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# The Linear Model in Focus

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I do business as a part-time activity and the sales I make can be represented through a linear equation. I can use a linear equation to know the profit I make for each item; moreover, I can estimate the amount of profit for different number of items sold. Among the stock I sell are bulbs. In a day, I usually have 30 bulbs to sell. If I sell 4 bulbs, I get a profit of \$60. If I sell 12 bulbs, I make a profit of \$100. The relationship between the number of bulbs and the profit made forms a linear equation. The independent variable is the number of bulbs sold while the dependent variable is the profit made. The amount of profit made depends on the number of bulbs sold.

The two data points for this relationship are (4, 60) and (12,100) and are used to calculate the slope

Slope: M= y2-y1/x2-x1

M= (100-60) / (12-4)

M= 40/8=5

The slope represents the cost of each bulb; one bulb costs \$5

In modeling the business situation, the equation is:

y = mx + b; y is the profit and x is the number of bulbs

60= 5(4) + b

60= 20 + b

b= 40

The y-intercept is (0, 40); this is the profit gained before a bulb is sold.

The business is thus modeled by y=5x + 40

From the above equation, I can estimate the profit I can make when I have sold different numbers of bulbs. For example:

6bulbs: 5(6) + 40 =\$70

10bulbs: 5(10) + 40 =\$90

13bulbs: 5(13) + 40 =\$105

15bulbs: 5(15) + 40 = 115

The table below shows a summary of the number bulbs and the profit gained when the bulbs are sold. I have presented this data in the linear graph below which explains my sales.

Number of bulbs

Amount of profit

4

60

6

70

10

90

12

100

13

105

15

115

September 18, 2023
Category:
Number of pages

1

Number of words

274