Question

Keiko measures the circumference of three fruits.
Complete the table to show Keiko's measurements, the actual measurements, the
difference between the measurement and the actual, and the percent error.
CLEAR
CHECK
6 cm
81 cm
10%
50%
20%
10 cm
Apple
Grapefruit
Watermelon
15 cm
24 cm
Keiko's measurement
DRAG AND DROP
AN ITEM HERE
DRAG AND DROP
AN ITEM HERE
30 cm
Actual measurement
90 cm
5 cm
Absolute difference
DRAG AND DROP
AN ITEM HERE
9 cm
Percent error
DRAG AND DROP
AN ITEM HERE
DRAG AND DROP
AN ITEM HERE
DRAG AND DROP
AN ITEM HERE
no

Answer 1

The absolute measurement specifies the magnitude of the difference

which may be positive or negative.

The completed table is presented as follows;

$\begin{tabular}{|l|c|c|c|}&\textbf{Apple}&\textbf{Grapefruit}&\textbf{Watermelon}\\\textbf{Keiko's measurement}&15 cm&24 cm& \underline{81 cm}\\\textbf{Actual measurement}&\underline{10 cm}&30 cm&90 cm\\\textbf{Absolute difference}&5 cm&\underline{6 cm}&9 cm\\\textbf{Percent error}&\underline{50\%}&\underline{20\%}&\underline{10\%}\end{array}\right]$

Reasons:

The formulas for the inputs in the table are;

Absolute difference = Keiko's measurement - Actual measurement

The difference between Keiko's measurement and the actual measurement can be positive or negative while the absolute difference is always positive

$\displaystyle Percent \ error = \mathbf{\frac{Keiko's \ measurement - Actual \ measurement}{Actual \ measurement} \times 100}$

Therefore;

$\displaystyle Percent \ error = \frac{ \mathbf{Absolute \ difference}}{Actual \ measurement} \times 100$

Which gives for Apple;

The actual measurement for Apple = 15 cm - 5 cm = 10 cm

$\displaystyle Percent \ error \ for \ Apple = \frac{5 \, cm}{10 \, cm} \times 100 = \mathbf{50\%}$

Grapefruit column of the table;

Absolute difference for Grapefruit = |24 cm - 30 cm| = |-6 cm| = 6 cm

The absolute difference for Grapefruit = 6 cm

$\displaystyle Percent \ error \ for \ Grapefruit = \mathbf{\frac{6 \, cm}{30 \, cm} \times 100} = 20\%$

Percent error for Grapefruit = 20%

Watermelon column of the table;

Keiko's measurement = Absolute difference + Actual measurement

Using the given options, we have, 90 - 9 = 81

Therefore, the difference = -9, while the absolute difference = 9

Keiko's measurement for Watermelon = 90 cm - 9 cm = 81 cm

$\displaystyle Percent \ error \ for \ Grapefruit= \frac{9 \, cm}{90 \, cm} \times 100 = 10\%$

The completed table is as follows;

$\begin{tabular}{|l|c|c|c|}&\textbf{Apple}&\textbf{Grapefruit}&\textbf{Watermelon}\\\textbf{Keiko's measurement}&15 cm&24 cm& \underline{81 cm}\\\textbf{Actual measurement}&\underline{10 cm}&30 cm&90 cm\\\textbf{Absolute difference}&5 cm&\underline{6 cm}&9 cm\\\textbf{Percent error}&\underline{50\%}&\underline{20\%}&\underline{10\%}\end{array}\right]$

Learn more about percentage error here:

https://brainly.com/question/5145821