Q. 17 - Precalculus Enhanced with Graphing Utilities

Question

Candy The following data represent the weight (in grams) of various candy bars and the corresponding number of calories.

(a) Draw a scatter diagram of the data treating weight as the independent variable.

(b) What type of relationship appears to exist between the weight of a candy bar and the number of calories?

(c) Select two points and find a linear model that contains the points.

(d) Graph the line on the scatter diagram drawn in part (a).

(e) Use the linear model to predict the number of calories in a candy bar that weighs 62.3 grams.

(f) Interpret the slope of the line found in part (c).

Step-by-Step Solution

Answer

(a). The scatter diagram of the data treating weight as the independent variable is,

(b). The relation between weight and calories is linear.

(c). Equation of the line is $y=2.59x+107.6432$

(d). The graph of the line on a scatter plot is 

(e). The required amount is 269 calories.

(f) If the candy bar is increased by 1 gm then the calories will increase by 2.59 calories. 

Step 1. Given information

The given data is 

Candy Bar Weight, x Calories, y Hershey's Milk Chocolate 44.28 230 Nestle's Crunch 44.84 230 Butterfinger 61.30 270 Baby Ruth 66.45 280 Almond Joy 47.33 220 Twix (With Caramel) 58.00 280 Snickers 61.12 280 Heath                                                           39.52                                                          210

Part (a) Step 2. Make the scatter plot graph.

Use the given data to draw the scatter plot diagram.

 

Part (b) Step 1. Explain the relationships.

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.

Part (c) Step 1. Equation of line.

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.

Part (d) Step 1. Graph the linear line.

The graph of the line on the scatter plot is,

Part (e) Step 1. Find the number of calories.

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

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

The required amount is 269 calories.

Part (f) Step 1. Best fit line

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.