Glencoe Math: Course 3, Volume 2
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Glencoe Math: Course 3, Volume 2 View details
Chapter Review
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Exercise 1 Page 729

Practice makes perfect

We are given a table that shows the number of customers that visit a store over a number of weeks.

Week Customers
1 542
2 601
3 589
4 610
5 648
6 670
7 631
8 620
9 723
10 754
11 885
12 910

We are asked to construct a scatter plot. We will do this by plotting each point from the table on the given coordinate plane. For example, let's sketch the second point from the table ( 2, 601).

Let's now graph the other points from the table.

We have created a scatter plot!

We are asked to interpret the scatter plot that we made in Part A.

Notice that the data points appear to lie close to an invisible line.

This suggests that the relationship is linear. We can see that as the number of weeks increases, the number of customers that visit the store also increases. This means that the association is positive!

Now that we know the scatter plot shows a linear and positive association, we can make a conjecture about the number of customers there will be in week 20. Let's try to follow the pattern until the x-value is 20.
We followed a pattern and found a point that could reasonably correspond to week 20. Let's try to find the y-value of this point.

The y-value of this point is about 1100. This means that we can make a conjecture that after 20 weeks, there will be approximately 1100 customers in the store. Notice that this is only our expectation based on the given data — the real value could be different.

We are asked to draw a line of best fit for the scatter plot that we made in Part A.

A line best of fit is a straight line that is close to most of the data points. A good line of best fit also has a similar number of points above and below the line. Let's try to draw one.

The points are close to the line and there are a similar number of points above and below it, so this is a reasonable line of best fit. Note that this is just one example of a line of best fit — we could sketch another line that is close to the data points and it would also be a line of best fit. Let's write an equation in slope-intercept form for our line! y=mx+b In this equation, m is the slope and b is the y-intercept. Let's find the slope of our line. We will choose any two points on the line to help us with this.

The first point has the coordinates x_1= 3 and y_1= 600 and the second point has the coordinates x_2= 10 and y_2= 800. Let's substitute these values into the slope formula.
m=y_2- y_1/x_2- x_1
m=800- 600/10- 3
m=200/7
m=28.571428...
m≈ 29
We found that the slope m is equal to about 29. This means that the number of customers in the store increases by 29 every week. y=mx+b ⇔ y= 29x+b The y-intercept b is the y-value when x=0. Let's find it on the graph.

We can see that the y-intercept is about 500. This means that at the beginning of the observation, approximately 500 customers visited the store. Finally, let's write the complete equation in slope-intercept form for our line of fit. y=29x+b ⇔ y=29x+ 500