Bob said a man sat on a pole for $3.8 \times 10^7$ seconds. We need to determine the most appropriate unit of time to use instead of seconds. The choices are minutes, hours, weeks, and decades.

1. Problem Description

Bob said a man sat on a pole for $3.8 \times 10^7$ seconds. We need to determine the most appropriate unit of time to use instead of seconds. The choices are minutes, hours, weeks, and decades.

2. Solution Steps

First, we convert seconds to minutes:

$3.8 \times 10^7 \text{ seconds} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{3.8 \times 10^7}{60} \text{ minutes} \approx 6.33 \times 10^5 \text{ minutes}$

Next, we convert seconds to hours:

$3.8 \times 10^7 \text{ seconds} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = \frac{3.8 \times 10^7}{3600} \text{ hours} \approx 1.06 \times 10^4 \text{ hours}$

Then, we convert seconds to weeks:

$3.8 \times 10^7 \text{ seconds} \times \frac{1 \text{ week}}{604800 \text{ seconds}} = \frac{3.8 \times 10^7}{604800} \text{ weeks} \approx 62.8 \text{ weeks}$

Finally, convert seconds to decades:

$3.8 \times 10^7 \text{ seconds} \times \frac{1 \text{ year}}{31536000 \text{ seconds}} \times \frac{1 \text{ decade}}{10 \text{ years}} = \frac{3.8 \times 10^7}{315360000} \text{ decades} \approx 0.12 \text{ decades}$

Since 62.8 weeks is easier to comprehend than $6.33 \times 10^5$ minutes, $1.06 \times 10^4$ hours or $0.12$ decades, it is more appropriate. Although 0.12 decades is small, weeks is likely the best option. However, notice that 10600 hours is also a more reasonable figure than $3.8 \times 10^7$ seconds.

Let's check the man's length of sitting on the pole:

62.8 weeks is approximately 1 year and 2 months.

10600 hours is approximately 442 days.

This means it is around 1 year and 2 months.

Therefore, hours is the most appropriate answer.

3. Final Answer

hours