Question

ty received test grades of 78%, 78%, 72%, 83%, and 72%.
part: 0 / 6
(a) what grade would he need to make on the sixth test to get a c if a c is at least 75% but less than 80?
let x represent the score on the sixth test.
the average of the six tests is given by
\\(\frac{78 + 78 + 72 + 83 + 72 + x}{6}\\)
verbal model:
to earn a c, ty requires
75 select (average of test scores) select 80

Explanation:

Step1: Calculate sum of existing scores

First, we sum up the given test scores: \(78 + 78 + 72 + 83 + 72\). Let's compute that: \(78+78 = 156\), \(156 + 72 = 228\), \(228 + 83 = 311\), \(311 + 72 = 383\). So the sum of the first five scores is \(383\).

Step2: Set up the inequality for average

The average of six tests is \(\frac{383 + x}{6}\). For a C, the average must be at least \(75\) and less than \(80\). So we set up the compound inequality: \(75 \leq \frac{383 + x}{6} < 80\).

Step3: Solve the left - hand side of the inequality

Multiply all parts of the inequality by \(6\) to eliminate the denominator: \(75\times6\leq\frac{383 + x}{6}\times6<80\times6\). This simplifies to \(450\leq383 + x<480\).

Step4: Solve for x

Subtract \(383\) from all parts of the inequality: \(450 - 383\leq383 + x-383<480 - 383\). Calculating the differences: \(67\leq x<97\). But we also need to consider the context of a test score (usually between \(0\) and \(100\), but here we are limited by the upper bound of the average). Wait, let's re - check the average formula. Wait, the average is \(\frac{78 + 78+72 + 83+72 + x}{6}=\frac{383 + x}{6}\). We need \(75\leq\frac{383 + x}{6}<80\).

Left inequality: \(\frac{383 + x}{6}\geq75\)

Multiply both sides by \(6\): \(383+x\geq450\)

Subtract \(383\): \(x\geq450 - 383=67\)

Right inequality: \(\frac{383 + x}{6}<80\)

Multiply both sides by \(6\): \(383+x<480\)

Subtract \(383\): \(x<480 - 383 = 97\)

But we also know that a test score \(x\) is a percentage, so \(0\leq x\leq100\). Combining with our inequality \(67\leq x<97\). But we also need to make sure that the average is in the range for a C. Let's verify with the lower bound: if \(x = 67\), the average is \(\frac{383+67}{6}=\frac{450}{6}=75\), which is good. If \(x = 96\) (the largest integer less than \(97\)), the average is \(\frac{383 + 96}{6}=\frac{479}{6}\approx79.83\), which is less than \(80\).

But let's check the sum of the first five scores again: \(78+78 = 156\), \(72+83=155\), \(72\), so \(156 + 155=311\), \(311+72 = 383\). That's correct.

So the inequality for the average is \(75\leq\frac{383 + x}{6}<80\), and solving it gives \(67\leq x<97\). But since we are dealing with a test score, and typically test scores are between \(0\) and \(100\), and we need the average to be at least \(75\) and less than \(80\), the value of \(x\) (the sixth test score) must satisfy \(67\leq x<97\). But if we consider that the question is likely expecting an inequality or a range, and also, if we made a mistake in the sum:

Wait, \(78+78 = 156\), \(72+72 = 144\), \(156+144 = 300\), \(300 + 83=383\). Yes, that's correct.

So the verbal model should have \(75\leq\) (average of test scores) \(< 80\). So the first select is "≤" and the second select is "<".

Answer:

For the first "Select" (between 75 and (average of test scores)): \(\leq\)

For the second "Select" (between (average of test scores) and 80): \(<\)

And the range of \(x\) (the sixth test score) is \(67\leq x<97\) (if we were to solve for \(x\) completely). But for the "Select" boxes in the verbal model:

First select: \(\leq\)

Second select: \(<\)