Question 161643: Can someone help please.
Assume that the sales of a certain appliance dealer are approximated by a linear function. Suppose that sales were $6500 in 1982 and $64,500 in 1987. Let x = 0 represent 1982. Find the equation giving yearly sales S(x).
a. S(x) = 11,600x + 6,500
b. S(x) = 58,000x + 64,500
c. S(x) = 58,000x + 6,500
d. S(x) = 11,600x + 64,500
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! let the years = x and the sales = y
make x = 0 for 1982.
then x = 5 for 1987 (1987 - 1982 = 5)
let y = 6500 for 1982, and y = 64500 for 1987.
you have two sets of data points.
they are in the form of (x,y) where x is the x coordinate on the graph and y is the y coordinate on the graph of the equation we're going to create.
(0,6500)
(5,64500)
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line you want to create is given by equation y = m*x + b where m is the slope and b is the y intercept.
slope is found by taking the change in y and dividing it by the change in x.
this is given by the formula
m = (y-y1) / (x-x1)
let y = 64500 and y1 = 6500
let x = 5 and x1 = 0
formula becomes
m = (64500 - 6500) / (5 - 0) which becomes 58000 / 5 = 11600.
m = 11600.
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next you want to find b.
the equation is y = 11600 * x + b
substitute one of the known points on the line and solve for b.
use 5,64500.
equation becomes
64500 = 11600 * (5) + b
64500 = 58000 + b
b = 64500 - 58000
b = 6500
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equation now becomes
y = 11600 * x + 6500
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looks like your answer is a.
to prove, substitute one of the known values in the equation and solve.
use (0,6500) first
6500 = 11600 * 0 + 6500 = 6500
first one is good.
use (5,64500) next.
64500 = 11600 * 5 + 6500 = 58000 + 6500 = 64500
second one is good.
answer is a: y = 11600 * x + 6500
graph of the equation shown below.
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