Assume that the sales of a certain appliance dealer can be approximaed y sraig were $6000 in 1982 and $ 64,000 in 1987. Let x - O represent 1982 Find the equation giving yearly alesSand then use it to predict the yearly sales in 1990.

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, [tex]x=0[/tex] then at 1987 x will be 5.

Now, we have two points (0,6000) and (5,64000).

[tex]\text{Slope}=\frac{64,000-6,000}{5-0}[/tex]

[tex]\text{Slope}=\frac{58,000}{5}[/tex]

[tex]\text{Slope}=11,600[/tex]

Now, we will represent this information in slope-intercept form of equation.

[tex]y=mx+b[/tex], where,

m = Slope,

b = Initial value or y-intercept.

We have been given that at [tex]x=0[/tex], the value of y is 6,000, so it will be y-intercept.

Substitute values:

[tex]y=11,600x+6,000[/tex]

Therefore, the equation [tex]S=11,600x+6,000[/tex] represents yearly sales.

Now, we will find difference between 1990 and 1982.

[tex]1990-1982=8[/tex]

To find yearly sales in 1990, we will substitute [tex]x=8[/tex] in the equation.

[tex]S=11,600(8)+6,000[/tex]

[tex]S=92,800+6,000[/tex]

[tex]S=98,800[/tex]

Therefore, the yearly sales in 1990 would be $98,800.