
a) The grades on the final to get A grade.
b) The grades on the final so as to lose B grade.
Solution:
a) With 100 marks in final exam, A is not possible.
b) To earn B, the score y in the final exam should be y ≥ 68 .
Explanation:
Given:
The given is that the two examination grades are 86 , 78 and 88 and in order to earn an A in the course, the final average should be atleast 90 . Also to earn a B in the course the final average should be atleast 8 0 .
Concept and Formula Used:
The concept used here is the concept of linear inequality. The concept of inequality includes two types of inequalities:
1) At most inequality: this inequality is less than equal to ( ≤ ) type inequality.
2) At least inequality: this inequality is more than equal to ( ≥ ) type inequality.
And the formula used here is the formula of an average.
Formula of the average = sum of the numbers whose average is to be taken out total number of observations
Calculation:
a) As to earn A, the final average should be atleast 90 , thus the inequality is of greater than type. As 100 is the score in the final, thus the average obtained is given by,
Formula of the average = sum of the numbers whose average is to be taken out total number of observations ⇒ Average = 88 + 86 + 78 + 100 4 = 352 4 = 88
As to earn A, the final average should be at least 90, but the average is 88 thus A is not possible.
b) As to earn B, the final average should be atleast 8 0 , thus the inequality is of greater than or equal to type.
Let y be the score in final exam, so according to the given situation the inequality is
Formula of the average = sum of the numbers whose average is to be taken out total number of observations ⇒ 86 + 88 + 78 + y 4 ≥ 80 ⇒ 4 ( 86 + 88 + 78 + y 4 ) ≥ 4 ( 80 ) (Multiplying both sides by 4) ⇒ 4 ( 86 + 88 + 78 + y ) 4 ≥ 320 ⇒ 86 + 88 + 78 + y ≥ 320 ⇒ 252 + y ≥ 320 ⇒ 252 + y − 252 ≥ 320 − 252 (Subtracting 252 from both the sides) ⇒ y ≥ 68
Thus to earn B, the score in the final exam should be greater than or equal to 68.
Conclusion:
a) With 100 marks in final exam, A is not possible.
b) To earn B, the score y in the final exam should be y ≥ 68 .
b) The grades on the final so as to lose B grade.
Solution:
- a) With 100 marks in final exam, A is not possible.
b) To earn B, the score
Explanation:
Given:
The given is that the two examination grades are
Concept and Formula Used:
The concept used here is the concept of linear inequality. The concept of inequality includes two types of inequalities:
1) At most inequality: this inequality is less than equal to
2) At least inequality: this inequality is more than equal to
And the formula used here is the formula of an average.
Calculation:
a) As to earn A, the final average should be
As to earn A, the final average should be at least 90, but the average is 88 thus A is not possible.
b) As to earn B, the final average should be
Let
Thus to earn B, the score in the final exam should be greater than or equal to 68.
Conclusion:
a) With 100 marks in final exam, A is not possible.
b) To earn B, the score

Want to see the full answer?
Check out a sample textbook solution
Chapter 2 Solutions
Introductory and Intermediate Algebra for College Students Access Card Package (5th Edition) (Blitzer Developmental Algebra Series)
- In simplest way, For each quadratic relation, find the zeros and the maximum or minimum. a) y = x 2 + 16 x + 39 b) y = 5 x2 - 50 x - 120arrow_forwardIn simplest terms and step by step Write each quadratic relation in standard form, then fi nd the zeros. y = - 4( x + 6)2 + 36arrow_forwardIn simplest terms and step by step For each quadratic relation, find the zeros and the maximum or minimum. 1) y = - 2 x2 - 28 x + 64 2) y = 6 x2 + 36 x - 42arrow_forward
- Write each relation in standard form a)y = 5(x + 10)2 + 7 b)y = 9(x - 8)2 - 4arrow_forwardIn simplest form and step by step Write the quadratic relation in standard form, then fi nd the zeros. y = 3(x - 1)2 - 147arrow_forwardStep by step instructions The path of a soccer ball can be modelled by the relation h = - 0.1 d 2 + 0.5 d + 0.6, where h is the ball’s height and d is the horizontal distance from the kicker. a) Find the zeros of the relation.arrow_forward
- Algebra and Trigonometry (6th Edition)AlgebraISBN:9780134463216Author:Robert F. BlitzerPublisher:PEARSONContemporary Abstract AlgebraAlgebraISBN:9781305657960Author:Joseph GallianPublisher:Cengage LearningLinear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning
- Algebra And Trigonometry (11th Edition)AlgebraISBN:9780135163078Author:Michael SullivanPublisher:PEARSONIntroduction to Linear Algebra, Fifth EditionAlgebraISBN:9780980232776Author:Gilbert StrangPublisher:Wellesley-Cambridge PressCollege Algebra (Collegiate Math)AlgebraISBN:9780077836344Author:Julie Miller, Donna GerkenPublisher:McGraw-Hill Education





