4. ron asked 18 classmates whether they prefer granola bars over muffins. he used a calculator to compare…

Explanation:

Step1: Let the decimal be $x = 0.666666667$. Assume $x=\frac{n}{m}$. Since $0.666\cdots=\frac{2}{3}$, and the given decimal is a close - approximation of the repeating decimal $0.\overline{6}$.

Let $x = 0.666666667=\frac{y}{z}$. We know that if $x = 0.\overline{6}=\sum_{i = 1}^{\infty}6\times10^{-i}$, using the formula for the sum of an infinite geometric series $S=\frac{a}{1 - r}$ where $a = 0.6$ and $r=0.1$, $S=\frac{0.6}{1 - 0.1}=\frac{6}{9}=\frac{2}{3}$. Since $0.666666667\approx0.\overline{6}$, we can say the fraction is $\frac{2}{3}$.

Step2: To find the number of students who prefer granola bars over muffins.

We know that the ratio of students who prefer granola bars over muffins to the total number of students surveyed is $0.666666667\approx\frac{2}{3}$, and the total number of students surveyed is $n = 18$. Let the number of students who prefer granola bars be $k$. Then $\frac{k}{18}=\frac{2}{3}$. Cross - multiply: $3k=2\times18$. $3k = 36$, so $k = 12$.

Answer:

Part A: $\frac{2}{3}$ Part B: $12$