QUESTION IMAGE
Question
- eight less than the product of a number n and
$\frac{1}{7}$ is no more than 95.
n $circ$ 8 $circ$ 95
n $circ$ 103
n $circ$
Step1: Translate the problem into an inequality
The problem says "Eight less than the product of a number \( n \) and \( \frac{1}{5} \) is no more than 95". The product of \( n \) and \( \frac{1}{5} \) is \( \frac{1}{5}n \), eight less than that is \( \frac{1}{5}n - 8 \), and "no more than" means \( \leq \), so the inequality is \( \frac{1}{5}n - 8 \leq 95 \).
Step2: Solve the inequality for \( n \)
First, add 8 to both sides of the inequality:
\( \frac{1}{5}n - 8 + 8 \leq 95 + 8 \)
\( \frac{1}{5}n \leq 103 \)
Then, multiply both sides by 5 to isolate \( n \):
\( 5\times\frac{1}{5}n \leq 103\times5 \)
\( n \leq 515 \)
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The inequality is \( \frac{1}{5}n - 8 \leq 95 \) and the solution is \( n \leq 515 \)