Question

order these numbers from least to greatest.
\\(\frac{79}{20}\\), 3.903, 3.49, \\(3\frac{10}{11}\\)
\\(\frac{79}{20}\\) 3.903 3.49 \\(3\frac{10}{11}\\)
\\(3\frac{10}{11}\\) < 3.49 < \\(\frac{79}{20}\\) < 3.903

Explanation:

Step1: Convert fractions to decimals

Convert \(\frac{79}{20}\) to decimal: \(\frac{79}{20}=79\div20 = 3.95\)
Convert \(3\frac{10}{11}\) to decimal: \(3\frac{10}{11}=3 + \frac{10}{11}\approx3 + 0.9091 = 3.9091\)

Step2: Compare decimals

Now we have the numbers as decimals: \(3.9091\) (for \(3\frac{10}{11}\)), \(3.49\), \(3.95\) (for \(\frac{79}{20}\)), and \(3.903\).
Compare the values:
\(3.49<3.903<3.9091<3.95\)

So the correct order from least to greatest is: \(3.49 < 3.903 < 3\frac{10}{11}<\frac{79}{20}\) (Wait, no—wait, let's re - check the decimal conversions:

• \(3.49\) is \(3.49\)
• \(3.903\) is \(3.903\)
• \(3\frac{10}{11}\approx3.9091\) (since \(\frac{10}{11}\approx0.9091\))
• \(\frac{79}{20}=3.95\)

Now order them: \(3.49<3.903<3\frac{10}{11}<\frac{79}{20}\)

Wait, the original incorrect order in the problem was \(3\frac{10}{11}<3.49<\frac{79}{20}<3.903\), which is wrong. Let's do the comparison properly:

1. \(3.49\) is the smallest.
2. Then \(3.903\) (since \(3.903 < 3.9091\) (which is \(3\frac{10}{11}\)) and \(3.903<3.95\) (which is \(\frac{79}{20}\))).
3. Then \(3\frac{10}{11}\approx3.9091\) (since \(3.9091<3.95\)).
4. Then \(\frac{79}{20}=3.95\).

So the correct order is \(3.49 < 3.903 < 3\frac{10}{11}<\frac{79}{20}\)

Answer:

The correct order from least to greatest is \(3.49 < 3.903 < 3\frac{10}{11}<\frac{79}{20}\)