Question

a career-placement test eliminates a profession when a person receives a score of 18 or less. the score is equal to the formula 2x - 3y, where x is the number of positive responses and y is the number of negative responses. which graph represents the range of test results that would be eliminated under this scenario (all points may not apply to the scenario)?

Explanation:

Step1: Formulate the inequality

The score is \(2x - 3y\) and elimination occurs when \(2x - 3y\leq18\) (since score \(\leq18\)). Rearrange this to slope - intercept form (\(y = mx + b\)):
Subtract \(2x\) from both sides: \(-3y\leq - 2x + 18\)
Divide both sides by \(-3\) (remember to reverse the inequality sign): \(y\geq\frac{2}{3}x - 6\)

Step2: Analyze the boundary line and shading

The boundary line for the inequality \(y\geq\frac{2}{3}x - 6\) has a slope of \(\frac{2}{3}\) and a \(y\) - intercept of \(-6\). Since the inequality is \(y\geq\) (greater than or equal to), the line should be dashed (if it were \(y\geq\) with a non - strict inequality, but wait, our original inequality after rearrangement: when we have \(2x-3y\leq18\), the boundary line \(2x - 3y=18\) or \(y=\frac{2}{3}x - 6\) should be dashed? Wait, no. Wait, the original condition is "score of 18 or less", so \(2x - 3y\leq18\). The boundary line \(2x - 3y = 18\) (or \(y=\frac{2}{3}x-6\)) is included (because of the "or equal to" in \(\leq\)), but in the graph, if the line is dashed, maybe there is a mis - step? Wait, no, let's check the slope and intercept. The slope of \(y=\frac{2}{3}x - 6\) is positive \(\frac{2}{3}\), and the \(y\) - intercept is \(-6\). Now, let's check the shading. For the inequality \(2x - 3y\leq18\), we can rewrite it as \(-3y\leq - 2x + 18\), then \(y\geq\frac{2}{3}x - 6\) (since we divided by a negative number, we flip the inequality). So we shade above the line \(y = \frac{2}{3}x-6\). Now, looking at the given graph: the line has a slope of \(\frac{2}{3}\) (rise over run: from \(y=-6\) (when \(x = 0\)) to, for example, when \(x = 9\), \(y=\frac{2}{3}(9)-6=6 - 6 = 0\), so the line passes through \((0,-6)\) and \((9,0)\), which matches the slope of \(\frac{2}{3}\). And the shading is above the line (since \(y\geq\)), which is what we need for \(2x - 3y\leq18\) (because \(y\geq\frac{2}{3}x - 6\) is equivalent to \(2x - 3y\leq18\)). So the graph with the line \(y=\frac{2}{3}x - 6\) (dashed or solid? Wait, in the problem statement, it says "all points may not apply to the scenario", so maybe the line is dashed. But the key is the slope and the intercept and the direction of shading. The line has a slope of \(\frac{2}{3}\), \(y\) - intercept at \(-6\), and shading above the line, which corresponds to the inequality \(2x - 3y\leq18\), which is the region where the score is 18 or less (eliminated region).

Answer:

The graph with the dashed line \(y = \frac{2}{3}x-6\) (or the line passing through \((0, - 6)\) and \((9,0)\)) and shading above the line (the blue - shaded region in the given graph that matches this description) is the correct one. If we assume the first graph (the one shown with the line from \((0,-6)\) going up with slope \(\frac{2}{3}\) and shading above) is the correct one, then the answer is the graph as described (with the line \(y=\frac{2}{3}x - 6\) and shading above it).