Four people have found the distance in kilometers across a small bridge using different methods.

Question

Their results are given in the table. Order the distances from least to greatest.

5 1/2, 28−−√, 5.5¯¯¯, 23/4
5 1/2, 28−−√, 5.5¯¯¯, 23/4

28−−√, 5 1/2, 5.5¯¯¯, 23/4
28−−√, 5 1/2, 5.5¯¯¯, 23/4

28−−√, 5.5¯¯¯, 23/4, 5 1/2
28−−√, 5.5¯¯¯, 23/4, 5 1/2

23/4 , 5.5¯¯¯, 28−−√, 5 1/2

Answer 1

To order the given distances from least to greatest, let's first convert all expressions into decimal form:

1. \(5 \frac{1}{2}\): This is a mixed number, which can be converted to decimal as follows: \[ 5 \frac{1}{2} = 5 + 0.5 = 5.5 \]

2. \(\sqrt{28}\): We can approximate this by calculating: \[ \sqrt{28} \approx 5.2915 \] (Since \( \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \approx 2 \times 2.64575 \approx 5.2915\)).

3. \(5.5\overline{5}\): This notation represents the repeating decimal \(5.5555...\). So, \[ 5.5\overline{5} = 5.5555... \quad (\text{which is greater than } 5.5) \]

4. \(\frac{23}{4}\): To convert this to decimal form: \[ \frac{23}{4} = 5.75 \]

Now, we have the following decimal equivalents:

• \(5 \frac{1}{2} \approx 5.5\)
• \(\sqrt{28} \approx 5.2915\)
• \(5.5\overline{5} \approx 5.5555...\)
• \(\frac{23}{4} = 5.75\)

Now we can order them from least to greatest:

1. \(\sqrt{28} \approx 5.2915\)
2. \(5 \frac{1}{2} = 5.5\)
3. \(5.5\overline{5} \approx 5.5555...\)
4. \(\frac{23}{4} = 5.75\)

Therefore, the final ordered list from least to greatest is: \[ \sqrt{28}, 5 \frac{1}{2}, 5.5\overline{5}, \frac{23}{4} \]

So the answer is: \(\sqrt{28}, 5 \frac{1}{2}, 5.5\overline{5}, \frac{23}{4}\).