The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $505 to drive 460 mi and in June it cost her $565 to drive 700 mi.(a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.C(d) = (b) Use part (a) to predict the cost of driving 1100 miles per month.$ (c) Draw the graph of the linear function.
Question
The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her 565 to drive 700 mi.(a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.C(d) = (b) Use part (a) to predict the cost of driving 1100 miles per month.$ (c) Draw the graph of the linear function.
Solution
(a) To express the monthly cost C as a function of the distance driven d, we first need to find the rate of change (slope) of this linear relationship. The slope is given by the change in cost divided by the change in distance.
From May to June, the cost increased by 505 = 60 / 240 miles = $0.25 per mile.
Therefore, the cost function C(d) is of the form C(d) = md + b, where m is the slope and b is the y-intercept. To find b, we can substitute one of the given points into the equation. Using the point from May (d = 460 miles, C = $505), we get:
505 = 0.25*460 + b 505 = 115 + b b = 505 - 115 = $390
So the cost function is C(d) = 0.25d + 390.
(b) To predict the cost of driving 1100 miles per month, we substitute d = 1100 into the cost function:
C(1100) = 0.25*1100 + 390 = $665
(c) To draw the graph of the linear function, plot the y-intercept (0, 390) and use the slope to find another point. For example, if d = 460 miles, C = 0.25*460 + 390 = $505. Draw a line through these points to represent the function. The y-axis represents the cost and the x-axis represents the distance.
Similar Questions
The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $505 to drive 460 mi and in June it cost her $565 to drive 700 mi.(a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.C(d) = (b) Use part (a) to predict the cost of driving 1100 miles per month.$ (c) Draw the graph of the linear function. What does the slope represent?It represents the fixed cost (amount she pays even if she does not drive).It represents the cost (in dollars) of driving. It represents the cost (in dollars) per mile.It represents the distance (in miles) traveled.(d) What does the y-intercept represent?It represents the fixed cost (amount she pays even if she does not drive).It represents the distance (in miles) traveled. It represents the cost (in dollars) of driving.It represents the cost (in dollars) per mile.(e) Why does a linear function give a suitable model in this situation?A linear function is suitable because the monthly cost increases as the number of miles driven decreases.A linear function is suitable because the monthly cost increases even if the miles driven is constant. A linear function is suitable because the monthly cost is fixed despite the fact that the miles driven may vary.A linear function is suitable because the monthly cost increases as the number of miles driven increases.
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