1. **Problem statement:** We are given a histogram showing honey production by beehives. We know 400 beehives produced between 16 kg and 18 kg of honey. We need to find the frequency density value $x$ for the bar from 10 kg to 16 kg. 2. **Formula:** Frequency density is given by $$\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}$$ 3. **Given:** - Frequency for 16–18 kg = 400 - Class width for 16–18 kg = $18 - 16 = 2$ - Frequency density for 16–18 kg (from histogram) is the highest bar, but we don't need its exact value here. 4. **Find frequency density for 16–18 kg:** $$\text{Frequency density} = \frac{400}{2} = 200$$ 5. **Use the histogram info:** The bar from 10 to 16 kg has frequency density $x$ and class width $$16 - 10 = 6$$ 6. **Frequency for 10–16 kg:** $$\text{Frequency} = x \times 6 = 6x$$ 7. **From the histogram, the total frequency for all bars must be consistent.** However, since only the 16–18 kg bar frequency is given, and the histogram shows the bar from 10 to 16 kg with frequency density $x$, we can use the fact that the frequency density for 16–18 kg is 200. 8. **Assuming the histogram is scaled so that the frequency density for 10–16 kg is $x$, and the frequency density for 16–18 kg is 200, and the bar heights are proportional, the problem states to find $x$ given 400 beehives in 16–18 kg.** 9. **Therefore, the value of $x$ is the frequency density for the 10–16 kg bar, which can be calculated if the frequency for that bar is known. Since it is not given, the problem likely expects us to calculate $x$ using the frequency and class width for the 16–18 kg bar only.** 10. **Hence, the value of $x$ is:** $$x = \frac{\text{Frequency}}{\text{Class width}} = \frac{400}{2} = 200$$ 11. **Answer:** $$x = 200.0$$ This means the frequency density value $x$ on the axis is 200.0 to 1 decimal place.