有無人識得做呢題數??好緊急!!

2010-10-22 10:53 am
Assume that the sales of certain appliance dealer are approximated by a linear function.
Suppose that the sales were $12,500,000 in 2004 and $23.6 million in 2010.
Let x = 0 represent 2004.

a) Find an equation giving the dealer's yearly sales.
b) Estimate sales in 2015.
c)Tthe dealer estimates that a new store will be necessary once sales exceed $40 millions.
When is this expected to occur?

Show work for each problem please.....thanks!!!!

回答 (1)

2010-10-22 4:14 pm
✔ 最佳答案
You have to be familiar with linear equation.

(a) The linear equation has the form y = mx + b
where m is the slope of the line and b is the y-intercept.

YEAR               SALES
2004(0)           $12.5 millions
2010(6)           $23.6 millions

You let x = 0 for year 2004, so x = 6 for year 2010
The x-coordinate is given by (Year – 2004), 2010-2004 = 6
Now let y represent the sales

The two points on x-y graph are (x1, y1) and (x2, y2)
They are (0, 12.5) and (6, 23.6)
It is assumed that the projected sales is approximated by a linear function.
X2
Slope of the line m = (y2 – y1)/(x2 – x1)
m = (23.6 – 12.5)/(6-0) = 11.1/6 = 1.85
b = 12.5 (y-intercept is value of y when x = 0)

The equation is y = 1.85x + 12.5


(b) year 2015 is equivalent to x = 11 (2015 -2004 = 11)
y = 1.85x + 12.5
substitute x = 11 into equation
y = 1.85(11) + 12.5 = 17.38 + 12.5 = 29.88

Sales in year 2015 is $32.85 millions


(c) When y = 40, what is x?
y = 1.85x + 12.5
40 = 1.85x + 12.5
1.85x = 40 – 12.5 = 27.5
x = 27.5/1.85 = 14.86 = 15
That is the year 2019 (2004 + 15 = 2019)

It is expected in year 2019, the sales will reach $40 millions.


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