Answer:
![y=11,600x+6,000](https://tex.z-dn.net/?f=y%3D11%2C600x%2B6%2C000)
Yearly sales in 1990: $98,800.
Step-by-step explanation:
We have been given that the sales of a certain appliance dealer can be approximated by a straight line. Sales were $6000 in 1982 and $ 64,000 in 1987.
If at 1982,
then at 1987 x will be 5.
Now, we have two points (0,6000) and (5,64000).
![\text{Slope}=\frac{64,000-6,000}{5-0}](https://tex.z-dn.net/?f=%5Ctext%7BSlope%7D%3D%5Cfrac%7B64%2C000-6%2C000%7D%7B5-0%7D)
![\text{Slope}=\frac{58,000}{5}](https://tex.z-dn.net/?f=%5Ctext%7BSlope%7D%3D%5Cfrac%7B58%2C000%7D%7B5%7D)
![\text{Slope}=11,600](https://tex.z-dn.net/?f=%5Ctext%7BSlope%7D%3D11%2C600)
Now, we will represent this information in slope-intercept form of equation.
, where,
m = Slope,
b = Initial value or y-intercept.
We have been given that at
, the value of y is 6,000, so it will be y-intercept.
Substitute values:
![y=11,600x+6,000](https://tex.z-dn.net/?f=y%3D11%2C600x%2B6%2C000)
Therefore, the equation
represents yearly sales.
Now, we will find difference between 1990 and 1982.
![1990-1982=8](https://tex.z-dn.net/?f=1990-1982%3D8)
To find yearly sales in 1990, we will substitute
in the equation.
![S=11,600(8)+6,000](https://tex.z-dn.net/?f=S%3D11%2C600%288%29%2B6%2C000)
![S=92,800+6,000](https://tex.z-dn.net/?f=S%3D92%2C800%2B6%2C000)
![S=98,800](https://tex.z-dn.net/?f=S%3D98%2C800)
Therefore, the yearly sales in 1990 would be $98,800.