## College Algebra (6th Edition)

The student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A.
To solve this exercise, we must first model the average grade in mathematical terms: $$Grade_{average} = \frac{Grade_{1} + Grade_{2} +...Grade{(n)}}{n}$$ Since we have a total of 4 grades given, but the final examination counts as 2 exams, we have a total $n = 6$. We can now write the following: $$Grade_{average} = \frac{86 + 88 + 92 + 84 + 2E}{6}$$ where $E$ represents the final examination. Since an A is defined as 90% or greater, we can finally model this as so: $$Grade_{average}\geq 90$$ $$\frac{86 + 88 + 92 + 84 + 2E}{6} \geq 90$$ By solving for $E$: $$86 + 88 + 92 + 84 + 2E \geq 90\times 6$$ $$350 + 2E \geq 540$$ $$2E \geq 190$$ $$E \geq 95$$ This means that the student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A.