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Question
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. choose the correct scatter diagram below.
(b) what type of relation appears to exist between the weight of a candy bar and the number of calories?
- linear
- nonlinear
(c) find the equation of the line that passes through (46,312) and (64,348) in slope - intercept form.
y=
(type an expression using x as the variable. use integers or decimals for any numbers in the expression.)
(d) graph the line on the scatter diagram from part (a). choose the correct graph below.
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(a) To draw the scatter - diagram, we plot the points with weight (x) on the x - axis and calories (y) on the y - axis. Looking at the data points:
| Candy Bar | Weight (x) | Calories (y) |
|---|---|---|
| Bar B | 55.5 | 339 |
| Bar C | 46 | 312 |
| Bar D | 51.5 | 316 |
| Bar E | 57.5 | 341 |
| Bar F | 47.4 | 318.8 |
| Bar G | 62.5 | 349 |
| Bar H | 59.1 | 350.2 |
| Bar I | 64 | 348 |
| Bar J | 60.1 | 327.2 |
We match these points to the scatter - diagrams given. By plotting these points, we can see that the correct scatter - diagram is the one that represents these data pairs accurately. Without actually plotting on graph paper, we can visually inspect the given scatter - diagrams and match the general pattern of points.
(b) By observing the scatter - diagram, we can see that the points seem to follow a linear pattern. As the weight of the candy bar (x) increases, the number of calories (y) also seems to increase in a somewhat straight - line fashion. So the type of relation is Linear.
(c) The slope - intercept form of a line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points \((x_1,y_1)=(46,312)\) and \((x_2,y_2)=(64,348)\)
Step1: Calculate the slope
Step2: Use the point - slope form \(y - y_1=m(x - x_1)\) to find the y - intercept
Using the point \((46,312)\) and \(m = 2\)
(d) To graph the line \(y = 2x+220\) on the scatter - diagram:
- When \(x = 40\), \(y=2\times40 + 220=80 + 220=300\)
- When \(x = 70\), \(y=2\times70+220=140 + 220=360\)
We plot the points \((40,300)\) and \((70,360)\) and draw a straight line passing through them on the scatter - diagram.