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The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her 380 dollars to drive 480 mi and in June it cost her 460 dollars to drive 800 mi.

(a) Express the monthly cost $ C $ as a function of the distance driven $ d $, assuming that a linear relationship gives a suitable model.

(b) Use part (a) to predict the cost of driving 1500 miles per month.

(c) Draw the graph of the linear function. What does the slope represent?

(d) What does the C-intercept represent?

(e) Why does a linear function give a suitable model in this situation?

a. $C=\frac{1}{4} d+260$

b. $\$ 635$

c. The slope is $\frac{1}{4}$

d. C-intercept $=$ fixed cost

e. See explanation in the full solution.

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Johns Hopkins University

Campbell University

University of Michigan - Ann Arbor

Boston College

alright in this problem, we have some information about the cost of driving a car, and we're going to come up with the linear relationship. We're going to let d equal the number of miles driven and see equal the cost. And we have some information that we can translate into ordered pairs. So we know that for 480 miles driven, the cost that month was $380 and for 800 miles driven. The cost that month was $460. So we can use those to ordered pairs to find the equation of the line for part of the problem. So let's start by finding the slope changing. Why over change in X for 60 minus 3 80 over 800 minus 4 80 and that gives us 80 divided by 3 20 that reduces to 1/4. And then for the Y intercept, we can use point slope form. Why minus y one equals M times X minus X one, and we can select either one of our two points. It doesn't matter which, so I'm just going to use the 1st 1 Why minus 3 80 equals 1/4 times X minus 4 80 We can simplify that equation we get Why minus 3 80 equals 1/4 X minus 1 20 and then we'll add 3 80 to both sides. And we have y equals 1/4 x plus 2 60 now putting it in terms of the variables that we were using in the problem that would be cost C equals 1/4 times miles driven D plus 2 60 All right now for part B, we want to predict the cost of driving 1500 miles. And so we're going to let d equal 1500 substitute that into the model that we just came up with. So the cost will be 1/4 times 1500 plus 2 60 and that works out to be $635. Okay, never part C. We want to graph the function, so we know from the function itself at the Y. Intercept is to 60 so we can plot that point and we know that we have the point for 80 comma 3 80 from before, so we can plot that point and we also have the 0.804 60 so we can play at that point, and then we have our line. We're not gonna go into negative distances or negative costs, so we'll just focus on in quadrant one. So what's the slope of the line? The slope was 1/4 and that represents the cost per mile. $1 per four miles. Or, you could say 1/4 of a dollar per mile. You could say 25 cents per mile. Okay, Part D. What is the sea intercept? We would normally call it the Y intercept, but because we're using see on our Y axis will call it the Sea Intercept. So that was the 0.0 to 60 and that would represent driving zero miles. If you drive zero miles, it's still going to cost $260 that month because there's fixed costs. Maybe it's the cost of the car payment. Maybe it's the cost of insurance. Things like that, okay, and part D. Finally, it says, Why does a linear function give a suitable model? Well, it makes sense that you would have a fixed cost like your car payment or your insurance and those kinds of things combined. But it also makes sense that the more miles you drive, the more it's going to cost just because the cost of gas, a certain number of dollars per gallon and then you have your um, you're mile your gas mileage, that's what it's called. Gas mileage is so many ah miles per gallon. So even though for any particular car you might get better gas mileage or worse, gas mileage, depending on if you're driving on the freeway or around town in general, we could just sort of loosely say that you're paying a certain amount of money per gallon of gas, and you're getting a certain number of miles for that amount of money.