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Pre-Algebra HOMEWORK. Page 407 #17-27 & Spiral Review. Our Learning Goal.

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Our Learning Goal

Students will be able to determine and construct fractions, decimals, and percents by understanding their relationships, solving for the unknown, solving increase and decrease, solving sales tax/total cost contextual problems and compute simple interest.

Our Learning Goal Assignments

- Learn to relate decimals, fractions, & percents (8-1)
- Learn to find percents (8-2)
- Learn to find a number when the percent is known (8-3)
- Learn to find percent increase and decrease (8-4)
- Learn to estimate with percents (8-5)
- Learn to find commission, sales tax, and withholding tax (8-6)
- Learn to compute simple interest (8-7)

Learning Goal ScaleStudents are able to determine & construct fractions, decimals, & percents:

Finding Percents

24

50

4

1

25

4

3

8

Pre-Algebra

- Warm Up
- Rewrite each value as indicated.
- 1. as a percent
- 2. 25% as a fraction
- 3. as a decimal
- 4. 0.16 as a fraction

48%

0.375

A number between 1 and 10 is halved, and the result is squared. This gives an answer that is double the original number. What is the starting number?

8

Today’s Learning Goal Assignment

Learn to find percents.

Relative humidity is a measure of the amount of water vapor in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

88 in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

p =

Solve for p.

220

Additional Example 1A: Finding the Percent One Number Is of Another

A. What percent of 220 is 88?

Method 1: Set up an equation to find the percent.

p 220 = 88 Set up an equation.

p= 0.4 0.4 is 40%.

So 88 is 40% of 220.

number in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

part

100

=

whole

24

n

=

160

100

Additional Example 1B: Finding the Percent One Number Is of Another

B. Eddie weighs 160 lb, and his bones weigh 24 lb. Find the percent of his weight that his bones are.

Think:What number is to 100 as 24 is to 160?

Set up a proportion.

Substitute.

n 160 = 100 24

Find the cross products.

160n = 2400

2400 in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

n =

160

15

24

=

100

160

Additional Example 1 Continued

Solve for n.

n = 15

The proportion is reasonable.

So 15% of Eddie’s weight is bone.

11 in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

p =

Solve for p.

110

Try This: Example 1A

A. What percent of 110 is 11?

Method 1: Set up an equation to find the percent.

p110 = 11 Set up an equation.

p= 0.1 0.1 is 10%.

So 11 is 10% of 110.

number in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

part

100

=

whole

21

n

=

140

100

Try This: Example 1B

B. Jamie weighs 140 lb, and his bones weigh 21 lb. Find the percent of his weight that his bones are.

Think:What number is to 100 as 21 is to 140?

Set up a proportion.

Substitute.

n 140 = 100 21

Find the cross products.

140n = 2100

2100 in the air. When the relative humidity is 100%, the air has the maximum amount of water vapor. At this point, any additional water vapor would cause precipitation. To find the relative humidity on a given day, you would need to find a percent.

n =

140

15

21

=

100

140

Solve for n.

n = 15

The proportion is reasonable.

So 15% of Jamie’s weight is bone.

A. After a drought, a reservoir had only 66 % of the average amount of water. If the average amount of water is 57,000,000 gallons, how much water was in the reservoir after the drought?

2

2

2

2

3

3

3

3

Think:What number is 66 % of 57,000,000?

w = 66 % 57,000,000 Set up an equation.

w = 57,000,000 66 % is equivalent to .

2

2

3

3

Additional Example 2A: Finding a Percent of a Number

Choose a method: Set up an equation.

114,000,000 average amount of water. If the average amount of water is 57,000,000 gallons, how much water was in the reservoir after the drought?

3

w = = 38,000,000

Additional Example 2A Continued

The reservoir contained 38,000,000 gallons of water after the drought.

Set up a proportion. average amount of water. If the average amount of water is 57,000,000 gallons, how much water was in the reservoir after the drought?

=

110

a

100

550

Additional Example 2B: Finding Percents

B. Ms. Chang deposited $550 in the bank. Four years later her account held 110% of the original amount. How much money did Ms. Chang have in the bank at the end of the four years?

Choose a method: Set up a proportion.

110 550 = 100 a Find the cross products.

60,500 = 100a

Additional Example 2B Continued average amount of water. If the average amount of water is 57,000,000 gallons, how much water was in the reservoir after the drought?

605 = a Solve for a.

Ms. Chang had $605 in the bank at the end of the four years.

A. After a drought, a river had only 50 % of the average amount of water flow. If the average amount of water flow is 60,000,000 gallons per day, how much water was flowing in the river after the drought?

2

2

2

3

3

3

Think:What number is 50 % of 60,000,000?

w = 50 % 60,000,000 Set up an equation.

2

3

w = 0.506 60,000,000 50 % is equivalent to 0.506.

Try This: Example 2A

Choose a method: Set up an equation.

Try This amount of water flow. If the average amount of water flow is 60,000,000 gallons per day, how much water was flowing in the river after the drought?: Example 2A

w = 30,400,000

The water flow in the river was 30,400,000 gallons per day after the drought.

Set up a proportion. amount of water flow. If the average amount of water flow is 60,000,000 gallons per day, how much water was flowing in the river after the drought?

=

120

a

100

770

Try This: Example 2B

B. Mr. Downing deposited $770 in the bank. Four years later his account held 120% of the original amount. How much money did Mr. Downing have in the bank at the end of the four years?

Choose a method: Set up a proportion.

120 770 = 100 a Find the cross products.

92,400 = 100a

Try This amount of water flow. If the average amount of water flow is 60,000,000 gallons per day, how much water was flowing in the river after the drought?: Example 2B Continued

924 = a Solve for a.

Mr. Downing had $924 in the bank at the end of the four years.

Lesson Quiz amount of water flow. If the average amount of water flow is 60,000,000 gallons per day, how much water was flowing in the river after the drought?

Find each percent to the nearest tenth.

1. What percent of 33 is 22?

2. What percent of 300 is 120?

3. 18 is what percent of 25?

4. The volume of Lake Superior is 2900 mi3 and the volume of Lake Erie is 116 mi3. What percent of the volume of Lake Superior is the volume of Lake Erie?

66.7%

40%

72%

4%

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