1. **State the problem:** We have the numbers 45, 46, 41, h, 41, 41, 43, 43, 46 and the mean of these numbers is 43. We need to find which value of $h$ (either 41 or 59) satisfies this condition. 2. **Recall the formula for the mean:** $$\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}$$ 3. **Apply the formula:** There are 9 numbers in total. $$43 = \frac{45 + 46 + 41 + h + 41 + 41 + 43 + 43 + 46}{9}$$ 4. **Calculate the sum of known numbers:** $$45 + 46 + 41 + 41 + 41 + 43 + 43 + 46 = 346$$ 5. **Set up the equation:** $$43 = \frac{346 + h}{9}$$ 6. **Multiply both sides by 9:** $$9 \times 43 = 346 + h$$ $$387 = 346 + h$$ 7. **Solve for $h$:** $$h = 387 - 346$$ $$h = 41$$ 8. **Check the options:** - If $h = 41$, the mean is 43 (correct). - If $h = 59$, the mean would be: $$\frac{346 + 59}{9} = \frac{405}{9} = 45$$ (not 43). **Final answer:** $h = 41$