Question

On two​ examinations, you have grades of 88 and 86. There is an optional final​ examination, which counts as one grade. You decide to take the final in order to get a course grade of​ A, meaning a final average of at least 90.
a. What must you get on the final to earn an A in the​ course?
b. By taking the​ final, if you do​ poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than​ 80, you will lose your B in the course. Describe the grades on the final that will cause this to happen.
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Part 1
a. You must get at least a
  
enter your response here.

Answer 1

To find out what grade you need on the final exam to achieve an average of at least 90, you can set up the following equation:

Let \( x \) be the grade you need on the final exam.

Your grades on the two examinations are 88 and 86. The final exam counts as one additional grade. Therefore, you will have three grades in total: 88, 86, and \( x \).

**a. What must you get on the final to earn an A in the course?**

To find the average grade, you will calculate:
\[
\text{Average} = \frac{88 + 86 + x}{3}
\]

To earn an A, this average must be at least 90:
\[
\frac{88 + 86 + x}{3} \geq 90
\]

First, combine the grades:
\[
\frac{174 + x}{3} \geq 90
\]

Next, multiply both sides by 3 to eliminate the fraction:
\[
174 + x \geq 270
\]

Now, solve for \( x \):
\[
x \geq 270 - 174
\]
\[
x \geq 96
\]

So, **you must get at least a 96 on the final exam to earn an A in the course**.

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**b. Describe the grades on the final that will cause you to lose your B in the course.**

To lose a B, your final average must be less than 80. Using the same average formula:
\[
\frac{88 + 86 + x}{3} < 80
\]

Combine the grades:
\[
\frac{174 + x}{3} < 80
\]

Multiply both sides by 3:
\[
174 + x < 240
\]

Now, solve for \( x \):
\[
x < 240 - 174
\]
\[
x < 66
\]

So, **if you score less than 66 on the final exam, you will lose your B in the course**.