Question

shureka washburn has scores of 79, 79, 77, and 91 on her algebra tests.
a. use an inequality to find the scores she must make on the final exam to pass the course with an average of 78 or higher, given that the final exam counts as two tests.
b. explain the meaning of the answer to part (a).
a. the solution set is {x|□}. (type an inequality.)

Explanation:

Step1: Define total tests and scores

Let \( x \) be the score on the final exam. The final counts as 2 tests, so total tests: \( 4 + 2 = 6 \). Total scores: \( 79 + 79 + 77 + 91 + 2x \).

Step2: Set up average inequality

Average \( \geq 78 \), so \( \frac{79 + 79 + 77 + 91 + 2x}{6} \geq 78 \).

Step3: Simplify numerator

\( 79 + 79 + 77 + 91 = 326 \), so \( \frac{326 + 2x}{6} \geq 78 \).

Step4: Multiply both sides by 6

\( 326 + 2x \geq 78 \times 6 \), \( 326 + 2x \geq 468 \).

Step5: Subtract 326

\( 2x \geq 468 - 326 \), \( 2x \geq 142 \).

Step6: Divide by 2

\( x \geq 71 \).

Answer:

\( x \geq 71 \)