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Crank
3 years ago
9

$30 to $50 what is the markup percent

Mathematics
1 answer:
n200080 [17]3 years ago
8 0
The markup percent is $15
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A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is us
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Answer:

The Absolute Error is the difference between the actual and measured value.

Absolute \:error = |Actual \:value - Measured \:value|

The Relative Error is the Absolute Error divided by the actual measurement.

Relative \:error = \frac{Absolute \:error}{Actual \:value}

We know that the actual value is 102.0 mg/dL.

To find the absolute error and relative error for each measurement made by the glucose monitor you must use the above definitions.

a) For a concentration of 104.5 mg/dL the absolute error and relative error are

Absolute \:error = \left|102-104.5\right|\\Absolute \:error =\left|-2.5\right|\\Absolute \:error =2.5

Relative \:error = \frac{2.5}{102.0}=0.0245

b) For a concentration of 96.2 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-96.2\right|\\Absolute \:error =\left|5.8\right|\\Absolute \:error =5.8

Relative \:error = \frac{5.8}{102.0}=0.0569

c) For a concentration of 102.2 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-102.2\right|\\Absolute \:error =\left|-0.2\right|\\Absolute \:error =0.2

Relative \:error = \frac{0.2}{102.0}=0.00196

d) For a concentration of 98.3 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-98.3\right|\\Absolute \:error =\left|3.7\right|\\Absolute \:error =3.7

Relative \:error = \frac{3.7}{102.0}=0.0363

e) For a concentration of 101.8 mg/dL the absolute error and relative error are

Absolute \:error = \left|102.0-101.8\right|\\Absolute \:error =\left|0.2\right|\\Absolute \:error =0.2

Relative \:error = \frac{0.2}{102.0}=0.00196

4 0
3 years ago
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