Respuesta :
Answer:
- For 2011 the number of employees will be 17.33 millions.
- The linear equation that best models the number of employees (in Millions) is [tex]y(x) = 0.49 * x + 11.94 [/tex]
Step-by-step explanation:
If we wish to model the data as a straight line, we need to use the straight line formula:
[tex]y(x) = m * x + b[/tex]
where x is the years that have passed since the year 2000, m is the slope of the line and b the value of y when x=0, and y the numer (in millions) of employees.
For x=5 we know that y(5) = 14.4. So, we have:
[tex]y(5) = m * 5 + b = 14.4 [/tex]
And for x=14 we know that y(14)= 18.8
[tex]y(14) = m * 14 + b = 18.8 [/tex]
Subtracting the first equation from the second one:
[tex]y(14) - y(5) = m * 14 + b - m * 5 - b = 18.8 - 14.4 [/tex]
[tex] m * (14 - 5 ) + b - b = 4.4[/tex]
[tex] m * 9 = 4.4[/tex]
[tex] m = 4.4 / 9[/tex]
[tex] m = 0.49 [/tex]
Putting this in the second equation
[tex]y(14) = 0.49 * 14 + b = 18.8 [/tex]
[tex] 6.86 + b = 18.8 [/tex]
[tex] b = 18.8 - 6.86 [/tex]
[tex] b = 11.94 [/tex]
So, our equation will be:
[tex]y(x) = 0.49 * x + 11.94 [/tex]
For 2011 the number of employees will be
[tex]y(11) = 0.49 * 11 + 11.94 =17.33[/tex]
For 2011 the number of employees will be 17.33 millions.
The linear equation that best models the number of employees (in Millions) is
Step-by-step explanation:
If we wish to model the data as a straight line, we need to use the straight line formula:
where x is the years that have passed since the year 2000, m is the slope of the line and b the value of y when x=0, and y the numer (in millions) of employees.
For x=5 we know that y(5) = 14.4. So, we have:
And for x=14 we know that y(14)= 18.8
Subtracting the first equation from the second one:
Putting this in the second equation
So, our equation will be:
For 2011 the number of employees will be