linear mode calculated in part (c).

$\begin{array}{rcl}y\left(x\right)& =& 2.5993x+107.2757\\ y\left(x\right)& =& 2.5993\left(62.3\right)+107.2757\\ & & \begin{array}{l}=269.21\\ ≈269\mathrm{calories}\end{array}\end{array}$

The required amount is 269 calories.

From the linear model, the slope is 2.59, which is positive. This implies that if the candy bar is increased by 1 gm then the calories will increase by 2.59 calories.