To solve this problem, we need to calculate the relative error as a percentage.

Information given:

1. Weight of apples: $3.4 \, \text{kg} = 3400 \, \text{g}$
2. Smallest division of the scale: $18.0 \, \text{g}$

Formula for Relative Error:

$\text{Relative Error (\%)} = \left( \frac{\text{Smallest Division}}{\text{Measured Value}} \right) \times 100$

Substituting the values:

$\text{Relative Error (\%)} = \left( \frac{18.0}{3400} \right) \times 100$

Calculation:

$\text{Relative Error (\%)} = \left( \frac{18.0}{3400} \right) \times 100 = 0.5294\%$

Final Answer:

The relative error is approximately 0.53%.

Would you like further explanation or have any additional questions? 😊

Here are 5 related questions for deeper understanding:

1. How does the relative error change with larger measured values?
2. What is the importance of the smallest division in error calculations?
3. How would the relative error differ if the weight was 1.7 kg instead?
4. What is the difference between absolute and relative error?
5. How can you reduce measurement error in practical experiments?

Tip: Always check the units in a problem to ensure consistency throughout your calculations!