Transcript Lesson Title

Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Problem of the Day
Replace the question marks with the correct digits.
a. 8 ? 9 + 6. ? ? = 15.96
9, 9, 7
b. 13. ? 0 ? – ? . 4 ? 2 = 4.122
6, 4, 9, 8
Lesson
Main
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Check Skills You’ll Need
(For help, go to Lesson 2-2.)
1. Vocabulary Review A rational number is a number that
can be written in the form ? .
Write each fraction in simplest form.
2. 90
3. 80
4. 35
5. 25
100
100
100
100
Check Skills You’ll Need
Lesson
Main
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Check Skills You’ll Need
Solutions
1.
a , b
b
3.
80 ÷ 20
4
=
100 ÷ 20
5
5.
25 ÷ 25 = 1
100 ÷ 25 4
Lesson
Main
0
2.
90 ÷ 10
9
=
100 ÷ 10 10
4.
35 ÷ 5
= 7
100 ÷ 5 20
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Additional Examples
Use mental math to write
3
as a percent.
25
What you think
3
I can write
as an equivalent fraction with a denominator of 100.
25
4
12
3 = 12
I can rewrite
as 12%.
100
25
100
4
Why it works
3
3•4
=
25 25 • 4
12
=
100
= 12%
Lesson
Main
Multiply the numerator and denominator by 4.
Simplify.
Write the fraction as a percent.
Lesson
5-1
Quick Check
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Additional Examples
Write each decimal as a percent.
a. 2.5
25
10
25 • 10 250
=
=
10 • 10 100
= 250%
2.5 =
Write the decimal as a fraction.
Write as an equivalent fraction with a
denominator of 100.
Write the fraction as a percent.
b. 0.003
3
0.003 = 1,000
3 ÷ 10
0.3
=
=
1,000 ÷ 10 100
Write the decimal as a fraction.
Write as an equivalent fraction with a
denominator of 100.
Write the fraction as a percent.
= 0.3%
Quick Check
Lesson
Main
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Additional Examples
1
A brand of cereal supplied 7 2 % of the RDA of
sodium. Write this portion of the RDA as a fraction.
1
72
1
7 2 % = 100
Write the percent as a fraction with a
denominator of 100.
15
Rewrite the fraction as division.
Write the mixed number as an
improper fraction.
=
15 x 1
100
2
Multiply by the reciprocal of 100.
=
3
40
Simplify.
= 2 ÷ 100
Quick Check
Lesson
Main
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Additional Examples
Order 27%, 0.24, and 1 from least to greatest.
5
27% = 0.27
1
=0.20
5
0.24=0.24
Answer:1 < 0.24 < 27%
5
Quick Check
Lesson
Main
Lesson
5-1
Feature
Fractions, Decimals, and Percents
LESSON 5-1
Course 3
Lesson Quiz
1
1. A cereal supplies 1 14 % of the RDA for calcium. Write 1 4 %
as a fraction.
1
80
7
2. Write
as a percent.
20
35%
3. Write 1.5 as a percent.
150%
4. Order 60%, 0.58, and 5 .
8
0.58 < 60% < 5
8
Lesson
Main
Lesson
5-1
Feature
Estimating With Percents
LESSON 5-2
Course 3
Problem of the Day
The Jackson Country Bird Sanctuary has three times as many owls
as hawks. It has 40 hawks and owls in all. How many of each are in
the sanctuary?
30 owls, 10 hawks
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Check Skills You’ll Need
(For help, go to Lesson 2-5.)
1. Vocabulary Review The multiplicative inverse of 3 is
7
Find each product.
3
4
2.
36 •
4.
9
• 60
10
3.
2
• 12
3
5. 81 • 5
9
Check Skills You’ll Need
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Check Skills You’ll Need
Solutions
1. 7
9
36 • 3
36 3
36 • 3
2.
•
=
= 1 • 4 = 27
1 4
1•4
1
4
3. 2 • 12 = 2 • 12 = 2 • 12 = 8
1
3 • 1 13 • 1
3
6
9 60
4.
•
= 9 • 60 = 9 • 60 = 54
10 1
10 • 1 10 • 1
1
3
9
81 5
81 • 5
81 • 5
5.
• =
=
= 45
1
9
1•9
1 • 91
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Additional Examples
Estimate 74% of 158 using decimals.
74%
0.75
Use a decimal that is close to 74%.
158
160
Round 158 to a number that is compatible
with 0.75.
74% of 158
0.75 of 160
= 0.75 • 160
Multiply to find 0.75 of 160.
= 120
Simplify.
Quick Check
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Additional Examples
A video store rented 297 videos. The customers
returned 19% of the videos late. Estimate, using fractions,
how many videos were returned late.
19%
297
1
5
Use a fraction that is close to 19%.
300
Round to a number that is compatible with 5.
1
of 300
5
19% of 297
60
1 300
= 5 • 1 = 60
1
1
Multiply to find 5 of 300.
About 60 videos were returned late.
Lesson
Main
Lesson
5-2
Quick Check
Feature
Estimating With Percents
LESSON 5-2
Course 3
Additional Examples
Dion is saving for a coat that costs $59.95. She
has saved 45% of the cost. Estimate how much she has
saved.
What you think
The coat costs about $60.
1
I know 50% of 60 equals 2 of 60, or 30. Since 5% is one-tenth of 50%,
then 5% of 60 is one-tenth of 30, which is 3. So, the amount Dion has
saved is about $30 minus $3, which is $27.
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Additional Examples
(continued)
Why it works
50% of 60 = 0.5 • 60
= 30
1
5% of 60 = 10 (50% of 60)
1
= 10 (30)
=3
45% of 60 = 30 – 3
= 27
To find 50% of a number, multiply the
number by 0.5.
Simplify.
5% of a number is one-tenth of 50% of
that number.
Substitute.
Simplify.
45% = 50% – 5%
Simplify.
Quick Check
So, 45% of $59.95 is about $27.
Lesson
Main
Lesson
5-2
Feature
Estimating With Percents
LESSON 5-2
Course 3
Lesson Quiz
Show the numbers you use to estimate. Numbers used may vary,
Samples are given.
1. Use decimals to estimate 54% of 29.
0.5 • 30 = 15
2. Use fractions to estimate 74% of 38.
3
• 40 = 30
4
3. Estimate a 15% tip on $8.15.
0.1(8) + 0.1(4) = 0.8 + 0.4 = $1.20
4. About 60% of 27 students are in the play.
3 (30) = 18 students or 0.6 • 30 = 18
5
Lesson
Main
Lesson
5-2
Feature
Percents and Proportions
LESSON 5-3
Course 3
Problem of the Day
Use graph paper. Design and draw a diagram to determine which has
the greater area—a square with sides of 10 cm or a circle with a diameter
of 10 cm.
Draw a circle within the square to prove that the square has the greater area.
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Check Skills You’ll Need
(For help, go to Lesson 4-3.)
1. Vocabulary Review Two equal ratios form a
Solve each proportion.
2. 4 = 20
3.
5. s = 75
6. 6 = 24
b
4
100
100
8 = e
12 100
y
4. 240 = 12
n
5
100
Check Skills You’ll Need
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Check Skills You’ll Need
Solutions
1. proportion
2. 20b = 400; b = 20
2
3. 12e = 800; e = 66 3
4. 12n = 1,200; n = 100
5. 100s = 300; s = 3
6. 24y = 600; y = 25
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
Find 32% of 240.
n
32
=
240 100
Write a proportion.
100n = 240 • 32
Write the cross products.
100n = 7,680
Simplify.
100n 7,680
=
100
100
Divide each side by 100.
n = 76.8
Simplify.
Quick Check
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
Brenda saw a blender for $24 in a bargain store. In a second
store, the same blender was 160% of the cost of the blender in the
bargain store. Find 160% of $24.
A diagram can help
you understand the
problem.
n
160
=
24 100
Write a proportion.
100n = 24 • 160
Write the cross products.
100n = 3,840
Simplify.
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
(continued)
100n 3,840
=
100
100
Divide each side by 100.
n = 38.4
Simplify.
The price of the blender in the second store was $38.40.
Quick Check
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
Suppose 11,550 elementary students make up
14% of a city’s population. What is the population of the
city?
A diagram can help
you understand the
problem.
11,500 14
=
w
100
11,550 • 100 = 14w
Write a proportion.
Write the cross products.
1,155,000 = 14w
Simplify.
1,155,000 14w
=
14
14
82,500 = w
Lesson
Main
Divide each side by 14.
Use a calculator.
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
(continued)
The population of the city is 82,500 people.
Check for Reasonableness 82,500 is about 80,000. Since 14% of
80,000 is 11,200, which is close to 11,155, the answer is
reasonable.
Quick Check
Lesson
Main
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Additional Examples
26 is what percent of 80?
A diagram can help you
understand the
problem.
26
p
=
80
100
26 • 100 = 80p
Write a proportion.
Write the cross products.
2,600 = 80p
Simplify.
2,600
80p
=
80
80
32.5% = p
Lesson
Main
Divide each side by 80.
Simplify and insert a percent
sign.
Lesson
5-3
Feature
Percents and Proportions
LESSON 5-3
Course 3
Lesson Quiz
1. Find 25% of 160.
40
2. The price of a music CD is $12. If the store raises the price to 125%
of its current price, what will be the new price of the CD?
$15
3. So far, the sixth grade class has sold 32 tickets to their play. The
number represents 20% of the tickets that are available. How many
tickets are available?
160
4. 98 is what percent of 56?
175%
Lesson
Main
Lesson
5-3
Feature
Percents and Equations
LESSON 5-4
Course 3
Problem of the Day
Write the prime factorization of 364.
22  7  13
Lesson
Main
Lesson
5-4
Feature
Percents and Equations
LESSON 5-4
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-7.)
1. Vocabulary Review Is 2 • 8 = 16 an equation or an expression?
Explain.
Solve each equation.
2. 0.25p = 10
3. 12.25 = 9.8x
4. 24 = 1.6s
5. 0.64k = 0.02
Check Skills You’ll Need
Lesson
Main
Lesson
5-4
Feature
Percents and Equations
LESSON 5-4
Course 3
Check Skills You’ll Need
Solutions
1. Equation; it contains an = sign.
3.
12.25 9.8x
=
; x = 1.25
9.8
9.8
2. 0.25p = 10 ; p = 40
0.25
4.
0.25
24
= 1.6s ; s =15
1.6
1.6
5. 0.64k = 0.02 ; k = 0.03125
0.64
Lesson
Main
0.64
Lesson
5-4
Feature
Percents and Equations
LESSON 5-4
Course 3
Additional Examples
Quick Check
Misha got 84% correct on a 25 problem test. How
many did he answer correctly?
Words
number of problems correct is 84% of 25
Let c = the number of problems correct.
Equation
c
=
84% of 25
c = 0.84 • 25
c = 21
Simplify.
Check for Reasonableness 84% of 25 80% of 25. Since 80% of
25 is 20, which is close to 21, the answer is reasonable.
Lesson
Main
Lesson
5-4
Feature
Percents and Equations
LESSON 5-4
Course 3
Additional Examples
Use an equation. 12 is 8% of what number?
12 = 0.08 • w
12 0.08w
=
0.08 0.08
150 = w
Write a percent equation.
Divide each side by 0.08.
Simplify.
Quick Check
Lesson
Main
Lesson
5-4
Feature
Percents and Equations
LESSON 5-4
Course 3
Lesson Quiz
1. Find 81% of 110.
89.1
2. You buy a book for $17.80. Sales tax is 8%.
What is the sales tax cost of the book?
$1.42
3. 45 is 75% of what number?
60
4. Find what percent 68 is of 80.
85%
Lesson
Main
Lesson
5-4
Feature
Percent of Change
LESSON 5-5
Course 3
Problem of the Day
Dan, Susan, Monica, and Jose want to talk on the phone once to each of
the others. How many telephone calls will be made?
6 calls
Lesson
Main
Lesson
5-5
Feature
Percent of Change
LESSON 5-5
Course 3
Check Skills You’ll Need
(For help, go to Lesson 5-1.)
1. Vocabulary Review A
is a ratio that compares a number to 100.
Write each fraction as a percent. Round to the nearest tenth of a percent.
2.
9
8
3.
6
22
4.
4
15
5.
11
3
Check Skills You’ll Need
Lesson
Main
Lesson
5-5
Feature
Percent of Change
LESSON 5-5
Course 3
Check Skills You’ll Need
Solutions
1. percent
4.
4
= 0.26; 26.7%
15
Lesson
Main
2. 9 = 1.125; 112.5%
8
3.
6
= 0.27; 27.3%
22
11
5. 3 = 3.6; 366.7%
Lesson
5-5
Feature
Percent of Change
LESSON 5-5
Course 3
Additional Examples
Ten years ago, Max’s comic book was worth $2.50.
Now it is worth $13. Find the percent of increase in value.
amount of change = 13 – 2.50 = 10.50
P=
10.50
2.50
10.50
2.50
4.2
= 420%
amount of change
original amount
Use a calculator to divide.
Write the decimal as a percent.
The percent of increase in value is 420%.
Check for Reasonableness 420% of 2.5 400% of 3. Since 400%
of 3 = 12, which is close to 13, the answer is reasonable.
Quick Check
Lesson
Main
Lesson
5-5
Feature
Percent of Change
LESSON 5-5
Course 3
Additional Examples
Andre changed the height of his basketball hoop
from 8 ft 4 in. to 9 ft 2 in. Find the percent of increase.
8 ft 4 in. = 8 • 12 + 4 = 100 in.
9 ft 2 in. = 9 • 12 + 2 = 110 in.
Write measures in the
same units.
amount of change = 110 – 100 = 10
P=
10
100
amount of change
original amount
= 0.1
Simplify.
= 10%
Write the decimal as a percent.
The height of the basketball hoop increased by 10%.
Lesson
Main
Lesson
5-5
Quick Check
Feature
Percent of Change
LESSON 5-5
Course 3
Additional Examples
In 1980, the population of a city was 557,927.
In 1990, its population was 496,938. Find the percent
of decrease. Round to the nearest tenth.
amount of change = 557,927 – 496,938
= 60,989
P=
60,989
557,927
= 0.109313584
11%
amount of change
original amount
Use a calculator.
Write the decimal as a percent. Round to the
nearest tenth.
The population decreased by about 11%.
Lesson
Main
Lesson
5-5
Quick Check
Feature
Percent of Change
LESSON 5-5
Course 3
Lesson Quiz
Round to the nearest whole percent.
1. 81 people attended last year’s annual picnic, and 93 people attended
this year’s picnic. What is the percent of increase?
about 15%
2. The speed limit on a highway was 55 miles per hour last year. This year
the speed limit was increased to 65 miles per hour. What is the percent
increase in the speed limit?
about 18%
3. The population in Arthur County, Nebraska dropped from 462 in 1990 to
444 in 2000. What was the percent of decrease?
about 4%
Lesson
Main
Lesson
5-5
Feature
Markup and Discount
LESSON 5-6
Course 3
Problem of the Day
Round to the underlined place.
a. 0.09972
0.0997
Lesson
Main
b. 0.109
c. 17.51
0.1
18
Lesson
5-6
d. 0.998
1.00
Feature
Markup and Discount
LESSON 5-6
Course 3
Check Skills You’ll Need
(For help, go to Lesson 5-4.)
1. Vocabulary Review A
relates a part to the whole.
Use an equation to solve each problem.
2. What number is 16% of 25?
3. Find 80% of 250.
4. 33 is 3% of what number?
5. 0.55% of what number is 77?
Lesson
Main
Check Skills You’ll Need
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Check Skills You’ll Need
Solutions
1. percent
2. 0.16 • 25 = n; n = 4
3. 0.80 • 250 = n; n = 200
4. 0.03 • w = 33; w = 1,100
5. 0.0055 • w = 77; w = 14,000
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
Find the percent of markup for a stapler costing
the school store $2.10 and selling for $3.36.
markup = selling price – store’s cost
= $3.36 – $2.10
Substitute.
= $1.26
Subtract.
percent of markup =
1.26
2.10
markup
store’s cost
= 0.6
Write the fraction as a decimal.
= 60%
Write the decimal as a percent.
Quick Check
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
A store sells a skirt that costs the store $40 and
marks up the price 25%. What is the selling price for this
skirt?
Method 1 Find the markup first. Then find the selling price.
25% of $40 equals the markup.
0.25 • 40 = 10
Multiply to find the markup.
= $10
$40 + 10 = $50
store’s cost + markup = selling price
The store sells the skirt for $50.
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
(continued)
Method 2 Find the selling price directly.
The selling price equals 100% of the store’s cost plus a markup of
25% of the store’s cost.
So, the selling price of the skirt is 100% + 25%, or 125%, of $40.
125% of $40 equals the selling price.
1.25 • 40 = 50
= $50
Multiply to find the selling price.
The store sells the skirt for $50.
Quick Check
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
A shoe store advertises a 35%-off sale. What is the
sale price of shoes that regularly cost $94.99?
Method 1 Find the discount first. Then find the sale price.
35% of $94.99 equals the discount.
0.35 • 94.99 = 33.2465
= $33.25
94.99 – 33.25 = 61.74
Multiply to find the discount.
Round to the nearest cent.
regular price – discount = sale price
The sale price is $61.74.
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
(continued)
Method 2 Find the sale price directly.
The sale price equals 100% of the regular price minus a discount
of 35% of the regular price.
The sale price is 100% – 35%, or 65%, of $94.99.
65% of $94.99 equals the sale price.
0.65 • 94.99 = 61.744
= $61.74
Multiply to find the sale price.
Round to the nearest cent.
The sale price is $61.74.
Quick Check
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Additional Examples
You buy a CD at the sale price of $6. This is 25%
off the regular price. Find the regular price of the CD.
regular price – 25% of regular price = sale price
Let r = the regular price.
r – (0.25 • r ) = 6
Substitute. Write the percent as a decimal.
0.75r = 6
Combine like terms: r – 0.25r = 0.75r.
0.75r
6
=
0.75 0.75
Divide each side by 0.75.
r=8
Simplify.
Quick Check
The regular price of the CD is $8.
Lesson
Main
Lesson
5-6
Feature
Markup and Discount
LESSON 5-6
Course 3
Lesson Quiz
1. A pair of shoes costs the store $40. The store sells them for $65.
What is the percent markup?
62.5%
2. A school service club sells calendars. Each calendar
costs the club $5.50. The club marks up the price 80%.
What is the selling price of each calendar?
$9.90
3. A sweater regularly sells for $49. It is on sale for 20% off.
What is the sale price?
$39.20
4. You buy a baseball cap for $13. This price is 35% off the
regular price. Find the regular price.
$20.00
Lesson
Main
Lesson
5-6
Feature
Simple Interest
LESSON 5-7
Course 3
Problem of the Day
Luis’s mother has two older sisters who are twins and 6 yr older than she is.
She has a brother who is half her age. The sum of all their combined ages is
145 yr. How old is each?
mother 38, older twins 44, brother 19
Lesson
Main
Lesson
5-7
Feature
Simple Interest
LESSON 5-7
Course 3
Check Skills You’ll Need
(For help, go to Lesson 2-6.)
1.
Vocabulary Review A
quantities.
is a rule that shows a relationship between
Solve each formula for the underlined variable.
2. V = lwh
3. d = rt
4. y = x + b
5. V = 1 Bh
3
Check Skills You’ll Need
Lesson
Main
Lesson
5-7
Feature
Simple Interest
LESSON 5-7
Course 3
Check Skills You’ll Need
Solutions
1. formula
2. w = V
lh
3. r = d
t
4. b = y – x
5. B = 3V
h
Lesson
Main
Lesson
5-7
Feature
Simple Interest
LESSON 5-7
Course 3
Additional Examples
A student deposits $150 into a bank that pays 6%
simple interest. Find the interest earned in 4 years.
l=p•r•t
Use the simple interest formula.
= 150 • 0.006 • 4
Substitute: p = 150, r = 6% = 0.06, t = 4.
= 36
Multiply.
In 4 years, the interest earned is $36.
Quick Check
Lesson
Main
Lesson
5-7
Feature
Simple Interest
LESSON 5-7
Course 3
Additional Examples
Use the information from Example 1. Find the
final balance in the account after 5 years.
First, find the interest earned.
l=p•r•t
= 150 • 0.06 • 5
= 45
Substitute into the simple interest formula.
Multiply.
In 5 years, the interest earned is $45.
Next, find the final balance in the account.
principal + earned interest = balance
150
+
45
= balance
= 195
The final balance in the account is $195.
Lesson
Main
Lesson
5-7
Substitution.
Add.
Quick Check
Feature
Simple Interest
LESSON 5-7
Course 3
Lesson Quiz
1. You deposit $80 into an account that earns 6% simple
interest. Find the amount of interest earned in 3 years.
$14.40
2. You deposit $180 into an account that earns 6% simple
interest. How much will be in the account after 3 years?
$212.40
Lesson
Main
Lesson
5-7
Feature
Ratios and Probability
LESSON 5-8
Course 3
Problem of the Day
Which of the following numbers would give the smallest product?
4.9, 75.12, 15.02, 21.275, 8.61, 0.942, 47.38, 3.824
0.942, 3.824
Lesson
Main
Lesson
5-8
Feature
Ratios and Probability
LESSON 5-8
Course 3
Check Skills You’ll Need
(For help, go to Lesson 4-1.)
1. Vocabulary Review A
division.
is a comparison of two quantities by
Write each ratio in simplest form.
2. 3 : 6
3.
8h
100 h
4.
90 s
270 s
5.
20 cm
36 cm
Check Skills You’ll Need
6. 17 to 68
Lesson
Main
Lesson
5-8
Feature
Ratios and Probability
LESSON 5-8
Course 3
Check Skills You’ll Need
Solutions
2. 3 = 3 = 3 ÷ 3 = 1 , = or 1:2
1. ratio
6
3. 8 h = 8
100 h
5.
100
= 8÷4 = 2
20 cm = 20
36 cm 36
Lesson
Main
100 ÷ 4
25
20 ÷ 4 = 5
36 ÷ 4
9
Lesson
5-8
6
6÷3
2
4.
90 s = 90 = 90 ÷ 90 = 1
270 s
270 270 ÷ 90 3
6.
17 = 17 ÷ 17 = 1 , = or 1 to 4
68
68 ÷ 17 4
Feature
Ratios and Probability
LESSON 5-8
Course 3
Additional Examples
There are 3 red, 2 green, 5 yellow, and 1 blue
marker pens in a box. Suppose you choose one at random.
Find these probabilities.
a. P (yellow)
5
P (yellow) = 11
5 favorable outcomes
11 possible outcomes
5
The probability of choosing a yellow marker pen is 11 .
b. P (brown)
0
0 favorable outcomes
P (brown) = 11
11 possible outcomes
Write the fraction in simplest form.
=0
The probability of choosing a brown marker pen is 0.
Lesson
Main
Lesson
5-8
Quick Check
Feature
Ratios and Probability
LESSON 5-8
Course 3
Additional Examples
Quick Check
In a survey of the class, 13% of the students prefer
vanilla, 27% prefer chocolate, 10% prefer strawberry, and
the rest chose other flavors of ice cream. What is the
probability that a student randomly selected from the class
chose vanilla or chocolate?
13% of the students chose vanilla, and 27% of the students chose chocolate.
P(vanilla or chocolate) = P(vanilla) + P(chocolate)
= 13% + 27%
Substitute.
= 40%
Simplify.
The probability that the student chose vanilla or chocolate is 40%.
Lesson
Main
Lesson
5-8
Feature
Ratios and Probability
LESSON 5-8
Course 3
Additional Examples
Express as a fraction the probability that the
outcome for rolling two number cubes has a sum less than 7.
Make a table to find the sample space for rolling two number cubes.
1
2
3
4
5
6
1
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
2
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
3
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
4
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
5
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
6
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
Out of the 36 possible outcomes, the 15 outcomes shown in red
have a sum less than 7.
So, P(sum less than 7) =
Lesson
Main
15
5
, or
.
36
12
Lesson
5-8
Quick Check
Feature
Ratios and Probability
LESSON 5-8
Course 3
Additional Examples
What is the probability (as a fraction) of there being
at least 1 male kitten in a litter of 4 kittens? Draw
the sample space. Express the probability as a fraction.
Lesson
Main
Lesson
5-8
Feature
Ratios and Probability
LESSON 5-8
Course 3
Additional Examples
(continued)
P(at least one male) =
=
number of outcomes with at least one male kitten
total number of possible outcomes with four kittens
15
16
Substitute.
Quick Check
Lesson
Main
Lesson
5-8
Feature
Ratios and Probability
LESSON 5-8
Course 3
Lesson Quiz
1. A 6-sided number cube has the numbers 1, 2, 3, 4, 5, and 6 on its faces. What
is the probability of rolling a number less than 5? Write your answer as a
2
fraction.
3
2. A survey shows that 24% of people get their news from the internet, 48%
percent from TV, 22% from newspapers, and 6% from news magazines. If you
interviewed at random one person who answered the survey, what is the
probability that you would select someone who gets news from TV or the
internet?
72%
3.
A spinner has two equal sections, one yellow and one green. You spin 3 times
in a row. Make an organized list to show the sample space for spinning the
spinner 3 times. What is the probability of spinning green at least twice in a
row?
Sample space: YYY YYG YGY YGG GYY GYG GGY GGG
Probability of spinning green at least twice in a row is 3 .
8
Lesson
Main
Lesson
5-8
Feature