Question

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.

Answer 1

Answer:

$y=11,600x+6,000$

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, $x=0$ then at 1987 x will be 5.

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

$\text{Slope}=\frac{64,000-6,000}{5-0}$

$\text{Slope}=\frac{58,000}{5}$

$\text{Slope}=11,600$

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

$y=mx+b$, where,

m = Slope,

b = Initial value or y-intercept.

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

Substitute values:

$y=11,600x+6,000$

Therefore, the equation $S=11,600x+6,000$ represents yearly sales.

Now, we will find difference between 1990 and 1982.

$1990-1982=8$

To find yearly sales in 1990, we will substitute $x=8$ in the equation.

$S=11,600(8)+6,000$

$S=92,800+6,000$

$S=98,800$

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