To solve this, we must compare the fractions \(\frac{1}{3}\) and \(\frac{3}{8}\). The total number of students, 24, is extra information. To compare the fractions directly, we find a common denominator, which is 24. Dinosaur exhibit: \(\frac{1}{3} = \frac{8}{24}\). Space exhibit: \(\frac{3}{8} = \frac{9}{24}\). Since \(9 > 8\), \(\frac{9}{24} > \frac{8}{24}\), which means a larger fraction of students chose the space exhibit. Therefore, the space exhibit was chosen by more students.

The fractions refer to different-sized wholes (a 'small pizza' versus a 'large pizza'). Because the total sizes of the pizzas are not the same and are not specified, we cannot compare the absolute amounts of pizza eaten. For example, \(\frac{1}{3}\) of a very large pizza could be more than \(\frac{1}{2}\) of a very small pizza. Without more information about the sizes of the pizzas, no definitive comparison can be made.

To find the rope with the middle length, we must order the fractions \(\frac{5}{9}\), \(\frac{1}{2}\), and \(\frac{4}{7}\). One way is to convert them to decimals: \(\frac{1}{2} = 0.5\), \(\frac{5}{9} = 0.555...\), and \(\frac{4}{7} \approx 0.571\). Ordering these from least to greatest gives 0.5, 0.555..., 0.571. This corresponds to the order \(\frac{1}{2}\), \(\frac{5}{9}\), \(\frac{4}{7}\). The fraction in the middle is \(\frac{5}{9}\).

To compare \(\frac{9}{10}\) and \(\frac{13}{15}\), we can find a common denominator, which is 30. Convert the fractions: Candidate A: \(\frac{9}{10} = \frac{27}{30}\). Candidate B: \(\frac{13}{15} = \frac{26}{30}\). Since \(27 > 26\), we know that \(\frac{27}{30} > \frac{26}{30}\). Therefore, Candidate A received a greater fraction of the votes.

To find which fraction is closest to \(\frac{1}{2}\), we find the absolute difference between each fraction and \(\frac{1}{2}\). A) \(|\frac{3}{8} - \frac{4}{8}| = \frac{1}{8}\). B) \(|\frac{4}{7} - \frac{1}{2}| = |\frac{8}{14} - \frac{7}{14}| = \frac{1}{14}\). C) \(|\frac{5}{9} - \frac{1}{2}| = |\frac{10}{18} - \frac{9}{18}| = \frac{1}{18}\). D) \(|\frac{7}{12} - \frac{1}{2}| = |\frac{7}{12} - \frac{6}{12}| = \frac{1}{12}\). Now we must find the smallest of these differences: \(\frac{1}{8}, \frac{1}{14}, \frac{1}{18}, \frac{1}{12}\). When fractions have the same numerator (1 in this case), the one with the largest denominator is the smallest. The largest denominator is 18, so \(\frac{1}{18}\) is the smallest difference. Therefore, \(\frac{5}{9}\) is closest to \(\frac{1}{2}\).

First, determine the fraction of pizza each person has left. Marco has \(1 - \frac{2}{5} = \frac{3}{5}\) of his pizza left. Jada has \(1 - \frac{3}{8} = \frac{5}{8}\) of her pizza left. Next, compare these two fractions. To compare \(\frac{3}{5}\) and \(\frac{5}{8}\), find a common denominator, which is 40. Marco's remaining pizza is \(\frac{3}{5} = \frac{24}{40}\). Jada's remaining pizza is \(\frac{5}{8} = \frac{25}{40}\). Since \(\frac{25}{40} > \frac{24}{40}\), Jada has more pizza left over.

To find a fraction between \(\frac{1}{3}\) and \(\frac{1}{2}\), convert them to fractions with a common denominator. A common denominator for all the fractions is 60. \(\frac{1}{3} = \frac{20}{60}\) and \(\frac{1}{2} = \frac{30}{60}\). We need a fraction between \(\frac{20}{60}\) and \(\frac{30}{60}\). Let's convert the answer choices: A) \(\frac{1}{4} = \frac{15}{60}\) (too small). B) \(\frac{5}{12} = \frac{25}{60}\) (this is between \(\frac{20}{60}\) and \(\frac{30}{60}\)). C) \(\frac{2}{3} = \frac{40}{60}\) (too large). D) \(\frac{3}{5} = \frac{36}{60}\) (too large). Therefore, \(\frac{5}{12}\) is the correct amount.

To determine who is in the lead, we must find the largest fraction among \(\frac{3}{4}\), \(\frac{2}{3}\), \(\frac{5}{6}\), and \(\frac{7}{12}\). A common denominator for these fractions is 12. Convert each fraction: Liam: \(\frac{3}{4} = \frac{9}{12}\). Noah: \(\frac{2}{3} = \frac{8}{12}\). Olivia: \(\frac{5}{6} = \frac{10}{12}\). Emma: \(\frac{7}{12}\). Comparing the numerators, 10 is the largest. Therefore, Olivia has painted the most and is in the lead.

When comparing fractions through addition and subtraction, you need to find common denominators and calculate the actual values to determine which expression yields the largest result.

Let's work through each expression systematically. For choice A: $$\frac{1}{2} - \frac{1}{10}$$, convert to the common denominator 10: $$\frac{5}{10} - \frac{1}{10} = \frac{4}{10} = 0.4$$

For choice B: $$\frac{1}{5} + \frac{1}{20}$$, convert to the common denominator 20: $$\frac{4}{20} + \frac{1}{20} = \frac{5}{20} = 0.25$$

For choice C: $$\frac{3}{4} - \frac{1}{2}$$, convert to the common denominator 4: $$\frac{3}{4} - \frac{2}{4} = \frac{1}{4} = 0.25$$

For choice D: $$\frac{1}{3} + \frac{1}{12}$$, convert to the common denominator 12: $$\frac{4}{12} + \frac{1}{12} = \frac{5}{12} ≈ 0.417$$

Comparing the results: A gives 0.4, B gives 0.25, C gives 0.25, and D gives approximately 0.417. Choice D produces the largest value.

Choice A is close but falls short of D's value. Choices B and C both equal 0.25, making them tied for the smallest values. The key trap here is that addition doesn't automatically create larger results than subtraction—the actual fractional values matter more than the operations.

When comparing fraction expressions, always calculate the final decimal values rather than making assumptions based on whether you're adding or subtracting. Convert everything to a common form for easy comparison.

First, calculate the fraction of water used from each tank. For the first tank, the amount used is \(\frac{4}{5} - \frac{1}{3}\). The common denominator is 15: \(\frac{12}{15} - \frac{5}{15} = \frac{7}{15}\). For the second tank, the amount used is \(\frac{5}{6} - \frac{1}{2}\). The common denominator is 6: \(\frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}\). Now, compare the fractions of water used: \(\frac{7}{15}\) and \(\frac{1}{3}\). The common denominator is 15: \(\frac{1}{3} = \frac{5}{15}\). Since \(\frac{7}{15} > \frac{5}{15}\), the first tank had a greater fraction of its water used.