To determine which set contains an irrational number, let's evaluate each set individually.
### Set 1: \(\{2300, 0.48, \frac{13}{1}\}\)
- \(2300\) is an integer and hence a rational number.
- \(0.48\) is a decimal that can be written as \(\frac{48}{100} = \frac{12}{25}\), which is a rational number.
- \(\frac{13}{1} = 13\), which is an integer and therefore a rational number.
This set contains only rational numbers.
### Set 2: \(\{18, 0.1, \frac{12}{5}\}\)
- \(18\) is an integer and hence a rational number.
- \(0.1\) is a decimal that can be written as \(\frac{1}{10}\), which is a rational number.
- \(\frac{12}{5}\) is a ratio of two integers, making it a rational number.
This set contains only rational numbers.
### Set 3: \(\{\frac{3}{8}, 4, \sqrt{52}\}\)
- \(\frac{3}{8}\) is a ratio of two integers, which makes it a rational number.
- \(4\) is an integer and hence a rational number.
- \(\sqrt{52}\) is not a perfect square. Rational numbers have square roots that are either integers (if they are perfect squares) or irrational numbers (if they are not perfect squares).
- \(\sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13}\). Since \(\sqrt{13}\) is an irrational number, \(\sqrt{52}\) must be irrational as well.
This set contains an irrational number, \(\sqrt{52}\).
### Set 4: \(\{0.333 \ldots, \sqrt{4}, 10\}\)
- \(0.333 \ldots\) (repeating 3s) is a repeating decimal that can be written as \(\frac{1}{3}\), which is a rational number.
- \(\sqrt{4} = 2\), which is an integer and therefore a rational number.
- \(10\) is an integer and hence a rational number.
This set contains only rational numbers.
### Conclusion
Based on the analysis, the set containing an irrational number is:
[tex]\[
\{\frac{3}{8}, 4, \sqrt{52}\}
\][/tex]
