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Bell Ringer

Bell Ringer. In the standard ( x,y ) coordinate plane below, the points (0,0), ( 0,9), and (6, 0) are the vertices of a triangle. What is the area of the triangle?. A ∆ = ½ B • H. ( 0,9). A ∆ = ½ (6 • 9). 27 sq. units. A ∆ = 27 sq. units. 9 units. (6,0 ). (0,0). 6 units.

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Bell Ringer

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  1. Bell Ringer In the standard (x,y) coordinate plane below, the points (0,0), (0,9), and (6, 0) are the vertices of a triangle. What is the area of the triangle? A∆ = ½ B • H (0,9) A∆ = ½ (6 • 9) 27 sq. units A∆ = 27 sq. units 9 units (6,0) (0,0) 6 units

  2. Calculate the missing value when given the average and all other values! Focusing on Average

  3. The Average! • We will now focus specifically on average and how to find data points when given the average. • There are several types of problems that involve finding missing data points when you are given the average. • We will explore several of those scenarios today and then I’ll set you off on your own!

  4. Scenario 1 • On the first four tests, Chuck scored 84, 82, 79, and 90. If Chuck wanted a B (85 average), what must he score on the next test? • If you took 5 tests and had an 85 average, how many total points would you need? • How many points does Chuck have already? • How many points will he need to reach the total points required to have an 85 average? 85 • 5 = 425 points 84 + 82 + 79 + 90 = 335 points 425 – 335 = 90 points needed on the next test!

  5. Scenario 2 • Chuck has an 82 average after the first four tests. How many points must he score on the next test to reach his goal of an 85 average? • If he has an 82 average through four tests, how many points does he have? • How many points are required to have an 85 average? • How many points does he need to have an 85 average? 82 + 82 + 82 + 82 = 328 points 85 • 5 = 425 points 425 – 328 = 97 points needed!

  6. Scenario 3 • Chuck’s average is 82. He took five tests. His first four test scores were 74, 87, 79, and 78. What was his fifth test score? • If his average is an 82, how many points has he earned? • What is the total points earned on the four tests he knows about? • How many points did he score on the fifth test? 82 + 82 + 82 + 82 + 82 = 410 points 74 + 87 + 79 + 78 = 318 points 410 – 318 = 92 points!

  7. Grade Point Average • To calculate grade point average, each letter grade is assigned a point value. • This means an honors B is worth the same as a regulars A. You can’t achieve a GPA above 4.0 without honors courses. D’s are all worth 1 point, no matter the level!

  8. Grade Point Average • You will receive your first GPA after the 1st semester (2nd Quarter). • You are taking 7 classes. • All of your classes have equal importance. • Know which of your classes are honors classes. • Assign point values to each letter grade. • Divide by the number of classes!

  9. Student #1 What is student #1’s GPA? 1. Add up all of the points. 3 + 4 + 4 + 3 + 2 + 4 + 2 = 22 2. Divide by the number of classes. 22 ÷ 7 = 3.14 3. Use the regular point scale to determine your letter average. 3.14 is a little more than a B average. 3 4 4 3 2 4 2 Student #1 has a 3.14 GPA, which is a B average!

  10. Student #2 1. Add up all of the points. 2 + 3 + 1 + 2 + 4 + 3 + 1 = 16 2. Divide by the number of classes. 16 ÷ 7 = 2.29 3. Use the regular point scale to determine your letter average. 2.29 is a little more than a C average. 2 3 1 2 4 3 1 Student #2 has a 2.29 GPA, which is a C average!

  11. Student #3 1. Add up all of the points. 5 + 5 + 4 + 5 + 4 + 3 + 3 = 16 2. Divide by the number of classes. 29 ÷ 7 = 4.14 3. Use the regular point scale to determine your letter average. 4.14 is a little more than an A average. 5 5 4 5 4 3 3 Student #3 has a 4.14 GPA, which is an A average!

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