Answer :
From the information given in the problem we can create a linear model like this: The [tex]x-axis[/tex] will represent the months and the [tex]y-axis[/tex] the subscribers. From the problem we know that after 5 months, the number of subscribers to a newspaper was 5,730, so [tex] x_{1} =7[/tex] and [tex] y_{1} =5730[/tex]. We also know that after 7 months, the number of subscribers to the newspaper was 6,022, so [tex] x_{2} =7[/tex] and [tex] y_{2} =6022[/tex].
Remember that the slope of a line is [tex]m= \frac{y_{2}-y _{1} }{ x_{2} - x_{1} } [/tex]. since we already found those values, lets replace them to find our slope:
[tex]m= \frac{6022-5730}{7-5} [/tex]
[tex]m= \frac{292}{2} [/tex]
[tex]m=146[/tex]
To create our linear model we are going to use the point-slope equation [tex]y- y _{1} =m(x- x_{1} )[/tex]. Notice that we have all the values, so the only thing left is plug them in our equation:
[tex]y-6022=146(x-5)[/tex]
[tex]y-5730=146x-730[/tex]
[tex]y=146x-730+5730[/tex]
[tex]y=146x+5000[/tex]
We can conclude that the equation which models the number of newspaper subscribers [tex]y[/tex] after [tex]x[/tex] months is [tex]y=146x+5000[/tex]
Remember that the slope of a line is [tex]m= \frac{y_{2}-y _{1} }{ x_{2} - x_{1} } [/tex]. since we already found those values, lets replace them to find our slope:
[tex]m= \frac{6022-5730}{7-5} [/tex]
[tex]m= \frac{292}{2} [/tex]
[tex]m=146[/tex]
To create our linear model we are going to use the point-slope equation [tex]y- y _{1} =m(x- x_{1} )[/tex]. Notice that we have all the values, so the only thing left is plug them in our equation:
[tex]y-6022=146(x-5)[/tex]
[tex]y-5730=146x-730[/tex]
[tex]y=146x-730+5730[/tex]
[tex]y=146x+5000[/tex]
We can conclude that the equation which models the number of newspaper subscribers [tex]y[/tex] after [tex]x[/tex] months is [tex]y=146x+5000[/tex]