Help me find a trend line for the data.
1 answer:
A linear equation of the trend line that models the data points contained in the table is y = 0.09x + 16.27.
<h3>How to find a trend line for the data?</h3>In order to determine a linear equation of the trend line (line of best fit) that models the data points contained in the table, we would have to use a scatter plot.
In this scenario, the body weight (in lbs) of the high school students would be plotted on the x-axis of the scatter plot while the backpack weight (in lbs) would be plotted on the y-axis of the scatter plot.
On the Excel worksheet, you should right click on any data point on the scatter plot, select format trend line, and then tick the box to display an equation for the trend line (line of best fit) on the scatter plot.
From the scatter plot (see attachment) which models the relationship between data points in the table, a linear equation of the trend line is given by:
y = 0.09x + 16.27
Read more on scatter plot here: brainly.com/question/28605735
#SPJ1
You might be interested in
Answer:
y =
+ 
Explanation:
The slope-intercept form of the equation has the following formula:
y = mx + c
where:
m is the slope
c is the y-intercept
The given is:
17x + 15y = 5
To put it in the slope-intercept formula, we should isolate the y as follows:
17x + 15y = 5
15y = -17x + 5
y = (-17/15) x + (5/15)
y = +
where:
m is the slope = -17/15
c is the y-intercept = 1/3
Hope this helps :)
y =
Explanation:
The slope-intercept form of the equation has the following formula:
y = mx + c
where:
m is the slope
c is the y-intercept
The given is:
17x + 15y = 5
To put it in the slope-intercept formula, we should isolate the y as follows:
17x + 15y = 5
15y = -17x + 5
y = (-17/15) x + (5/15)
y =
where:
m is the slope = -17/15
c is the y-intercept = 1/3
Hope this helps :)