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2 years ago

I need the answer quickly;)

Mathematics

1 answer:

EleoNora [17]2 years ago

4 0

Answer:

$37$

Step-by-step explanation:

$\text{Apply Pythagorean theorem,}\\\\~~~~~~~\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\\ \\\implies c^2=1^2 +6^2\\\\\implies c^2 = 1 + 36\\\\\implies c^2 = 37\\\\\implies c =\sqrt{37}$

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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.

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

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

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

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

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

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

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

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