Transcript Purpose - Sage Middle School

Concept Review
Purpose: The purpose of the following set of slides is to review the major
concepts for the math course this year. Please look at each problem and how to
solve them. Then attempt to solve the second problem on your own. If you
have difficulty it would be good to come see me and ask questions.
Ratios
• You have a bag full of 500 yellow marbles and
1,000 blue marbles. What is the ratio of
yellow to blue marbles? Please simplify.
Solution
• The first thing to do is to identify the quantity of each color of marble.
• There are 500 yellow
• There are 1000 blue
• The ratio without simplifying is 500:1,000
• Now identify what each quantity is divisible by. In this case they are both
divisible by 100.
• This results in a ratio of 5:10.
• This can be simplified further, because both are divisible by 5. The final
simplified ratio is 1:2
• Extension: What is the probability of removing a yellow marble on one reach
into the bag of marbles.
Ratio Practice Problems
• You have a bag full of 200 red marbles and
600 white marbles. What is the ratio of red to
white marbles? Please simplify.
• You have a bag full of 350 turquoise marbles
and 625 pink marbles. What is the ratio of
turquoise to pink marbles? Please simplify.
Ratio Table
• Complete the ratio table
X
Y
3
9
6
9
27
Ratio Table
• Complete the ratio table Solution
• The rule is to multiply by 3
X
Y
3
9
6
18
9
27
Ratio Table Practice Problem
• Complete the ratio
X
Y
4
16
6
8
32
Ratios Continued
• John has 30 pens and 45 pencils. Sally has 47
pens and 47 pencils. Who has a higher ratio of
pens to pencils.
• Solution: In this case Sally has more pens per
every pencil, because she has a one to one
ratio whereas John has a less than one to one
ratio of pens to pencils.
Practice Ratio Problem Continued
• Jake has 40 red marbles to 40 blue marbles.
Jenny has 35 red marbles to 45 blue marbles.
Who has the higher red to blue marble ratio?
Functions
• Function n = b + 5
• Complete the table
B
n
4
9
11
13
15
Functions
• Function n = b + 5
• Complete the table Solution
B
n
4
9
11
16
13
18
15
20
Functions Practice Problem
• Function n = b + 12
• Complete the table
B
n
8
20
14
17
22
Division
• There are 49 pears that need to go in 7 crates.
How many pears will end up in each box?
Division
• There are 49 pears that need to go in 7 crates.
How many pears will end up in each box?
• Solution
7
7 49
crate =
- 49
0
7
7
7
7
7
7
7
*Remember the mathematical relationship between
division and multiplication which is linked to
addition and subtraction.
Division Practice Problem
• There are 64 apples. Please evenly distribute
the apples into 8 baskets.
Division of Fractions Review
• Key concepts and terms
• Improper fraction Complex Fraction, and
Mixed Number
Improper Fraction
• A fraction in which the numerator is greater than
the denominator.
12
4
Hint: 12/4 is equal to 3 wholes. The different ways
to express fractions are representing the same
amounts in different forms.
Complex Fraction
• A fraction in which either the numerator,
denominator, or both contain fractions.
½
¾
2
Denominator ¾
Numerator
½
3
Mixed Numbers
• A mixed number is a whole number and a
fraction expressed together.
1 whole
1/4
• 1 ¼
1/4
1/4
• Hint: as an improper fraction this would be
5/4.
Mixed Numbers
• A mixed number is a whole number and a
fraction expressed together.
1 whole
1/4
• 1 ¼
1/4
1/4
• Hint: as an improper fraction this would be
5/4.
Division of Fraction problems
• The procedure when dividing by fractions is to invert the second fraction and
multiply, but keep in mind the models that were presented on the first three
slides. They help to explain why this is the case.
•
1½
3/4
Solution:
Convert 1 and ½ to an improper fraction. To do this multiply the whole number
time the denominator and add the numerator
3/2
¾
(1 ½ is equal to 3/2, because there are 3 halves in 1 ½)
Now invert ¾ so the problem reads 3/2 X 4/3
You now have 12/6 which means you have 2 wholes.
For further explanation please refer back to the garden problem on the website
under the trimester 2 section.
Practice Problems for Division of
Fractions
• 3/4
6/2 =
• 1¼
3/4
=
• 2/3 5/6
=
Area, Surface Area and Volume
• Our goal this year is to identify how to find the
area, surface area and volume of basic shapes,
squares, cubes, rectangle, rectangular prisms,
triangles, and triangular prisms. We can then
use these basic shapes to break down more
complex shapes to find the area, surface area,
and volume.
Area
• When working with area we must remember that we are referring
to the inside of a shape or figure. This is different than perimeter
that looks at the distance around the outside of a shape or figure.
The area of the rectangle is
9 x 4 or 36 units squared,
because there are 36 1 X 1
smaller squares inside. The formula is L X W
4 units
9 units
The perimeter would be 9 + 4 +9 +4 =
26units. Notice it is not squared. This
would be like building a fence around the
rectangle. The formula is L + W + L + W
Area
• Math contains many patterns and relationships. This
slide addresses the relationship between rectangles
and triangles.
• How many triangles are inside of the rectangle?
• Knowing that a triangle is ½ of a rectangle or square
we can deduce that the formula for area of a triangle is
L X W . Once again we are talking about square area.
2
Surface area
• When we represent 3 dimensional shapes and
objects on paper we have to let our eyes
imagine in 3 dimensions. For this slide looking at
a 3 dimensional figure may help you see what
the figure is conveying. Pick up a shoe box. How
many flat surfaces can you identify?
Surface area
Continued
• To find the surface area you need to identify all
six sides. The front side is equal to the back and
the side to the opposing side. Finally the top
and bottom should have equal dimensions. To
find the surface area simply find the area of
each side and then add them together.
Surface Area
4 units
There are two squares on the front and back
4 X 4 = 16, now multiply it by 2 = 32 units squared
There are four 4 X 9 rectangles which are the two
sides and top and bottom. 4 X9 = 36 units
squared. Multiply 36 X 4 and you have 144 units
units squared. Now add 144 to 32
for a total surface area of 176
units squared.
4 units
Volume
• Volume address the space inside of a shape or
figure. For example how many ice cubes could
you fit in the rectangle that are 1 X 1 X 1 units
cubed? To solve this you multiply the L X W X H.
• Therefore 4 X 4 X 9 = 144 units cubed, or there
are 144 1 X 1 X1 little cubes inside the rectangle.
•
•
4 units
A rectangular prism would be
half. In other words divide it by two
4 units
Practice
7 inches squared
4 inches squared
4 inches squared
• Find the area
7 inches squared
3 inches
Find the Surface area
3inches
3 inches
3 inches
Find the Volume
3 inches
3inches
More Complex Shapes
• Knowing how to find the area and volume of
simple shapes can help us break down more
complex shapes. For example a trapezoid.
• We can see that we have two triangles and a
rectangle.
More Complex Shapes
• There are different ways to approach solving
this problem. One would be to find the area of
the rectangle and the two triangles and add
them. If the triangles have equal bases you
could combine the triangle into a square.
Measures of Central Tendency
Mean Median and Mode
Measures of Central Tendency
x
• 0 10
x
x
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x
x
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x
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x
x
x
x
20
30
40
50
60
70
80
Mean Median and Mode
Continued
The mode is the data point that occurs the most. The mode of this data is 40.
x
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20
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80
Mean Median and Mode
Continued
The median is the middle of the data. It helps to string out the data.
10, 20 , 20, 30, 30, 30, 40, 40, 40, 40, 50, 50, 50, 60, 60, 70. The median is 40.
x
• 0 10
x
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20
30
40
50
60
70
80
Mean Median and Mode
Continued
The mean is related to the distance between the data points. Imagine that each data point
is a sticky note. You could divide the distance between 10 and 70 and arrive at forty. If you
this over and over you would eventually arrive at the mean. The algorithm to solve it simply
States , add all the data points and divide by the number of data points.
x
• 0 10
x
x
x
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x
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20
30
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50
60
70
80
Mean Median and Mode
Continued
Mean
10 + 20 + 20+30+30+30+40+40+40+40+50+50+50+60+60+70 = 640
Dividing by 640 by 16 you get the mean of 40.
x
• 0 10
x
x
x
x
x
x
x
x
x
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x
x
x
x
x
20
30
40
50
60
70
80
Mean Median and Mode
Continued
Challenge what will happen to the mean if we changed 70 to 90? Would the mean move
To the right or the left? How do you know?
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20
30
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Mean Median and Mode
Continued
The total of the data points now is 660 and 660 divided by 16 is 41.25 so the curve is skewed
to the right.
x
• 0 10
x
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x
x
20
30
40
Mean = 41.25
x
x
x
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x
50
60
70
x
80
90
Mean Median and Mode
Continued
The spread of the data deals with the distance between data points and the relation to the
mean. Think about which of the two data distributions are more spread out. Is it the one
with a balanced curve of the one that is skewed to the right? Take a moment to think
about it.
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• 0 10
x
x
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x
x
20
30
40
50
60
70
80
Mean Absolute Deviation
• First we must find the mean. 10 + 30 +70 =
110 divided by 3 = 36.6 repeating.
x
x
x
10
30
70
Mean Absolute Deviation
• Once we have the mean we then find the
distance between the mean and the data
points. I rounded the mean to 36.7.
36.7-10 = 26.7 36.7-30 =16.7 70- 36.7 = 33.3
x
x
x
10
30
70
Mean Absolute Deviation
• Now add the distances up and divide by the
number of distances.
26.7 + 16.7 + 33.3 = 76.7
76.7 divided by 3 = 25.56 repeating or 25.57
There for the MAD is 25.57
x
x
x
10
30
70
Mean Absolute Deviation
Application
• Making comparisons between data sets and making meaning out of a
single data set is inferential statistics or drawing inferences from the data.
If the following two schools both had the following data sets and are
teaching the same curriculum. Which one is getting more consistent
results even if the averages are the same. MAD would be a very useful
statistic in this case.
X
x
x
x
x
x x
*** Be careful not to make judgments based on one statistic i.e. mean,
median, mode, or MAD. Look at the data as many ways as you can to include
models like above. Your looking for the story that the data is trying to tell you.
Mean Absolute Deviation
• What this allows us to do is compare the spread of a
set of data to another. If a set of data has a mad of
25.57 versus a second data distrubution with a MAD of
5.6 which one is clumped more closely together?
X
x
x
x
x
x x
The larger the mad the more spread out the data.
Unit Rate Problems
• When you are driving in your car and look at
your speedometer it gives you a unit rate. For
example a car traveling at 60 miles per hour
literally means that you can travel 60 miles in
1 hour. Unit rate can be applied in many
situations.
60 miles
60 miles
1 hour
1 hour
Unit Rate Problems
• What if you were bottling soda at a factory
and new that you could bottle 60 sodas in 1
hours. What is your rate per hour? What is
your rate per minute? What is your rate per
second?
60 sodas
60 sodas
1 hour
1 hour
Unit Rate Problems
• There are 60 minutes in 1 hour. Therefore you
can divide 60 by 60 and your rate per minute is 1
soda per minute.
60 sodas
1 hour
60 sodas
1 hour
Inside the red line represents 1/60th of an hour or 1
minute.
Unit Rate Problems
• There are 60 seconds in 1 minute. Therefore
you can divide 1 soda by 60 and your rate per
second is .016 repeating of a soda per second.
1 soda
1 soda
1 minute
1 minute
Inside the red line represents 1/60th of a minute
or 1 second.