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Weighted Averages Lesson 2-9A Mixture Problems. Splash Screen. Learning Goal. You translated sentences into equations. Solve mixture problems. Then/Now. Vocabulary.

daryl-bright
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  1. Weighted Averages Lesson 2-9A Mixture Problems Splash Screen

  2. Learning Goal You translated sentences into equations. Solve mixture problems. Then/Now

  3. Vocabulary • weighted average – The sum of the product of the number of units and the value per unit divided by the sum of the number of units, represented by M.

  4. Mixture Problems • mixture problems – problems where two or more parts are combined into a whole. • In mixture problems, the units are usually the number of gallons or pounds and the value is the cost, value, or concentration per unit. • It is usually helpful to set up a table for these problems.

  5. Mixture Problem PETSMandisha feeds her cat gourmetcat food that costs $1.75 per pound. She combines it with cheaper food that costs $0.50 per pound. How many pounds of cheaper food should Mandisha buy to go with 5 pounds of gourmet food, if she wants the average price to be $1.00 per pound? Let w = the number of pounds of cheaper cat food. Make a table. x = This is your equation Add column Price she wants Example 1

  6. Price of gourmet cat food plus price of cheaper cat food equals price of mixed cat food. 8.75 + 0.5w = 1.00(5 + w) Mixture Problem Write and solve an equation using the information in the table. 8.75 + 0.5w = 1.00(5 + w) Original equation 8.75 + 0.5w = 5 + 1w Distributive Property 8.75 + 0.5w– 0.5w= 5 + 1w – 0.5w Subtract 0.5w from each side. 8.75 = 5 + 0.5w Simplify. Example 1

  7. Mixture Problem 8.75 – 5 = 5 + 0.5w– 5 Subtract 5 from each side. 3.75 = 0.5w Simplify. Divide each side by 0.5. 7.5 = w Simplify. Answer: Mandisha should buy 7.5 pounds of cheaper cat food to be mixed with the 5 pounds of gourmet cat food so that the average price is $1.00 per pound of cat food. Example 1

  8. Cheryl bought 3 ounces of glass beads that cost $1.79 an ounce. The seed beads cost $0.99 an ounce. How many ounces of seed beads can she buy if she only wants the beads to be $1.29 an ounce for her craft project? x = Equation : 5.37 + 0.99s = 1.29(3 + s) Example 1

  9. Solve : 5.37 + 0.99s = 1.29(3 + s) 5.37 + 0.99s = 3.87 + 1. 29s Distributive Property 1.5 + 0.99s = 1.29s Subtraction Property of Equality 1.5 = 0.3s Subtraction Property of Equality 5 = s Division Property of Equality Cheryl would have to buy 5 ounces of seed beads for her project to make the total bead cost $1.29 per ounce .

  10. Percent Mixture Problem AUTO MAINTENANCEA car’s radiator should contain a solution of 50% antifreeze. Bob has 2 gallons of a 35% antifreeze. How many gallons of 100% antifreeze should Bob add to his solution to produce a solution of 50% antifreeze? Let g = the number of gallons of 100% antifreeze to be added. Make a table. What Bob has What Bob adds What Bob wants Example 2

  11. Amount of antifreeze in 35% solution plus amount of antifreeze in 100% solution equals amount of antifreeze in 50% solution. 0.35(2) + 1.0(g) = 0.50(2 + g) Percent Mixture Problem Write and solve an equation using the information in the table. 0.35(2) + 1.0(g) = 0.50(2 + g) Original equation 0.70 + 1g = 1 + 0.50g Distributive Property 0.70 + 1g– 0.50g= 1 + 0.50g– 0.50g Subtract 0.50g from each side. Example 2

  12. Percent Mixture Problem 0.70 + 0.50g = 1 Simplify. 0.70 + 0.50g– 0.70 = 1 – 0.70 Subtract 0.70 from each side. 0.50g = 0.30 Simplify. Divide each side by 0.50. g = 0.6 Simplify. Answer: Bae should add 0.6 gallon of 100% antifreeze to produce a 50% solution. Example 2

  13. A recipe calls for mixed nuts with 50% peanuts. pound of 15% peanuts has already been used. How many pounds of 75% peanuts needs to be addedto obtain the required 50% mix? x = Equation: 0.075 +.75x = .5(½ + x) Example 2

  14. Homework Mixture Problem worksheet Chapter 2 Review due Wednesday

  15. Weighted Averages Lesson 2-9B Uniform Motion Problems

  16. Learning Goal You translated sentences into equations. Solve uniform motion problems.

  17. Uniform Motion Problems • uniform motion problem – problems in which an object moves at a certain speed or rate. Also referred to as a rate problem. • Uniform Motion problems use the formula • d = rt • d represents distance • r represents rate • t represents time Vocabulary

  18. Speed of One Vehicle AIR TRAVELNita took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip? This is a Round Trip Problem. Nita traveled 750 mile to her grandmother’s and 750 miles back so her round trip was1500 miles. She traveled a 3.75 mi/hr to get there and 4.75 mi/hr on her return, for a round trip time of 8.5 hours. Note: you have to change minutes to hours. Example 3

  19. Speed of One Vehicle Use the formula M = Round Trip Distance ÷ Round Trip Time Answer: The average speed was about 176 miles per hour. CheckThe solution of 176 miles per hour is between the going portion rate 200 miles per hour, and the return rate, 157.9 miles per hour. So, the answer is reasonable. Example 3

  20. In the morning, when traffic is light, it takes 30 minutes to get to work. The trip is 15 miles through towns. In the afternoon, when traffic is a little heavier, it takes 45 minutes. What is the average speed for the round trip? Convert minutes to hours since we are looking for speed in mi/hr The average speed for the trip to and from work is 24 miles per hour M = 24 mi/hr Example 3

  21. Takes 5 miles to stop Takes 3 miles to stop Speeds of Two Vehicles RESCUEA railroad switching operator has discovered that two trains are heading toward each other on the same track. Currently, the trains are 53 miles apart. One train is traveling at 75 miles per hour and the other 40 miles per hour. The faster train will require 5 miles to stop safely, and the slower train will require 3 miles to stop safely. About how many minutes does the operator have to warn the train engineers to stop their trains? Step 1 Draw a diagram. 53 miles apart 53 – (5 + 3) = 45 miles Example 4

  22. Distance traveled by fast train plus distance traveled by other train equals 45 miles. 75m + 40m = 45 Speeds of Two Vehicles Step 2 Let m = the number of hours that the operator has to warn the train engineers to stop their trains safely. Make a table. Step 3 Write and solve an equation using the information in the table. Example 4

  23. Speeds of Two Vehicles 75m + 40m = 45 Original equation 115m = 45 Simplify. Divide each side by 115. m≈ 0.39 Round to the nearest hundredth. 0.39 × 60 = 23.4 Convert to minutes by multiplying by 60. Answer: The operator has about 23 minutes to warn the engineers. Example 4

  24. Two students left the school on their bicycles at the same time, one heading north and the other heading south. The student heading north travels 15 miles per hour, and the one heading south travels at 17 miles per hour. After about how many minutes will they be 7.5 miles apart? Let t = time 17t + 15t = 7.5 add the distances together to equal 7.5 32t = 7.5 t = .23 hrs convert the hours to minutes .23 x 60 = 14 min After 14 minutes the students will be 7.5 miles apart. Example 4

  25. Homework • Page 136 #6-21 all • Chapter 2 review due tomorrow

  26. End of the Lesson

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