Example of linear equations, Algebra

Assignment Help:

In a certain Algebra class there is a total 350 possible points. These points come through 5 homework sets which are worth 10 points each and 3 hour exams that are worth 100 points each.  A student has attained homework scores of 4, 8, 7, 7, & 9 and the first two exam scores are 78 & 83.  Supposing that grades are assigned according to the standard scale and there are no weights assigned to any of the grades is it probable for the student to attain an A in the class and if so what is the minimum score on the third exam which will give an A? What about a B?

Solution

Let's begin by defining p to be the minimum needed score on the third exam.

Now, let's remember how grades are set.  As there are no weights or anything on the grades, the grade will be set by first calculating the following percentage.

                            actual points            / total possible points  =  grade percentage

As we are using the standard scale if the grade percentage is 0.9 or higher the student will get an A.  Similarly if the grade percentage is among 0.8 & 0.9 the student will get a B.

We know that the overall possible points is 350 and the student contain a total points (by including the third exam) of,

                                 4 + 8 + 7 +7 +9 + 78 + 83 + p = 196 + p

The smallest possible percentage for an A is 0.9 and thus if  p is the minimum needed score on the third exam for an A we will have the given equation.

                                                  196 + p/350 = 0.9

It is a linear equation which we will need to solve for p.

196 + p = 0.9 (350)= 315                  ⇒          p = 315 -196 = 119

Thus, the minimum needed score on the third exam is 119.  It is a problem as the exam is worth only 100 points.  In other terms, the student will not be getting an A in the Algebra class.

Now let's verify if the student will get a B.  In this case the minimum percentage is 0.8.  Thus, to determine the minimum required score on the third exam for a B we will have to solve,

                                   196 + p /350 = 0.8

Solving out this for p gives,

                                 196 + p = 0.8 (350) =280           ⇒        p = 280 -196 =84

Thus, it is possible for the student to get a B in the class. All that the student will have to do is get at least an 84 on the third exam.


Related Discussions:- Example of linear equations

Equations, if a-b equals 73 what is a

if a-b equals 73 what is a

Use the quadratic formula to solve the equation, Example :   Use the quadra...

Example :   Use the quadratic formula to solve following equation.                            x 2 + 2x = 7 Solution Here the important part is to ensure that before we b

Math, how do i do the fundamental counting principe

how do i do the fundamental counting principe

Write the equation for the perimeter, The image it''s a regtangle with y+2 ...

The image it''s a regtangle with y+2 length and x+4 width

Boolean logic-truth table and digital circuit diagram, Part A. relates to d...

Part A. relates to data representation and Part B. relates to Boolean logic.  Part A. Data Representation The very first thing you need to do to begin Part A is to make

Translating functions, f(x)=x square. graph g(x) by translating the graph ...

f(x)=x square. graph g(x) by translating the graph of f. g(x) = x square + 1

Sum of cubes, x= sum of 2 perfect cubes in two ways

x= sum of 2 perfect cubes in two ways

Standard test problems, a national park keeps track of how many people per ...

a national park keeps track of how many people per car enter the park. today, 57 cars had 4 people, 61 cars had 2 people, 9 cars had 1 person, and 5 cars had 5 people. what si the

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd