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

Word problem.., In a certain class, one half of the male students and two t...

In a certain class, one half of the male students and two thirds of the female students speak Freanch. If there are three fourths as many girls as boys in the class. What fractio

Evaluate, 9-4x, for x=5 what is the answer

9-4x, for x=5 what is the answer

Homework question help, Karen is 62 1/2 Inches tall . Her mother is 5 3/4 ...

Karen is 62 1/2 Inches tall . Her mother is 5 3/4 inches taller. How tall is Karen''s mother?

Basic Algebra, Find the following product (-4)(-2)(5)=

Find the following product (-4)(-2)(5)=

Process for graphing a polynomial, 1. Find out all the zeroes of the polyno...

1. Find out all the zeroes of the polynomial and their multiplicity.  Utilizes the fact above to find out the x-intercept which corresponds to each zero will cross the x-axis or on

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