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actor jerry o'connell - Malaeb

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Thus, the ratio is $ \boxed{\dfrac{9}{5}} $.Question: A museum curator is cataloging a collection of 48 ancient tablets. If the ratio of inscribed tablets to plain tablets is $5:3$, and all inscribed tablets must be displayed in groups of 7, what is the greatest number of inscribed tablets that can be grouped without leaving any out?

📌 Solution: The ratio of inscribed to plain tablets is $5:3$, so the total number of parts is $5 + 3 = 8$. Since there are 48 tablets, each part represents $ \frac{48}{8} = 6 $ tablets. Thus, the number of inscribed tablets is $5 \times 6 = 30$. We are told that inscribed tablets must be displayed in groups of 7, so we seek the greatest multiple of 7 that is less than or equal to 30. The multiples of 7 below 30 are $7, 14, 21, 28$. The greatest is $28$. Therefore, the largest number of inscribed tablets that can be grouped in sevens is $28$.

📌 $\boxed{28}$

📌 Question: A palynologist collects pollen samples from three layers of sediment. The first layer contains $ \frac{3}{8} $ kg of pollen, the second $ \frac{5}{12} $ kg, and the third $ \frac{7}{16} $ kg. What is the total mass of pollen collected, expressed as a simplified fraction of a kilogram?

📌 Solution: We compute $ \frac{3}{8} + \frac{5}{12} + \frac{7}{16} $. The least common denominator of 8, 12, and 16 is 48. Convert each fraction:

📌 \frac{3}{8} = \frac{18}{48}, \quad \frac{5}{12} = \frac{20}{48}, \quad \frac{7}{16} = \frac{21}{48}

📌 \frac{18 + 20 + 21}{48} = \frac{59}{48}

📌 Since $ \frac{59}{48} > 1 $, the total is $ \frac{59}{48} $ kg, which is an improper fraction. It cannot be simplified further.

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