The data in the table shows the relationship between the time of day and the total number of calories that a teenager consumes throughout the day.
Number of Calories Consumed 525 675 1,425 1,675 1,675 2,195 2,195 2,395       Time                                            8am 10am 12am 2pm 4pm 6pm 8pm 10pm
 Write the equation of the best fit line in slope-intercept form. Include all of your calculations in your final answer. Hint: On the plot, the time is represented using a 12-hour clock. To get an accurate equation, you will want to represent the time using a 24-hour clock instead. For example, 2 pm can be represented as 12 + 2 = 14.

The data in the table shows the relationship between the time of day and the total number of calories that a teenager consumes throughout the day Number of Calo class=

Respuesta :

To create our linear equation, we are going to use the tow points. Our first point will be the first point in our data graph, so (8, 525), and our second point will be the last point in our data graph, so  (22, 2395). The [tex]x-axis[/tex] will represent the time, and the [tex]y-axis[/tex] will represent the calories.

The first thing we are going to do is find the slope of our linear equation. To do that we'll use the slope formula: [tex]m= \frac{y_{2}-y_{1} }{x_{2}-x_{1}} [/tex]. For our two points we can infer that [tex]x_{1}=8[/tex], [tex]y_{1}=525[/tex], [tex]x_{2}=22[/tex], and [tex]y_{2}=2395[/tex]. So lets replace those values in our slope formula to find [tex]m[/tex]:
[tex]m= \frac{2395-525}{22-8} [/tex]
[tex]m= \frac{1870}{14} [/tex]
[tex]m= \frac{935}{7} [/tex]

Now that we have the slope, we are going to use the point slope formula, [tex]y-y_{1}=m(x-x_{1})[/tex], to complete our equation:
[tex]y-525= \frac{935}{7} (x-8)[/tex]
[tex]y-525= \frac{935}{7} x- \frac{7480}{7} [/tex]
[tex]y= \frac{935}{7} x- \frac{7480}{7} +525[/tex]
[tex]y= \frac{935}{7} x- \frac{3805}{7} [/tex]

We can conclude that the linear equation that best fits our data is [tex]y= \frac{935}{7} x- \frac{3805}{7} [/tex].