Question: The website also states: "Almost a billion people on the planet don't have access to clean drinking water." If there were $7.8 \times 10^9$ people in the world in 2020, then approximately what percent of them lived without clean drinking water? Round to the nearest tenth of a percent.
1. **State the problem:** We want to find what percent of the world's population lived without clean drinking water in 2020.
2. **Identify the given values:**
- Total population = $7.8 \times 10^9$
- People without clean water = $1 \times 10^9$ (almost a billion)
3. **Formula for percentage:**
$$\text{Percentage} = \left( \frac{\text{Part}}{\text{Whole}} \right) \times 100$$
4. **Apply the formula:**
$$\text{Percentage} = \left( \frac{1 \times 10^9}{7.8 \times 10^9} \right) \times 100$$
5. **Simplify the fraction:**
$$= \left( \frac{\cancel{1 \times 10^9}}{7.8 \times \cancel{10^9}} \right) \times 100 = \frac{1}{7.8} \times 100$$
6. **Calculate the division:**
$$\frac{1}{7.8} \approx 0.128205$$
7. **Multiply by 100 to get percentage:**
$$0.128205 \times 100 = 12.8205$$
8. **Round to the nearest tenth:**
$$12.8\%$$
**Final answer:** Approximately $12.8\%$ of the world's population lived without clean drinking water in 2020.