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Unformatted text preview: or an A is 0.9 and so if p is the minimum required score on the
third exam for an A we will have the following equation. 196 + p
350 This is a linear equation that we will need to solve for p. 196 + p = 0.9 ( 350 ) = 315 Þ p = 315 - 196 = 119 So, the minimum required score on the third exam is 119. This is a problem since the exam is
worth only 100 points. In other words, the student will not be getting an A in the Algebra class.
Now let’s check if the student will get a B. In this case the minimum percentage is 0.8. So, to
find the minimum required score on the third exam for a B we will need to solve, 196 + p
350 Solving this for p gives, 196 + p = 0.8 ( 350 ) = 280 Þ p = 280 - 196 = 84 So, it is possible for the student to get a B in the class. All that the student will need to do is get
at least an 84 on the third exam. © 2007 Paul Dawkins 72 http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 2 We want to build a set of shelves. The width of the set of shelves n...
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- Spring '12