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
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?
1 answer
1 answer