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Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Model Data Storage on an MP3 Player and a DVD Using Linear Equations
and Graphs
Model Data Storage on an MP3 Player and a DVD Using Linear Equations
and Graphs
Chapter 7Chapter 7
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Part 1Part 1
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Model Data on an MP3 PlayerModel Data on an MP3 Player
MP3 Players can be purchased with different amounts of memory, usually measured in Gigabytes (GB).
Investigate Storage
The number of songs that can be stored depends on how much memory is available.
Determine a linear equation that models the relation between the number of songs stored and the memory available.
1GB = 1 000 000 000 bytes (B)
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
An advertisement for a popular MP3 player says that with 8 GB of memory, the device can store 1750 songs.
With 32 GB, it stores 7000 songs.
With 64 GB, it stores 14 000 songs.
Songs vary in length. Claims such as these assume that an average song takes up about 4.5 Megabytes (4.5 MB) or 4 500 000 B of memory.
Model Data on an MP3 PlayerModel Data on an MP3 PlayerInvestigate Storage
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
The data can be represented in a table of values. Create a table of values for the relation between the amount of memory
and the number of songs in the MP3 player.
Memory (GB) (x)
Number of Songs (y)
Model Data on an MP3 PlayerModel Data on an MP3 PlayerCreate a Table of Values
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Memory (GB)(x)
Number of Songs
(y)8 1750
32 7000
64 14000
Model Data on an MP3 PlayerModel Data on an MP3 PlayerCreate a Table of Values
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Represent the data on a graph.
Model Data on an MP3 PlayerModel Data on an MP3 PlayerCreate a Graph
Song Storage on an MP3 Player
0
2000
4000
6000
8000
10000
12000
14000
16000
0 10 20 30 40 50 60 70
Memory (GB)
Nu
mb
er o
f S
on
gs
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Represent the data on a graph.
Model Data on an MP3 PlayerModel Data on an MP3 PlayerCreate a Graph
Song Storage on an MP3 Player
0
2000
4000
6000
8000
10000
12000
14000
16000
0 10 20 30 40 50 60 70
Memory (GB)
Nu
mb
er o
f S
on
gs
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
risem
runy
mx
14000 1750
64 8m
218.75m
Model Data on an MP3 PlayerModel Data on an MP3 PlayerDetermine the Slope and y-intercept
218.75y x b
218.71 50 57 8 b
0b
y mx b
The slope is 218.75. The y-intercept is 0.
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
The slope is 218.75. The y-intercept is 0.
The equation is:
218.75 0
218.75
y mx b
y x
y x
Where x is amount of memory in gigabytes and y is the number of songs.
Model Data on an MP3 PlayerModel Data on an MP3 PlayerDetermine the Equation in Slope-Intercept Form
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Compare Storage MethodsCompare Storage Methods
An MP3 player can store approximately 218.75 songs per GB of storage space. The equation that represents this situation is
where x represents the storage space in GB, and y represents the number of songs stored.
DVDs are another method of file storage.
218.75y x
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Compare Storage MethodsCompare Storage MethodsUse the Slope-Intercept Form of an Equation
A standard DVD has about 4.7 GB of storage space.
Determine the number of songs that can be stored on a standard DVD using the equation from the previous problem.
218.75y x
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Compare Storage MethodsCompare Storage MethodsUse the Slope-Intercept Form of an Equation
The amount of memory, x, is 4.7 GB. Substitute this value into the equation.
The equation is:
218.75
218.75(4.7)
1028.125
1028
y x
y
y
y
A DVD can store approximately 1028 songs.
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Part 2Part 2
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Recordable DVDs hold 4.7 GB of data.
The track starts 2.4 cm from the centre and finishes 5.8 cm from the centre.
The data is written on a spiral track with a total length of about 12 km.
Model Data on a DVDModel Data on a DVDInvestigate Data on a DVD
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Change 12 km to cm:
Divide the number of bytes by the length in cm:
There are 3917 bytes of data per cm of track.
Change 4.7 GB to bytes:
12 km = 12 000 m = 1 200 000 cm
4.7 GB = 4 700 000 000 B
4 700 000 000 = 3917
1 200 000
Model Data on a DVDModel Data on a DVDCalculate the Amount of Data
1 km = 1 000 m
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
The inside loop has an average radius
of 2.4 cm.
The data stored on the inside loop is:
There are 59 147 bytes of data on the inside loop.
2 2.4C
=15.1 3917
Calculate the circumference:
15.1 cmC
Model Data on a DVDModel Data on a DVDCalculate the Amount of Data on the Inside Loop
amount of data per = circumference amount of data per cm
59 147
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
The outside loop has an average radius
of 5.8 cm.
The data stored on the outside loop is:
There are 142 579 bytes of data on the outside loop.
2 5.8C
= 36.4 3917
Calculate the circumference:
36.4 cmC
Model Data on a DVDModel Data on a DVDCalculate the Amount of Data on the Outside Loop
amount of data = circumference amount of data per cm
= 142 579
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Distance from Centre (cm)
(x)
Data Stored (B)(x)
Represent the distance from centre and the amount of data stored in a table of values.
Model Data on a DVDModel Data on a DVDCreate a Table of Values.
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Distance from Centre (cm)
(x)
Data Stored (B)(y)
2.4 59 147
5.8 142 579
Model Data on a DVDModel Data on a DVDCreate a Table of Values.
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
y
mx
142579 59147
5.8 2.4m
24 539m
24539y x b
2453959147 2.4 b
1.0043...
1
b
b
Model Data on a DVDModel Data on a DVDDetermine the Slope and the y-intercept
risem
run
The slope is 24 539 and the y-intercept is approximately 1.
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
24539 + 1.00
y mx b
y x
The distance from the centre in centimetres is x
and the amount of data in bytes is y. This function is only defined for values of x from 2.4 to 5.8 cm.
Model Data on a DVDModel Data on a DVDDetermine the Equation in Slope-Intercept Form
The slope is 24 539. The y-intercept is approximately 1.
The equation is:
Copyright © 2010 McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved.
Model Data on a DVDModel Data on a DVDUse the Equation
The amount of data stored in a loop on a DVD depends on the distance of the loop from the centre of the DVD.
Where the distance from the centre in centimetres is x and the amount of data in bytes is y. The function is defined only for values of x from 2.4 to 5.8 cm. How much data is stored on the loop that is 3.5 cm from the centre?
24539 + 1.00
24539(3.5) 1.00
85886.5 1
85887.5
.00
y x
y
y
y
24539 + 1.00y x
In the loop 3.5 cm from the centre, 85 887.5 B of data is stored.
