Question

A student measures the length of a magnesium ribbon three times and records the data as: (a). 23.486 cm, (b). 23.491 cm, (c). 23.511 cm, The labeled value of the magnesium ribbon is 23.490 cm. Calculate the precent error for ALL the data recorded by the student. Which has the least percentage error? (Show all work or an explanation)

Answer 1

Step-by-step explanation:

A student measures the length of a magnesium ribbon three times and records the data as: (a). 23.486 cm, (b). 23.491 cm, (c). 23.511 cm

The labeled value of the magnesium ribbon is 23.490 cm

We need to find the precent error for all the data recorded by the student.

`\text{Percentage error} = \frac{|\text{calculated value-actual value}|}{\text{acutal value}}* 100`

(a) When the calculated value = 23.486 cm

`\text{Percentage error} = (23.486 -23.490 )/(23.490 )* 100\\\\=0.017\%`

(b) When the calculated value = 23.491 cm

`\text{Percentage error} = (23.491 -23.490 )/(23.490 )* 100\\\\=0.0042\%`

(c) When the calculated value = 23.511 cm

`\text{Percentage error} = (23.511 -23.490 )/(23.490 )* 100\\\\=0.0893\%`

When the calculated value is 23.486 cm, it has least error of 0.017%.