Pre-Algebra Objectives: 1. To round decimals 2. To estimate sums and differences Rounding and...

Pre-AlgebraPre-Algebra

Objectives:1. To round decimals2. To estimate sums and differences

Rounding and EstimatingRounding and Estimating

Lesson 3-1


Tip: ≈ means approximately equal to


Lesson 3-1



Lesson 3-1

a. Round 8.7398 to the nearest tenth.

8.7398

tenths place

less than 5

8.7

Round down to 7.

Additional Examples


(continued)


Lesson 3-1

b. Round 8.7398 to the nearest integer.

8.7398

nearest integer is ones place

5 or greater

9

Round up to 9.

Additional Examples


Estimate to find whether each answer is reasonable.


Lesson 3-1

a. Calculation

+$ 59.98

$ 83.21

$115.67

$258.86+$ 60

$ 80

$120

Estimate

$260

The answer is close to the estimate. It is reasonable.

Additional Examples


(continued)


Lesson 3-1

b. Calculation

–$ 39.34

$176.48

$137.14

Estimate

–$ 40

$180

$140

The answer is not close to the estimate. It is not reasonable.

Additional Examples


Add thefront-end digits.

You are buying some fruit. The bananas cost $1.32,

the apples cost $2.19, and the avocados cost $1.63. Use

front-end estimation to estimate the total cost of the fruit.


Lesson 3-1

5.10=

The total cost is about $5.10.

+

Estimate byrounding.

.60

.20

.30

1.10

Additional Examples


Estimate the total electricity charge: March: $81.75; April: $79.56; May: $80.89.


Lesson 3-1

240=

The total electricity charge is about $240.00.

The values cluster around $80. 80

3 months

• 3

Additional Examples


Objectives:1.To estimate products2. To estimate quotients

Estimating Decimal Products and QuotientsEstimating Decimal Products and Quotients

Lesson 3-2


Tips:On multiple choice questions, sometimes you can eliminate answers by estimating.


Lesson 3-2


Estimate 6.43 • 4.7.


Lesson 3-2

Multiply.6 • 5 = 30

6.43 6 4.7 5 Round to the nearest integer.

6.43 • 4.7 30

Additional Examples


Joshua bought 3 yd of fabric to make a flag. The fabric

cost $5.35/yd. The clerk said his total was $14.95 before tax.

Did the clerk make a mistake? Explain.


Lesson 3-2

Multiply 5 times 3, the number ofyards of fabric.

5 • 3 = 15

5.35 5 Round to the nearest dollar.

The sales clerk made a mistake. Since 5.35 > 5, the actual cost should be more than the estimate. The clerk should have charged Joshua more than $15.00 before tax.

Additional Examples


The cost to ship one yearbook is $3.12. The total cost

for a shipment was $62.40. Estimate how many books were in

the shipment.


Lesson 3-2

3.12 3 Round the divisor.

The shipment is made up of about 20 books.

62.40 60 Round the dividend to a multiple of 3 that is close to 62.40.

60 ÷ 3 = 20 Divide.

Additional Examples


Is 3.29 a reasonable quotient for 31.423 ÷ 5.94?


Lesson 3-2

Since 3.29 is not close to 5, it is not reasonable.

5.94 6 Round the divisor.

31.423 30 Round the dividend to a multiple of 6 that is close to 31.423.

30 ÷ 6 = 5 Divide.

Additional Examples


Objectives:1.To find mean, median, mode, and range of a set of data.

2. To choose the best measure of central tendency.

Mean, Median, and ModeMean, Median, and Mode

Lesson 3-3


New Terms:1 Measures of Central Tendency – mean, median, mode of a collection of data.

2. Mean – is the sum of the data values divided by the number of data values, average.

3. Median – is the middle number when data values are written in order and there is an odd number of data values. For an even number of data values, the median is the mean of the two middle numbers.

4. Mode – is the data item that occurs most often. There can be one mode, more than one, or none.

5. Range – the difference between the greatest and least values in the data set.

6. Outlier – a data value that is much greater or less than the other data values.


Lesson 3-3


Six elementary students are participating in a one-week

Readathon to raise money for a good cause. Use the graph. Find

the (a) mean, (b) median, and (c) mode of the data if you leave

out Latana’s pages.


Lesson 3-3

a. Mean:

40 + 45 + 48 + 50 + 505=

2335=

46.6=

The mean is 46.6.

sum of data valuesnumber of data values

Additional Examples


(continued)


Lesson 3-3

b. Median: 40 45 48 50 50 Write the data in order.

The median is the middle number, or 48.

c. Mode: Find the data value that occurs most often.

The mode is 50.

Additional Examples



Lesson 3-3

a. $1.10 $1.25 $2.00 $2.10 $2.20 $3.50

No values are the same, so there is no mode.

b. 1 3 4 6 7 7 8 9 10 12 12 13

How many modes, if any, does each have? Name them.

c. tomato, tomato, grape, orange, cherry, cherry, melon, cherry, grape

There is one mode.

Both 7 and 12 appear more than the other data values.

Cherry appears most often.

Since they appear the same number of times, there are two modes.

Additional Examples



Lesson 3-3

a. Which data value is an outlier?

Use the data: 7%, 4%, 10%, 33%, 11%, 12%.

The data value 33% is an outlier. It is an outlier because it is 21% away from the closest data value.

b. How does the outlier affect the mean?

The outlier raises the mean by about 4 points.

12.8 – 8.8 = 4

Find the mean with the outlier.776 12.8

Find the mean without the outlier.445

8.8

Additional Examples



Lesson 3-3

a. the monthly amount of rain for a year

Which measure of central tendency best describes

each situation? Explain.

since the average monthly amount of rain for a year is not likely to have an outlier, mean is the appropriate measure.

Mean;

b. most popular color of shirt

Mode;

When the data have no outliers, use the mean.

When determining the most frequently chosen item, or when the data are not numerical, use the mode.

since the data are not numerical, the mode is the appropriate measure.

Additional Examples



Lesson 3-3

c. times school buses arrive at school

(continued)

since one bus may have to travel much farther than other buses, the median is the appropriate measure.

Median;

When an outlier may significantly influence the mean, use the median.

Additional Examples


Objectives: 1.To substitute into formulas2. To use the formula for the perimeter of a rectangle

Using FormulasUsing Formulas

Lesson 3-4


New Terms:1. Formula – an equation that shows a relationship between quantities that are represented by variables.

2. Perimeter – the distance around a figure.


Lesson 3-4



Lesson 3-4

Suppose you ride your bike 18 miles in 3 hours.

Use the formula d = r t to find your average speed.

d = r t Write the formula.

18 = (r )(3) Substitute 18 for d and 3 for t.

Divide each side by 3.183 =

3r

3

Simplify.6 = r

Your average speed is 6 mi/h.

Additional Examples


Use the formula F = + 37, where n is the number

of chirps a cricket makes in one minute, and F is the

temperature in degrees Fahrenheit. Estimate the temperature

when a cricket chirps 76 times in a minute.


Lesson 3-4

n4

The temperature is about 56°F.

F = + 37n4 Write the formula.

F = + 37764 Replace n with 76.

F = 19 + 37 Divide.

F = 56 Add.

Additional Examples


Find the perimeter of a rectangular tabletop with a

length of 14.5 in. and width of 8.5 in. Use the formula for the

perimeter of a rectangle, P = 2 + 2w.


Lesson 3-4

The perimeter of the tabletop is 46 in.

P = 2 + 2w Write the formula.

P = 46 Add.

P = 2(14.5) + 2(8.5) Replace with 14.5 and w with 8.5.

P = 29 + 17 Multiply.

Additional Examples


Solving Equations by Adding or Subtracting DecimalsSolving Equations by Adding or Subtracting Decimals

Lesson 3-5

Objectives:

1. To solve one-step decimal equations involving addition

2. To solve one-step decimal equations involving subtraction


Solve 6.8 + p = –9.7.


Lesson 3-5

6.8 + p = –9.7

6.8 – 6.8 + p = –9.7 – 6.8 Subtract 6.8 from each side.

p = –16.5 Simplify.

Check: 6.8 + p = –9.7

6.8 + (–16.5) –9.7 Replace p with –16.5.

–9.7 = –9.7

Additional Examples


Ping has a board that is 14.5 ft long. She saws off a

piece that is 8.75 ft long. Use the diagram below to find the

length of the piece that is left.


Lesson 3-5

The length of the piece that is left is 5.75 ft.

x = 5.75 Simplify.

x + 8.75 – 8.75 = 14.5 – 8.75 Subtract 8.75 from each side.

x + 8.75 = 14.5

Additional Examples


Solve –23.34 = q – 16.99.


Lesson 3-5

–23.34 = q – 16.99

–23.34 + 16.99 = q – 16.99 + 16.99 Add 16.99 to each side.

–6.35 = q Simplify.

Additional Examples


Alejandro wrote a check for $49.98. His new account

balance is $169.45. What was his previous balance?


Lesson 3-5

Alejandro had $219.43 in his account before he wrote the check.

Simplify.p = 219.43

Add 49.98 to each side.p – 49.98 + 49.98 = 169.45 + 49.98

p – 49.98 = 169.45

Equation p – 49.98 = 169.45

Let p = previous balance.

Words previous balance minus check is new balance

Additional Examples


Objectives:

1. To solve one-step decimal equations involving multiplication

2. To solve one-step decimal equations involving division

Solving Equations by Multiplying or Dividing DecimalsSolving Equations by Multiplying or Dividing Decimals

Lesson 3-6


Solve –6.4 = 0.8b.


Lesson 3-6

–6.4 = 0.8b

–6.40.8

0.8b0.8= Divide each side by 0.8.

–8 = b Simplify.

Check: –6.4 = 0.8b

–6.4 0.8(–8) Replace b with –8.

–6.4 = –6.4

Additional Examples


Every day the school cafeteria uses about 85.8

gallons of milk. About how many days will it take for the

cafeteria to use the 250 gallons in the refrigerator?


Lesson 3-6

Let x = number of days.

numberof days

Words times is 250 gallonsdaily milk

consumption

Equation • =85.8 x 250

Additional Examples


(continued)


Lesson 3-6

85.8x = 250

The school will take about 3 days to use 250 gallons of milk.

x = 2.914 . . . Simplify.

x 3 Round to the nearest whole number.

85.8x85.8 =

25085.8 Divide each side by 85.8.

Additional Examples


Solve –37.5 = .


Lesson 3-6

c–1.2

–37.5 =c

–1.2

Multiply each side by –1.2.–37.5(–1.2) = (–1.2)c

–1.2

45 = c Simplify.

Additional Examples


A little league player was at bat 15 times and had

a batting average of 0.133 rounded to the nearest thousandth.

The batting average formula is batting average (a) = .

Use the formula to find the number of hits made.


Lesson 3-6

hits (h)times at bat (n)

a =hn

0.133 =h

15 Replace a with 0.133 and n with 15.

Additional Examples


(continued)


Lesson 3-6

0.133(15) = (15)h

15 Multiply each side by 15.

Simplify.1.995 = h

2 h Since h (hits) represents an integer, round to the nearest integer.

The little league player made 2 hits.

Additional Examples


Using the Metric SystemUsing the Metric System

Lesson 3-7

Objectives:1.To identify appropriate metric measures2. To convert metric units


Choose an appropriate metric unit. Explain your choice.


Lesson 3-7

a. the width of this textbook

b. the mass of a pair of glasses

c. the capacity of a thimble

Milliliter; a thimble will hold only a small amount of water.

Gram; glasses have about the same mass as many paperclips, but less than this textbook.

Centimeter; the width of a textbook is about two hands, or ten thumb widths, wide.

Additional Examples


Choose a reasonable estimate. Explain your choice.


Lesson 3-7

a. capacity of a drinking glass: 500 L or 500 mL

b. length of a hair clip: 5 m or 5 cm

c. mass of a pair of hiking boots: 1 kg or 1 g

1 kg; the mass is about one half the mass of your math book.

5 cm; the length of a hair clip would be about 5 widths of a thumbnail.

500 mL; a drinking glass holds less than a quart of milk.

Additional Examples


Complete each statement.


Lesson 3-7

a. 7,603 mL = L

7,603 ÷ 1,000 = 7.603 To convert from milliliters to liters, divide by 1,000.

7,603 mL = 7.603 L

4.57 m = 457 cm

To convert meters to centimeters, multiply by 100.

4.57 100 = 457 cm

b. 4.57 m = cm

Additional Examples


A blue whale caught in 1931 was about 2,900 cm

long. What was its length in meters?


Lesson 3-7

Wordscentimeters per meter

length in centimeters ÷

length in meters=

2,900Equation 100 29÷ =

The whale was about 29 m long.

Additional Examples


Objectives:1.To solve complex problems by first solving simpler cases

Problem Solving Strategy: Act It OutProblem Solving Strategy: Act It Out

Lesson 3-8


Marta gives her sister one penny on the first day of October,

two pennies on the second day, and four pennies on the third day. She

continues to double the number of pennies each day. On what date will

Marta give her sister $10.24 in pennies?


Lesson 3-8

1 2 2 • 2 = 4 4 • 2 = 8 8 • 2 = 1616 • 2 = 32

Number ofpennies

Days after the first

012345

Amount

$0.01$0.02$0.04$0.08 $0.16$0.32

Additional Examples


(continued)


Lesson 3-8

2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 • 2 = 1024

You can tell from the pattern in the chart that you just need to count the number of 2’s multiplied until you reach 1,024, which is $10.24 in pennies.

Marta will give her sister $10.24 in pennies on October 11.

10 twos = 10 days after the first penny is given

Additional Examples

Pre-Algebra Objectives: 1. To round decimals 2. To estimate sums and differences Rounding and...

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