Question

Asked Sep 11, 2019

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Measure the area of the rectangular box with length of 11.5 cm (plus or minus) 0.1 cm, and a height of 1 cm(plus or minus) 0.1 cm. Use a two-repetition approach: the highest and lowest possible bounds. The standard or accepted value is 9.80 cm^2. Give the percent relative and absolute errors. Discuss the sources of these errors. Which dimension contributes greatest to the relative error?

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Step 1

**Introduction:**

Absolute error = | Observed value – Expected value |.

Relative error = (Observed value – Expected value)/ Expected value.

Relative absolute error = | Observed value – Expected value |/ Expected value.

Percent relative error = (Observed value – Expected value)/ Expected value × 100%.

Step 2

**Calculation:**

*Length:*

The length is given as (11.5 ± 0.1) cm.

So, the **absolute error for length is 0.1 cm**.

Relative error percentage

= (0.1/11.5) × 100%

= 0.87%.

So, the **percent relative absolute error for length is 0.87%**.

*Height:*

The height is given as (1 ± 0.1) cm.

So, the **absolute error for height is 0.1 cm**.

Relative error percentage

= (0.1/1) × 100%

= 10%.

So, the **percent relative absolute error for height is 10%**.

*Area:*

Lowest possible bound of length = (11.5 – 0.1) cm = 11.4 cm.

Highest possible bound of length = (11.5 + 0.1) cm = 11.6 cm.

Lowest possible bound of height = (1 – 0.1) cm = 0.9 cm.

Highest possible bound of height = (1 + 0.1) cm = 1.1 cm.

Thus, lowest possible bound of area = (11.4 × 0.9) cm = 10.26 cm.

Highest possible bound of area = (11.6 × 1.1) cm = 12.76 cm.

The standard or expected value of area = 9.8 cm2.

With respect to the lowest possible bound of area, absolute error = (10.26 – 9.8) cm = 0.46 cm.

So, the **absolute error for area with respect to the** **lowest possible bound of area**** is...**

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