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svet-max [94.6K]
2 years ago
5

Grayson measured a line to be 10.9 inches long. If the actual length of the line is 13.4 inches, then what was the percent error

of the measurement, to the nearest tenth of a percent?
Mathematics
1 answer:
aksik [14]2 years ago
7 0

The percent error of the measurement, to the nearest tenth of a percent is 18.6%

<h3>Calculating percent error</h3>

The formula for calculating the percent error is expressed as:

%error = actual - measure/actual * 100

Given the following

actual length = 13.4 in

Measured =10.9in

Substitute

%error = 13.4-10.9/13.4 * 100
%error = 2.5/13.4 * 100
%error = 18.65%

Hence the percent error of the measurement, to the nearest tenth of a percent is 18.6%

Learn more on percent error here; brainly.com/question/1600371

#SPJ1

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The percentage of the bank's customers carry daily balances between $700 and $1,000 is 65.7%.

The minimum daily balance on which it should be willing to pay interest is $1,198.

Step-by-step explanation:

We have a normal distribution with mean = $800 and standard deviation = $150.

a) We can calculate this value with the standard normal distribution, calculating the z-value for $700 and $1,000.

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The percentage of the bank's customers carry daily balances between $700 and $1,000 is 65.7%.

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