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

Change of base formula - logarithms, The last topic that we have to discuss...

The last topic that we have to discuss in this section is the change of base formula. Most of the calculators these days are able of evaluating common logarithms & natural logar

Quarditic, what is quarditic equation

what is quarditic equation

., what the hell is the problem to this solution .

what the hell is the problem to this solution .

Quadratic equations, In the earlier section we looked at using factoring & ...

In the earlier section we looked at using factoring & the square root property to solve out quadratic equations. The problem is that both of these methods will not always work. Not

Problem solving .use multiplacation, ellie uses 12.5 pounds of potatoes to ...

ellie uses 12.5 pounds of potatoes to make mashed potatoes. she uses one-tenth as many pounds of butter as potatoes. how many pounds of butter does ellie use

Math, Two sides of an isosceles triangle are 7and 3. The perimeter of the t...

Two sides of an isosceles triangle are 7and 3. The perimeter of the triangle is?

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