Ask Question
4 March, 06:47

Christopher measured in line to be 7.2 inches long if the actual length of the line is 7.1 inches then what is the percent error of the measurement to the nearest 10th of a percent

+4
Answers (2)
  1. 4 March, 08:06
    0
    The percent error in his measurement was 1.41%

    Step-by-step explanation:

    In order to compute the percentage error we first need to know the absolute error of his measurement. This error is given by what Christopher measured sutracted by the actual length, wich is:

    absolute error = 7.2 - 7.1 = 0.1 inches

    To know this value in percentage we need to know how many percentage 0.1 inches is from 7.1 inches. So we can use a rule of three such as 7.1 inches relates to 100% as 0.1 inches relates to x%. We have:

    7.1 inches - > 100%

    0.1 inches - > x

    x = (100*0.1) / 7.1 = 1.41 %

    The percent error in his measurement was 1.41%
  2. 4 March, 09:31
    0
    1.4%

    Step-by-step explanation:

    To calculate the error, we first need to subtract Christopher measurement of the line by the actual length of the line to find the absolute error:

    7.2 - 7.1 = 0.1

    The absolute error is 0.1 inches.

    Now, to know the percent error, we need to divide this value of error by the actual length of the line:

    0.1 / 7.1 = 0.01408 = 1.408% = 1.4%

    So the percent error of the measurement is 1.4%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Christopher measured in line to be 7.2 inches long if the actual length of the line is 7.1 inches then what is the percent error of the ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers