|
1
|
|
|
2
|
- Walter R. Hunter whunter@mc3.edu
- Roseanne S. Hofmann rhofman@mc3.edu
|
|
3
|
- Philosophy of the the course:
- Applications should not only
motivate, but also give concrete examples to the abstract mathematical
concepts of the course.
- AMATYC Standards:
- Problems should be solved
numerically, graphically, symbolically, and
verbally.
|
|
4
|
- Statistics:
- Day students taught by full time
faculty.
- p-value 0.000000121 for average
grade.
|
|
5
|
- Statistics:
- Day students taught by full time
faculty.
- p-value 0.0047 for average
grade.
|
|
6
|
- Statistics: Intermediate Algebra
- Day students taught by full time faculty and passed Beginning Algebra the previous
semester.
|
|
7
|
- Keep current on the “standards”
- Textbook provides flexible teaching modes
- Lecture materials
- Group Work
- Homework Exercises
- Multimedia matches textbook
- Smartboard lectures ready made
|
|
8
|
|
|
9
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
|
|
10
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
- 10 - 25 = -15
|
|
11
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
- 10 - 25 = -15
- 2. You loose $5 a day for the
next three days, how much money will you loose?
|
|
12
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
- 10 - 25 = -15
- 2. You loose $5 a day for the
next three days, how much money will you loose?
- -5 ( 3 ) = - 15
|
|
13
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
- 10 - 25 = -15
- 2. You loose $5 a day for the
next three days, how much money will you loose?
- -5 ( 3 ) = - 15
- 3. You lost $7 for the last six
days, how much more money did you have six days ago?
|
|
14
|
- 1. You have $10 and you owe the
IRS $25, what is your net worth?
- 10 - 25 = -15
- 2. You loose $5 a day for the
next three days, how much money will you loose?
- -5 ( 3 ) = - 15
- 3. You lost $7 for the last six
days, how much more money did you have six days ago?
- ( - 7 )( - 6 ) = 42
|
|
15
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
16
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
17
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
18
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
19
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
20
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
|
|
21
|
- A truck rental company, WE-HAUL,
charges $21.95 plus 41 cents per mile.
- The equation relating cost and miles driven:
- C = .41m + 21.95
|
|
22
|
|
|
23
|
|
|
24
|
|
|
25
|
|
|
26
|
|
|
27
|
|
|
28
|
|
|
29
|
|
|
30
|
|
|
31
|
|
|
32
|
|
|
33
|
|
|
34
|
|
|
35
|
- What is the equation that relates weight and
height? Simplify the equation.
- W = 5( h - 60 ) + 100
- W = 5h - 300 +
100
- W = 5h - 200
|
|
36
|
- How tall should you be if you
weigh 135 lbs?
- Find h if W = 135
- 135
= 5h - 200
- 135 + 200 = 5h
- 200 + 200
- 335
= 5h
- 335/5 =
5h/5
- 67
= h
- You should be 5 feet 7 inches
tall if you weigh 135 lbs.
|
|
37
|
- How tall should you be if you weigh
0 lbs?
- Find h if W = 0
- 0 = 5h
- 200
- 0 + 200 = 5h -
200 + 200
- 200 = 5h
- 200/5 = 5h/5
- 40 = h
- Does this answer makes sense?
|
|
38
|
- Solve for h from the formula in part 1b.
- W =
5h - 200
- W + 200 = 5h -
200 +200
- W + 200 = 5h
- (W + 200)/5 = 5h/5
- (W + 200)/5 = h
|
|
39
|
|
|
40
|
|
|
41
|
- Two companies offer you very similar positions. Random House will pay you $10,000 a
year plus 7% commission on the dollar amount of book sales. Moore Publishing Co. will pay you
$8,000 a year plus 11% commission. Random House SALES CALCULATION INCOME
|
|
42
|
- Two companies offer you very similar positions. Random House will pay you $10,000 a
year plus 7% commission on the dollar amount of book sales. Moore Publishing Co. will pay you
$8,000 a year plus 11% commission. Random House SALES CALCULATION INCOME 100,000 .07(100,000) + 10,000 17,000
|
|
43
|
- Two companies offer you very similar positions. Random House will pay you $10,000 a
year plus 7% commission on the dollar amount of book sales. Moore Publishing Co. will pay you
$8,000 a year plus 11% commission. Random House SALES CALCULATION INCOME 100,000 .07(100,000) + 10,000 17,000 S .07S + 10,000 I
|
|
44
|
- For what dollar amount of book sales does Random House pay more than
Moore Publishing Co.?
|
|
45
|
- For what dollar amount of book sales does Random House pay more than
Moore Publishing Co.?
- Income from R.H. > Income from Moore
- .07S + 10000 > .11S
+ 8000
|
|
46
|
- For what dollar amount of book sales does Random House pay more than
Moore Publishing Co.?
- Income from R.H. > Income from Moore
- .07S + 10000 > .11S
+ 8000
- .07S + 2000 > .11S
- 2000 >
.04S
- 50,000 > S
- Random House will pay more for sales less than $50,000.
|
|
47
|
- When NASA sends a rocket into space, they monitor the temperature of
certain gases. Time is the independent variable and temperature is the
dependent variable. Notice that
negative time is used to denote time before lift off.
Time Temp Time Temp -20 -20 10 80 -15 -20 15 120 -10 -10 20 90 -5 0 25 25 0 15 30 5 5 40 35 -15
|
|
48
|
- MIN
- What was the temperature at take off?
|
|
49
|
- MIN
- When was the temperature zero?
|
|
50
|
- An appliance repair shop charges an initial fee of $30 plus $15 per
hour to fix an appliance.
- Complete the table to find the cost of repairing an appliance.
- Hours Calculation Costs 3 15(3) + 30 75 5 15(5) + 30 105 10 15(10) + 30 180 t 15t + 30 C
|
|
51
|
- Use the results in part a. to graph the equation in part b. Choose an appropriate scale and only
graph the portion that makes sense to the problem. Label the axis.
|
|
52
|
- Use the results in part a. to graph the equation in part b. Choose an appropriate scale and only
graph the portion that makes sense to the problem. Label the axis.
|
|
53
|
- Sally has a lawn-mowing business for the summer. She bought a lawn mower for $200 and
she charges $5 an hour. The
equation that relates profit and hours worked is p = 5h - 200.
- How many hours does she have to work to break even?
- Find h, if p = 0.
- 0 = 5h - 200
- 40 = h
|
|
54
|
- Sally has a lawn-mowing business for the summer. She bought a lawn mower for $200 and
she charges $5 an hour. The
equation that relates profit and hours worked is p = 5h - 200.
- How much money will she lose if she doesn’t work any hours?
- Find p, if h = 0.
- p = 5( 0) - 200
- p = -200
|
|
55
|
- Graph the line p = 5h - 200 by plotting the points obtained in parts a.
and b. Choose an appropriate
scale and only graph the portion that makes sense to the problem. Label the axis.
- Break even point (40, 0 ).
- Starting costs (0, -200 ).
|
|
56
|
|
|
57
|
|
|
58
|
|
|
59
|
|
|
60
|
|
|
61
|
|
|
62
|
- Which grow faster: Hybrid A corn seedlings, which grew 14.6 centimeters
in 15 days, or Hybrid B, which grew 11.2 centimeters in 12 days.
|
|
63
|
- Which grow faster: Hybrid A corn seedlings, which grew 14.6 centimeters
in 15 days, or Hybrid B, which grew 11.2 centimeters in 12 days.
- Find how fast they grow per day.
|
|
64
|
- Which grow faster: Hybrid A corn seedlings, which grew 14.6 centimeters
in 15 days, or Hybrid B, which grew 11.2 centimeters in 12 days.
- Find how fast they grow per day.
- Hybrid A: 14.6/15 =
|
|
65
|
- Which grow faster: Hybrid A corn seedlings, which grew 14.6 centimeters
in 15 days, or Hybrid B, which grew 11.2 centimeters in 12 days.
- Find how fast they grow per day.
- Hybrid A: 14.6/15 =
- 14.6/15 = .9733
- Hybrid A grew an average of .9733 centimeters per day.
|
|
66
|
- You are going to rent a car for a day.
You have two choices, Wrecker Car company or Caddie Car
Rental. Wrecker charges $20 a day
and $0.75 a mile, while Caddie charges $27 a day and $0.40 a mile.
|
|
67
|
- You are going to rent a car for a day.
You have two choices, Wrecker Car company or Caddie Car
Rental. Wrecker charges $20 a day
and $0.75 a mile, while Caddie charges $27 a day and $0.40 a mile.
- Write an equation for the cost of renting a car from each company.
- Wrecker: C = .75m + 20
- Caddie: C = .40m + 27
|
|
68
|
- Graph both equations on the same set of axis. Label each axis and choose an
appropriate scale. Only graph the
portion that is relevant to the problem.
|
|
69
|
- Find where the two lines intersect.
Label the point.
|
|
70
|
- Find where the two lines intersect.
Label the point.
- How far do you have drive for the two companies to charge the same?
|
|
71
|
- Find where the two lines intersect.
Label the point.
- How far do you have drive for the two companies to charge the same?
- .75M + 20 = .40M + 27
|
|
72
|
- Find where the two lines intersect.
Label the point.
- How far do you have drive for the two companies to charge the same?
- .75M + 20 = .40M + 27
- .35M + 20 = 27
- .35M = 7
- M = 20
|
|
73
|
- Find where the two lines intersect.
Label the point.
- If M = 20, find C.
|
|
74
|
- Find where the two lines intersect.
Label the point.
- If M = 20, find C.
- C = .75(20) + 20
|
|
75
|
- Find where the two lines intersect.
Label the point.
- If M = 20, find C.
- C = .75(20) + 20
- C = 35
- The two lines intersect at (20, 35).
- Both companies charge $35 when you drive 20 miles.
|
|
76
|
- When does Wrecker charge more than Caddie?
|
|
77
|
- What are the cost intercepts for Wrecker and Caddie? What do they mean?
|
|
78
|
- What does the slope of each line mean in terms of the problem?
|
|
79
|
- The population of Shanghai was 10,820,000 in 1974. If the population increased by 1% each
year, complete the table below.
|
|
80
|
- The population of Shanghai was 10,820,000 in 1974. If the population increased by 1% each
year, complete the table below.
- For 1975:
|
|
81
|
- The population of Shanghai was 10,820,000 in 1974. If the population increased by 1% each
year, complete the table below.
- For 1975: 10820000 + (.01)10820000 =
|
|
82
|
- The population of Shanghai was 10,820,000 in 1974. If the population increased by 1% each
year, complete the table below.
- For 1975: 10820000 + (.01)10820000 =
-
(1.01)10820000 =
|
|
83
|
- The population of Shanghai was 10,820,000 in 1974. If the population increased by 1% each
year, complete the table below.
- For 1975: 10820000 + (.01)10820000 =
-
(1.01)10820000 =
-
10928200
|
|
84
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76
- 77
- 78
- n
|
|
85
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)10928200
- 77
- 78
- n
|
|
86
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)10928200
- (1.01)(1.01)10820000
- 77
-
78
- n
|
|
87
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)10928200
- (1.01)(1.01)10820000
- (1.01)210820000 11,037,482
- 77
- 78
- n
|
|
88
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)210820000 11,037,482
- 77 (1.01)310820000 11,147,857
- 78 (1.01)410820000 11,259,335
- n
|
|
89
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)210820000 11,037,482
- 77 (1.01)310820000 11,147,857
- 78 (1.01)410820000 11,259,335
- n (1.01)n10820000 P
|
|
90
|
- YR CALCULATION POP
- 75 (1.01)10820000 10,928,200
- 76 (1.01)210820000 11,037,482
- 77 (1.01)310820000 11,147,857
- 78 (1.01)410820000 11,259,335
- n (1.01)n10820000 P
- What is the equation that relates Shanghai’s population and year?
- P = (1.01)n10802000, n is the number of years since 1974.
|
|
91
|
- Use the equation to find Shanghai’s population in 1985.
- P = (1.01)n10802000, n is the number of years since 1974.
- n = 1985 - 1974 = 11
- P = (1.01)1110820000
- P = (1.116)(1082000)
- P = 12075120
- The formula estimates that Shanghai’s population in 1985 was
12,075,120.
|
|
92
|
- Use the following formula to find the monthly payments of a
loan. P = A i
1 - ( 1 + i )-n P
is the monthly payment A is the amount of the loan n is the number
of payments i is the interest rate per month
- Eddie Nerder can afford a $250 car payment at 6% annual interest rate
for 36 months. How expensive a
car can he afford? ( Hint: I =.06 / 12 )
|
|
93
|
- 250 = A ( 0.0304
)
- 8223 = A
- Eddie can afford an $8,223 car.
|
|
94
|
- Suppose the cost of removing p percent of the particulate pollution
from the exhaust gases at an industrial site is given by c = 6800p 100 - p
|
|
95
|
- c = 6800p 100 - p
- Find the cost for p = 75
- c = 6800(75) 100 - 75
- c = 510000 25
- c = 20400
- It would cost $20,400 to remove 75% of the particulate pollution.
|
|
96
|
- c = 6800p 100 - p
- Find the cost for p = 100
- c = 6800(100) 100 - 100
- c = 680000 0
- c = undefined
- The formula cannot compute the cost of removing all of the pollution.
|
|
97
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t.
|
|
98
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- Find the height of the arrow for t = 2.
- h = -16(22) + 160(2)
- h = -16(4) + 160(2)
- h = -64 + 320
- h = 256
|
|
99
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- Compute the height for:
- t = 2, h = 256
- t = 5, h = 400
- t = 8, h = 256
- t = 10, h = 0
|
|
100
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- According to the calculations above, when will the arrow reach its
maximum height?
- t = 5
- The arrow reaches the maximum height in 5 seconds. The maximum height is 400 feet.
|
|
101
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- According to the calculations above, when will the arrow hit the
ground?
- When the arrow hits the ground the height, h, is zero.
- t = 10
- The arrow reaches the ground in 10 sec.
|
|
102
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- Graph the points obtained in a through d.
|
|
103
|
- William Tell shoots an arrow straight up with an initial velocity of
160 feet per second. The height
of the arrow is given by the equation h = -16t2 + 160t
- Graph the points obtained in a through d.
- The points are: ( 2, 256)
- ( 5,
400)
- ( 8,
256)
- (10,
0)
|
|
104
|
|
|
105
|
- Earl Black makes tea bags. The
cost and revenue of making and selling x million tea bags per month
is C = x2 -
38x + 400 R = -x2
+ 78x. C and R are in thousands of dollars.Find the equation
for profit.
|
|
106
|
- Earl Black makes tea bags. The
cost and revenue of making and selling x million tea bags per month
is C = x2 -
38x + 400 R = -x2
+ 78x. C and R are in thousands of dollars.Find the equation
for profit.
- P = R - C
- P = (-x2 + 78x) - (x2 - 38x + 400)
- P = -x2 + 78x - x2 + 38x - 400
- P = -2x2 + 116x - 400
|
|
107
|
- Graph the profit equation.
Explain what the vertex, x and p intercepts mean in terms of
making tea bags. Make sure you
label the axis and use an appropriate scale.
|
|
108
|
|
|
109
|
- P = -2x2 + 116x - 400
- 1. The Vertex, (29, 1282), means;
If Earl Black sells 29 million tea bags
they will get a maximum profit of 1.282 million dollars.
- 2. The x-intercepts, (3.675, 0) (54.33, 0),
mean; If Earl Black sells 3.675 million or 54.33 million tea bags they
will break even.
- 3. The P-intercept, (0, -400), means;
The company’s start up costs.
|
|
110
|
- Suppose Earl Black needs to make $500,000 in profit (P = 500). Graph this line on the graph above and
find out where the line intersects the graph. Explain what the answers mean.
|
|
111
|
|
|
112
|
- Suppose Earl Black needs to make $500,000 in profit (P = 500). Graph this line on the graph above and
find out where the line intersects the graph. Explain what the answers mean.
- Find x, if P = 500.
- 500 = -2x2 + 116x - 400
- 0 = -2x2 + 116x -
900
- a = -2, b = 116, c = -900
- x =9.228, 48.77
|
|
113
|
- x = 9.228, 48.77
- The line intersects the graph at (9.228, 500) and (48.77, 500).
- If Earl Black needs to make $500,000 in Profit then they need to sell
9,228,000 or 48,770,000 tea bags.
|
|
114
|
|
|
115
|
- When does Earl Black make
more than
- $500,000 in profit?
|
|
116
|
|
|
117
|
|
|
118
|
|
|
119
|
|
|
120
|
|
|
121
|
- in the broadest sense has always been important in mathematics
|
|
122
|
- Expectations by students
- To present visualizations of concepts
- To present animations
- To require interactive participation of learner
|
|
123
|
- Web Page
- ToolBook
- PowerPoint
- Excel
- Flash
- Merlot
- ScreenWatch
|
|
124
|
|
|
125
|
|
|
126
|
|
|
127
|
|
|
128
|
|
|
129
|
|
|
130
|
|
|
131
|
|
|
132
|
|
|
133
|
|
|
134
|
|
|
135
|
|
|
136
|
|
|
137
|
|
|
138
|
|
|
139
|
|
|
140
|
|
|
141
|
|
|
142
|
|
|
143
|
|
|
144
|
|
|
145
|
|
|
146
|
|
|
147
|
- small bits of reusable materials
- learning modules
|
|
148
|
|
|
149
|
|
|
150
|
|
|
151
|
|
|
152
|
|
|
153
|
- Ongoing project
- Living book
- Flash
|
|
154
|
|