To calculate the percent error in the volume of the rectangular prism, we need to understand how measurement errors propagate. If the prism's dimensions are $7$, $15$, and $8$, and we assume a small measurement error $\Delta l$ applies to each dimension, the volume and its error propagation are calculated as follows:

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Step 1: Volume Formula

The volume of a rectangular prism is given by: $V = l \cdot w \cdot h$ Substituting the given dimensions: $V = 7 \cdot 15 \cdot 8 = 840$

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Step 2: Error Propagation Formula for Volume

If $\Delta l$, $\Delta w$, and $\Delta h$ are the absolute errors in length ($l$), width ($w$), and height ($h$), the formula for the percent error in volume is: $\text{Percent Error in Volume} = \left( \frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta h}{h} \right) \cdot 100$

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Step 3: Assume an Absolute Error

Unless otherwise stated, we typically assume the measurement error ($\Delta l$, $\Delta w$, $\Delta h$) to be $\pm 0.5$ units (common for approximate measurements). Thus: $\Delta l = \Delta w = \Delta h = 0.5$

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Step 4: Calculate Relative Errors

The relative error for each dimension is: $\frac{\Delta l}{l} = \frac{0.5}{7} \approx 0.0714, \quad \frac{\Delta w}{w} = \frac{0.5}{15} \approx 0.0333, \quad \frac{\Delta h}{h} = \frac{0.5}{8} \approx 0.0625$

Adding these together: $\frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta h}{h} = 0.0714 + 0.0333 + 0.0625 = 0.1672$

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Step 5: Convert to Percent Error

$\text{Percent Error in Volume} = 0.1672 \cdot 100 \approx 16.72\%$

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Final Answer:

The percent error in the calculated volume of the rectangular prism is approximately 16.72%.

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Would you like further clarification or details on this process? Let me know! Here are some related questions for deeper understanding:

1. What happens to the percent error if the measurement error is increased to $\pm 1$?
2. How does the percent error change if only one dimension has an error (e.g., width)?
3. Can you explain how relative errors combine in other 3D shapes like cylinders or spheres?
4. What if the absolute error is instead $\pm 0.1$? How would this affect the result?
5. How does this method generalize to higher dimensions, like hypercubes?

Tip:

When calculating percent errors, always ensure you know the precision of the measuring tools used—it directly impacts your calculations!