1. The problem states that there are 10 bags of sweets with a mean of 42 sweets per bag. 2. The mean formula is: $$\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}$$ 3. We know the mean is 42 and the number of bags is 10, so: $$42 = \frac{\text{Sum of sweets in 10 bags}}{10}$$ 4. Multiply both sides by 10 to find the total sweets in all 10 bags: $$42 \times 10 = \text{Sum of sweets in 10 bags}$$ $$420 = \text{Sum of sweets in 10 bags}$$ 5. Calculate the total sweets in the first 9 bags using the frequency table: - 39 sweets \times 1 bag = 39 - 40 sweets \times 2 bags = 80 - 41 sweets \times 5 bags = 205 - 42 sweets \times 0 bags = 0 - 43 sweets \times 1 bag = 43 Sum for 9 bags: $$39 + 80 + 205 + 0 + 43 = 367$$ 6. Let $x$ be the number of sweets in the 10th bag. Then: $$367 + x = 420$$ 7. Solve for $x$: $$x = 420 - 367$$ $$x = 53$$ **Answer:** The 10th bag contains **53** sweets.