Mean of Graph:
Table:
Graph:
Equation of Line of Best Fit:
Y= 0.5x + 45.5
Y=mx + c
Y=mx + 45.5
Gradient= Y/X = 34.5/76 = 0.5 (1.d.p)
Y= 0.5x + 45.5
E.g.:
Assume a car has done 40,000 miles. How much is its current price if the car was bought new for £8,000?
%age loss = 0.5 x 40 + 45.5
= 66%
Money lost = 66% of £8,000
= 5,280
New value = 8,000 – 5,280
= £2,720
Mean of Graph:
Table:
Graph:
Equation of Line of Best Fit:
Y= 0.6x + 37
Y= mx + c
Y= mx + 37
Gradient = Y/X = 43/72 = 0.6 (1.d.p)
Y= 0.6x + 37
E.g.:
Assume a car has done 50,000 miles. How much is its current price if the car was bought new for £15,000
% Age loss = 0.6 x 50 + 37
= 67%
Money lost = 67% of £15,000
= 10,050
New value = 15,000 – 10,050
= £4,950
Mean of Graph:
Table:
GRAPH:
Equation of Line of Best Fit:
Y= 5x + 35
Y= mx + c
Y= mx + 34.5
Gradient = Y/X = 44.5/8.8 = 5
Y = 5x + 35
E.g.:
Assume a car is 3 years old. How much is its current price if the car was bought new for £19,250?
%age loss = 5 x 3 + 34.5
= 49.5%
money lost = 49.5% of 19,250
= £9,529
new value = 19,250 - 9,529
= £9,72
Mean of Graph:
Note Teacher:
FOR THE GRAPHS ABOVE (MEAN OF GRAPH) I WILL DRAW THE MEAN POINT BUT WITH A PEN BEACAUSE I DON’T KNOW HOW TO DO IT ON COMPUTER, IF YOU DO KNOW TELL ME HOW TO DO IT PLEASE, THAT IS WHY YOU DON’T SEE IT ON THE GRAPH. I HOPE YOU UNDERSTAND WHAT I MEAN.
I WOULD LIKE TO GET THE BEST POSSIBLE MARK BEACAUSE I HAVE WORKED HARD ON THIS COURSEWORK. IF YOU HAVE ANY EXAMINERS TIPS PLEASE HELP ME.
THANK YOU.
P.S THIS IS ONLY THE MIDDLE PART OF MY COURSEWORK SINCE I HAVE TO DO THE PLAN AND CONCLUSION, COULD YOU HELP ME ON THAT AS WELL THANK YOU.