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Concept Review

Concept Review.

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Concept Review

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  1. Concept Review Purpose: The purpose of the following set of slides is to review the major concepts for the math course this year. Please look at each problem and how to solve them. Then attempt to solve the second problem on your own. If you have difficulty it would be good to come see me and ask questions.

  2. Ratios • You have a bag full of 500 yellow marbles and 1,000 blue marbles. What is the ratio of yellow to blue marbles? Please simplify.

  3. Solution • The first thing to do is to identify the quantity of each color of marble. • There are 500 yellow • There are 1000 blue • The ratio without simplifying is 500:1,000 • Now identify what each quantity is divisible by. In this case they are both divisible by 100. • This results in a ratio of 5:10. • This can be simplified further, because both are divisible by 5. The final simplified ratio is 1:2 • Extension: What is the probability of removing a yellow marble on one reach into the bag of marbles.

  4. Ratio Practice Problems • You have a bag full of 200 red marbles and 600 white marbles. What is the ratio of red to white marbles? Please simplify. • You have a bag full of 350 turquoise marbles and 625 pink marbles. What is the ratio of turquoise to pink marbles? Please simplify.

  5. Ratio Table • Complete the ratio table

  6. Ratio Table • Complete the ratio table Solution • The rule is to multiply by 3

  7. Ratio Table Practice Problem • Complete the ratio

  8. Ratios Continued • John has 30 pens and 45 pencils. Sally has 47 pens and 47 pencils. Who has a higher ratio of pens to pencils. • Solution: In this case Sally has more pens per every pencil, because she has a one to one ratio whereas John has a less than one to one ratio of pens to pencils.

  9. Practice Ratio Problem Continued • Jake has 40 red marbles to 40 blue marbles. Jenny has 35 red marbles to 45 blue marbles. Who has the higher red to blue marble ratio?

  10. Functions • Function n = b + 5 • Complete the table

  11. Functions • Function n = b + 5 • Complete the table Solution

  12. Functions Practice Problem • Function n = b + 12 • Complete the table

  13. Division • There are 49 pears that need to go in 7 crates. How many pears will end up in each box?

  14. Division • There are 49 pears that need to go in 7 crates. How many pears will end up in each box? • Solution 7 • 49 crate = - 49 0 7 7 7 7 7 7 7 *Remember the mathematical relationship between division and multiplication which is linked to addition and subtraction.

  15. Division Practice Problem • There are 64 apples. Please evenly distribute the apples into 8 baskets.

  16. Division of Fractions Review • Key concepts and terms • Improper fraction Complex Fraction, and Mixed Number

  17. Improper Fraction • A fraction in which the numerator is greater than the denominator. 12 4 Hint: 12/4 is equal to 3 wholes. The different ways to express fractions are representing the same amounts in different forms.

  18. Complex Fraction • A fraction in which either the numerator, denominator, or both contain fractions. ½ Numerator 2 ½ ¾ Denominator ¾ 3

  19. Mixed Numbers • A mixed number is a whole number and a fraction expressed together. 1 whole 1/4 • 1 ¼ • Hint: as an improper fraction this would be 5/4. 1/4 1/4

  20. Mixed Numbers • A mixed number is a whole number and a fraction expressed together. 1 whole 1/4 • 1 ¼ • Hint: as an improper fraction this would be 5/4. 1/4 1/4

  21. Division of Fraction problems • The procedure when dividing by fractions is to invert the second fraction and multiply, but keep in mind the models that were presented on the first three slides. They help to explain why this is the case. • 1 ½ 3/4 Solution: Convert 1 and ½ to an improper fraction. To do this multiply the whole number time the denominator and add the numerator 3/2 ¾ (1 ½ is equal to 3/2, because there are 3 halves in 1 ½) Now invert ¾ so the problem reads 3/2 X 4/3 You now have 12/6 which means you have 2 wholes. For further explanation please refer back to the garden problem on the website under the trimester 2 section.

  22. Practice Problems for Division of Fractions • 3/4 6/2 = • 1 ¼ 3/4 = • 2/3 5/6 =

  23. Area, Surface Area and Volume • Our goal this year is to identify how to find the area, surface area and volume of basic shapes, squares, cubes, rectangle, rectangular prisms, triangles, and triangular prisms. We can then use these basic shapes to break down more complex shapes to find the area, surface area, and volume.

  24. Area • When working with area we must remember that we are referring to the inside of a shape or figure. This is different than perimeter that looks at the distance around the outside of a shape or figure. The area of the rectangle is 9 x 4 or 36 units squared, because there are 36 1 X 1 smaller squares inside. The formula is L X W The perimeter would be 9 + 4 +9 +4 = 26units. Notice it is not squared. This would be like building a fence around the rectangle. The formula is L + W + L + W 9 units 4 units

  25. Area • Math contains many patterns and relationships. This slide addresses the relationship between rectangles and triangles. • How many triangles are inside of the rectangle? • Knowing that a triangle is ½ of a rectangle or square we can deduce that the formula for area of a triangle is L X W . Once again we are talking about square area. 2

  26. Surface area • When we represent 3 dimensional shapes and objects on paper we have to let our eyes imagine in 3 dimensions. For this slide looking at a 3 dimensional figure may help you see what the figure is conveying. Pick up a shoe box. How many flat surfaces can you identify?

  27. Surface areaContinued • To find the surface area you need to identify all six sides. The front side is equal to the back and the side to the opposing side. Finally the top and bottom should have equal dimensions. To find the surface area simply find the area of each side and then add them together.

  28. Surface Area There are two squares on the front and back 4 X 4 = 16, now multiply it by 2 = 32 units squared There are four 4 X 9 rectangles which are the two sides and top and bottom. 4 X9 = 36 units squared. Multiply 36 X 4 and you have 144 units units squared. Now add 144 to 32 for a total surface area of 176 units squared. 9 units 9 units 4 units 4 units

  29. Volume • Volume address the space inside of a shape or figure. For example how many ice cubes could you fit in the rectangle that are 1 X 1 X 1 units cubed? To solve this you multiply the L X W X H. • Therefore 4 X 4 X 9 = 144 units cubed, or there are 144 1 X 1 X1 little cubes inside the rectangle. • A rectangular prism would be • half. In other words divide it by two 9 units 9 units 4 units 4 units

  30. Practice • Find the area 7 inches squared 4 inches squared 4 inches squared 7 inches squared

  31. Find the Surface area 9 inches 3 inches 3inches

  32. Find the Volume 9 inches 9 inches 3 inches 3 inches 3 inches 3inches

  33. More Complex Shapes • Knowing how to find the area and volume of simple shapes can help us break down more complex shapes. For example a trapezoid. • We can see that we have two triangles and a rectangle.

  34. More Complex Shapes • There are different ways to approach solving this problem. One would be to find the area of the rectangle and the two triangles and add them. If the triangles have equal bases you could combine the triangle into a square.

  35. Measures of Central Tendency

  36. Mean Median and Mode Measures of Central Tendency x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80

  37. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 The mode is the data point that occurs the most. The mode of this data is 40.

  38. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 The median is the middle of the data. It helps to string out the data. 10, 20 , 20, 30, 30, 30, 40, 40, 40, 40, 50, 50, 50, 60, 60, 70. The median is 40.

  39. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 The mean is related to the distance between the data points. Imagine that each data point is a sticky note. You could divide the distance between 10 and 70 and arrive at forty. If you this over and over you would eventually arrive at the mean. The algorithm to solve it simply States , add all the data points and divide by the number of data points.

  40. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 Mean 10 + 20 + 20+30+30+30+40+40+40+40+50+50+50+60+60+70 = 640 Dividing by 640 by 16 you get the mean of 40.

  41. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 Challenge what will happen to the mean if we changed 70 to 90? Would the mean move To the right or the left? How do you know?

  42. Mean Median and Mode Continued x Mean = 41.25 x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 90 The total of the data points now is 660 and 660 divided by 16 is 41.25 so the curve is skewed to the right.

  43. Mean Median and Mode Continued x x x x x x x x x x x x x x x x • 0 10 20 30 40 50 60 70 80 The spread of the data deals with the distance between data points and the relation to the mean. Think about which of the two data distributions are more spread out. Is it the one with a balanced curve of the one that is skewed to the right? Take a moment to think about it.

  44. Mean Absolute Deviation • First we must find the mean. 10 + 30 +70 = 110 divided by 3 = 36.6 repeating. x x x 10 30 70

  45. Mean Absolute Deviation • Once we have the mean we then find the distance between the mean and the data points. I rounded the mean to 36.7. 36.7-10 = 26.7 36.7-30 =16.7 70- 36.7 = 33.3 x x x 10 30 70

  46. Mean Absolute Deviation • Now add the distances up and divide by the number of distances. 26.7 + 16.7 + 33.3 = 76.7 76.7 divided by 3 = 25.56 repeating or 25.57 There for the MAD is 25.57 x x x 10 30 70

  47. Mean Absolute DeviationApplication • Making comparisons between data sets and making meaning out of a single data set is inferential statistics or drawing inferences from the data. If the following two schools both had the following data sets and are teaching the same curriculum. Which one is getting more consistent results even if the averages are the same. MAD would be a very useful statistic in this case. x X x x x x x *** Be careful not to make judgments based on one statistic i.e. mean, median, mode, or MAD. Look at the data as many ways as you can to include models like above. Your looking for the story that the data is trying to tell you.

  48. Mean Absolute Deviation • What this allows us to do is compare the spread of a set of data to another. If a set of data has a mad of 25.57 versus a second data distrubution with a MAD of 5.6 which one is clumped more closely together? x X x x x x x The larger the mad the more spread out the data.

  49. Unit Rate Problems • When you are driving in your car and look at your speedometer it gives you a unit rate. For example a car traveling at 60 miles per hour literally means that you can travel 60 miles in 1 hour. Unit rate can be applied in many situations. 60 miles 60 miles 1 hour 1 hour

  50. Unit Rate Problems • What if you were bottling soda at a factory and new that you could bottle 60 sodas in 1 hours. What is your rate per hour? What is your rate per minute? What is your rate per second? 60 sodas 60 sodas 1 hour 1 hour

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