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8-2

8-2. Finding Percents. Course 3. Warm Up. Problem of the Day. Lesson Presentation. 8-2. Finding Percents. 24. 50. 4. 1. 25. 4. 3. 8. Course 3. 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%.

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8-2

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  1. 8-2 Finding Percents Course 3 Warm Up Problem of the Day Lesson Presentation

  2. 8-2 Finding Percents 24 50 4 1 25 4 3 8 Course 3 • 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

  3. 8-2 Finding Percents Course 3 Problem of the Day 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

  4. 8-2 Finding Percents Course 3 Learn to find percents.

  5. 8-2 Finding Percents Course 3 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.

  6. 8-2 Finding Percents 88 p = Solve for p. 220 Course 3 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.

  7. 8-2 Finding Percents number part 100 = whole 24 n = 160 100 Course 3 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. Method 2: Set up a proportion to find the percent. 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

  8. 8-2 Finding Percents 2400 n = 160 15 24 = 100 160 Course 3 Additional Example 1 Continued Solve for n. n = 15 The proportion is reasonable. So 15% of Eddie’s weight is bone.

  9. 8-2 Finding Percents 11 p = Solve for p. 110 Course 3 Insert Lesson Title Here Try This: Example 1A A. What percent of 110 is 11? Method 1: Set up an equation to find the percent. p110 = 11 Set up an equation. p= 0.1 0.1 is 10%. So 11 is 10% of 110.

  10. 8-2 Finding Percents number part 100 = whole 21 n = 140 100 Course 3 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. Method 2: Set up a proportion to find the percent. 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

  11. 8-2 Finding Percents 2100 n = 140 15 21 = 100 140 Course 3 Solve for n. n = 15 The proportion is reasonable. So 15% of Jamie’s weight is bone.

  12. 8-2 Finding Percents 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 Course 3 Additional Example 2A: Finding a Percent of a Number Choose a method: Set up an equation.

  13. 8-2 Finding Percents 114,000,000 3 w = = 38,000,000 Course 3 Additional Example 2A Continued The reservoir contained 38,000,000 gallons of water after the drought.

  14. 8-2 Finding Percents Set up a proportion. = 110 a 100 550 Course 3 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

  15. 8-2 Finding Percents Course 3 Additional Example 2B Continued 605 = a Solve for a. Ms. Chang had $605 in the bank at the end of the four years.

  16. 8-2 Finding Percents 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. Course 3 Try This: Example 2A Choose a method: Set up an equation.

  17. 8-2 Finding Percents Course 3 Try This: Example 2A w = 30,400,000 The water flow in the river was 30,400,000 gallons per day after the drought.

  18. 8-2 Finding Percents Set up a proportion. = 120 a 100 770 Course 3 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

  19. 8-2 Finding Percents Course 3 Try This: Example 2B Continued 924 = a Solve for a. Mr. Downing had $924 in the bank at the end of the four years.

  20. 8-2 Finding Percents Course 3 Insert Lesson Title Here Lesson Quiz 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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