Question

approximately what portion of the beaker is filled? a. $\frac{5}{6}$ b. $\frac{2}{3}$ c. $\frac{6}{10}$

Explanation:

Step1: Analyze the beaker's filling level

Visually, the liquid in the beaker is close to the top but not completely full. Let's compare the fractions:

• $\frac{5}{6} \approx 0.833$, which would be very close to full.
• $\frac{2}{3} \approx 0.666$, a moderate fill.
• $\frac{6}{10} = 0.6$, also moderate but less than $\frac{2}{3}$.

Looking at the beaker, the liquid seems to fill most of it but not 83% (which $\frac{5}{6}$ is). It's more than half but less than very close to full. $\frac{5}{6}$ is too high, $\frac{6}{10}$ is 0.6, and $\frac{2}{3} \approx 0.666$ fits better visually. Wait, no—wait, maybe I misjudged. Wait, the beaker: let's see, the liquid is up to near the top, but the beaker's shape—wait, maybe the correct one is $\frac{5}{6}$? Wait, no, let's re - evaluate. Wait, the beaker in the image: the liquid is almost to the brim? Wait, no, the drawing shows the liquid level below the rim but quite high. Wait, $\frac{5}{6}$ is 5 out of 6 parts, so 83% full. $\frac{2}{3}$ is 66% full, $\frac{6}{10}$ is 60% full. Looking at the beaker, the liquid is more than 2/3? Wait, maybe the answer is A? Wait, no, maybe I made a mistake. Wait, let's check the fractions:

$\frac{5}{6} \approx 0.83$, $\frac{2}{3} \approx 0.67$, $\frac{6}{10}=0.6$. The beaker's liquid level: from the image, the liquid is almost filling the beaker, with a small space at the top. So $\frac{5}{6}$ is a better approximation than $\frac{2}{3}$ or $\frac{6}{10}$.

Step2: Compare the fractions to the visual

The beaker's liquid appears to fill most of the volume, with only a small portion empty. $\frac{5}{6}$ means only 1/6 is empty, which matches the visual (a small empty space at the top). $\frac{2}{3}$ would mean 1/3 is empty, which looks like more empty space than in the image. $\frac{6}{10}$ (or 3/5) would mean 2/5 is empty, which is even more. So the best approximation is $\frac{5}{6}$.

Answer:

A. $\frac{5}{6}$