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Weighted Averages. An average is a direct measurement of the sum of the measurements divided by the number of things being measured. A weighted average.

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Presentation on theme: "Weighted Averages. An average is a direct measurement of the sum of the measurements divided by the number of things being measured. A weighted average."— Presentation transcript:

1 Weighted Averages

2 An average is a direct measurement of the sum of the measurements divided by the number of things being measured. A weighted average is skewed in one direction.

3 Price Problems How many pounds of mixed nuts selling for $4.75 per pound should be mixed with 10 pounds of dried fruit selling for $5.50 per pound to obtain a trail mix that sells for $4.95 per pound? Step 1: Draw a picture that models an equation. Step 2: Fill in the equation with the corresponding numbers. Remember to change percentages to decimals.

4 Solution 4.75(n) + 5.50(10) = (10 + n)(4.95) 4.75n + 55 = 49.5 + 4.95n 5.5 =.2n 27.5 = n It would take 27.5 lbs of mixed nuts

5 Mixtures A chemistry experiment calls for a 30% solution of copper sulfate. Kendra has 40 mL of 25% solution. How many mL of 60% solution should she add to make a 30% solution? 40(.25) + n(.6) = (40 + n)(.3) 10 +.6n = 12 +.3n.3n = 2 n ≈ 6.67 mL of the 60% solution

6 Uniform Motion The formula d = rt is used for these problems involving speed and movement. On Alberto’s drive to his aunt’s house, the traffic was light and he drove the 45 mile trip in one hour. However, the return trip took him two hours. What was his average speed for the round trip? We are looking for the rate of speed here, so r = d/t. Going there r = 45/1Coming home r = 45/2 Round Trip = 45(1) + 22.5(2) = 90 = 30 miles/hour 3 hours total3

7 Speed of Two Vehicles A car and an emergency vehicle are heading toward each other. The car is traveling at a speed of 30 miles per hour or about 44 feet per second. The emergency vehicle is traveling at a speed of 50 miles per hour or about 74 feet per second. If the vehicles are 1000 feet apart and the conditions are ideal, in how many seconds will the driver of the car first hear the siren? (Note: A siren can be heard from 440 feet under ideal conditions.) How far must the two vehicles travel in order to be 440 feet apart? The sum of the two distances must = 560. Car d = rt d = 44t EVd = rt d = 74t 44t + 74t = 560 118t = 560 t ≈ 4.75 seconds when the driver of the car will hear the siren

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