We can see from the question that for Andre's entry-level job a company offered him:
• $46,000 per year plus 2.8% of his total sales.
,
• We need to determine the total sales for Andre to be for his pay to be ,at least (as high as the average),, that is, ,equal to or greater,, than the average pay for this job ($46,000).
1. Now, we can translate this situation as follows -if we have that n represents Andre's sales amount:
[tex]46000+\frac{2.8}{100}n\ge69000[/tex]
2. To solve the total sales, we can proceed as follows:
• Subtract 46000 from both sides of the inequality:
[tex]\begin{gathered} 46000-46000+\frac{2.8}{100}n\ge69000-46000 \\ \\ \frac{2.8}{100}n\ge23000 \end{gathered}[/tex]
3. Now, we can multiply both sides by 100/2.8 as follows:
[tex]\begin{gathered} \frac{100}{2.8}\frac{2.8}{100}n\ge23000*\frac{100}{2.8} \\ \\ \\ n\ge821428.571429 \end{gathered}[/tex]
If we round our value for the total sales, n, to two decimal places, then we have:
[tex]n\ge\$821428.57[/tex]
Therefore, in summary, Andre's total sales would need to be greater or equal to $821,428.57 for his pay to be at least as high as the average pay for this job ($69,000), and we can write it - algebraically - as follows:
[tex]n\ge\$821428.57[/tex]
