The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $380 to drive 480 mi. and in June it cost her $460 to drive 800 mi.
a) Express the montly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.
I got C= 1/4 a + 260
b) Use part (a) to predict the cost of driving 1500 mi. per month.
I got C= 1/4(1500) + 260= 635
c) Draw the graph of the linear function and what does the slope represent?
Not sure but, I think the slope represents the relationship between the number of mi. driven and the montly cost of drving a car. The more miles driven, the higher the monthly cost.
d) What does the y-intercept represent?
I am not sure, other than it is a constant?
e) Why does a linear function give a suitable model in this situation?
Is it because the cost to drive a mile will always be constant and therefore, the rate of change will be constant?
a) Express the montly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.
I got C= 1/4 a + 260
slope = changeincost/change in miles
slope= 90/320=9/32
intercept: when miles is zero.
cost= 9/32 x+b
380=9/32*480+b
b= 135 check that.
cost= 9/32 miles + 135
b) Use part (a) to predict the cost of driving 1500 mi. per month.
I got C= 1/4(1500) + 260= 635 redo
c) Draw the graph of the linear function and what does the slope represent?
Not sure but, I think the slope represents the relationship between the number of mi. driven and the montly cost of drving a car. The more miles driven, the higher the monthly cost. almost, I answered it above
a) The monthly cost C can be expressed as a function of the distance driven d using a linear relationship.
We can find the equation of the line using the given data points (480, 380) and (800, 460). The formula for a linear relation is y = mx + b, where m is the slope and b is the y-intercept.
First, let's find the slope:
m = (y2 - y1) / (x2 - x1)
m = (460 - 380) / (800 - 480)
m = 80 / 320
m = 1/4
Now, let's substitute the values of one of the points and the calculated slope into the equation to find the y-intercept:
380 = (1/4) * 480 + b
380 = 120 + b
b = 380 - 120
b = 260
Therefore, the function that expresses the monthly cost C as a function of the distance driven d is:
C = (1/4)d + 260
b) Now let's use the equation from part (a) to predict the cost of driving 1500 mi per month:
C = (1/4)(1500) + 260
C = 375 + 260
C = 635
Therefore, the predicted cost of driving 1500 miles per month is $635.
c) The graph of the linear function C = (1/4)d + 260 would have distance driven (d) on the x-axis and the monthly cost (C) on the y-axis. The slope (1/4) represents the rate at which the cost increases per unit distance. In this case, it means that for every additional mile driven, the monthly cost increases by $0.25.
d) The y-intercept (260) represents the fixed monthly cost or the cost that Lynn incurs even if she doesn't drive any miles. It includes costs such as insurance, registration, maintenance, etc., which are incurred regardless of the distance driven.
e) A linear function gives a suitable model in this situation because it assumes a constant rate of increase in the monthly cost per unit distance driven. It is a reasonable assumption to make since the cost per mile driven is assumed to be constant. However, it's worth mentioning that in reality, there may be other factors or variables that can contribute to the cost besides the distance driven, which may warrant a more complex model.
a) To express the monthly cost C as a function of the distance driven d, we can use the formula for a linear equation: C = md + b, where m represents the slope and b represents the y-intercept.
First, we need to find the values of m and b. Let's use the information provided in May and June:
In May, Lynn drove 480 miles and the cost was $380. So we have the equation:
380 = m * 480 + b
In June, Lynn drove 800 miles and the cost was $460. We have another equation:
460 = m * 800 + b
We can use these two equations to solve for m and b. Subtracting the second equation from the first, we get:
380 - 460 = m * 480 - m * 800
-80 = -320m
m = -80 / -320
m = 1/4
Substituting this value of m back into either of the two equations, we can find the y-intercept b. Let's use the first equation:
380 = (1/4) * 480 + b
380 = 120 + b
b = 380 - 120
b = 260
Therefore, the equation expressing the monthly cost C as a function of the distance driven d is:
C = (1/4)d + 260
b) To predict the cost of driving 1500 miles per month using the linear equation, we substitute d = 1500 into the equation:
C = (1/4)(1500) + 260
C = 375 + 260
C = 635
So, the predicted cost of driving 1500 miles per month is $635.
c) The graph of the linear function C = (1/4)d + 260 would be a straight line with a positive slope of 1/4. The slope represents the rate of change, which in this case means the increase in monthly cost for each additional mile driven. Therefore, for each additional mile driven, the monthly cost increases by 1/4 dollar.
d) The y-intercept represents the value of C when d is zero. In this case, when there are zero miles driven, the y-intercept is 260. This means that even if you don't drive any miles, there will still be a fixed monthly cost of $260. It could include factors like insurance, registration fees, or other fixed costs.
e) A linear function gives a suitable model in this situation because it assumes a constant rate of change in the monthly cost as the number of miles driven increases. This assumes that there are no sudden increases in costs for additional miles, and that the relationship between distance and cost is relatively consistent. It provides a straightforward and simple approximation that can help in making predictions and understanding the general trend.