Question

Shureka Washburn has scores of 77, 87, 85, and 55 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 77 or higher, given
that the final exam counts as two tests.
b. Explain the meaning of the answer to part (a).

Answer 1

Part (a)

x = score on the final exam

The final exam counts as two regular tests. So what we really have are two copies of x in the set of scores like this

Set of scores = {77, 87, 85, 55, x, x}

We can think of this set of scores as basically the set of regular test scores.

Add up the scores and divide by the sample size 6 to compute the average. We want this average to be 77 or higher, so,

`\text{average} \ge 77\\\\\frac{77+87+85+55+\text{x}+\text{x}}{5} \ge 77\\\\\frac{304+2\text{x}}{6} \ge 77\\\\304+2\text{x} \ge 6*77\\\\304+2\text{x} \ge 462\\\\2\text{x} \ge 462-304\\\\2\text{x} \ge 158\\\\\text{x} \ge 158/2\\\\\text{x} \ge 79\\\\`

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Part (b)

The result
`\text{x} \ge 79` means that she needs an 79 or better on the final exam so that her overall course average is 77 or better.