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For an activity in class, a team of students was given the line of best fit, y = - 0.9x + 18.11 one of the actual data points i...Asked by What
For an activity in class, a team of students was given the line of best fit, y=−0.9x+18
. If one of the actual data points is (1,16.88)
, what is the residual of the data point? Round the answer to the nearest hundredths.(1 point)
The residual of the data point (1,16.88)
is
.A team of students collected data on the growth of a plant. They plotted the height of the plant every day and found the line of best fit y=0.23x+9.9
for the growth of the plant. If the actual data point for day six is (6,7.82)
, what is the residual of the data point for day six? Round the answer to the nearest hundredths.(1 point)
The residual of the data point (6,7.82)
is
.
. If one of the actual data points is (1,16.88)
, what is the residual of the data point? Round the answer to the nearest hundredths.(1 point)
The residual of the data point (1,16.88)
is
.A team of students collected data on the growth of a plant. They plotted the height of the plant every day and found the line of best fit y=0.23x+9.9
for the growth of the plant. If the actual data point for day six is (6,7.82)
, what is the residual of the data point for day six? Round the answer to the nearest hundredths.(1 point)
The residual of the data point (6,7.82)
is
.
Answers
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Answered by
GPT-4o mini
To find the residual of a data point, we need to calculate the difference between the actual y-value of the data point and the predicted y-value from the line of best fit.
1. For the first data point (1, 16.88) with the line of best fit \( y = -0.9x + 18 \):
- Calculate the predicted y-value when \( x = 1 \):
\[
y_{\text{pred}} = -0.9(1) + 18 = -0.9 + 18 = 17.1
\]
- The actual y-value is 16.88. Now we calculate the residual:
\[
\text{Residual} = \text{Actual} - \text{Predicted} = 16.88 - 17.1 = -0.22
\]
So, the residual of the data point (1, 16.88) is **-0.22**.
2. For the second data point (6, 7.82) with the line of best fit \( y = 0.23x + 9.9 \):
- Calculate the predicted y-value when \( x = 6 \):
\[
y_{\text{pred}} = 0.23(6) + 9.9 = 1.38 + 9.9 = 11.28
\]
- The actual y-value is 7.82. Now we calculate the residual:
\[
\text{Residual} = \text{Actual} - \text{Predicted} = 7.82 - 11.28 = -3.46
\]
So, the residual of the data point (6, 7.82) is **-3.46**.
In summary:
- The residual for (1, 16.88) is **-0.22**.
- The residual for (6, 7.82) is **-3.46**.
1. For the first data point (1, 16.88) with the line of best fit \( y = -0.9x + 18 \):
- Calculate the predicted y-value when \( x = 1 \):
\[
y_{\text{pred}} = -0.9(1) + 18 = -0.9 + 18 = 17.1
\]
- The actual y-value is 16.88. Now we calculate the residual:
\[
\text{Residual} = \text{Actual} - \text{Predicted} = 16.88 - 17.1 = -0.22
\]
So, the residual of the data point (1, 16.88) is **-0.22**.
2. For the second data point (6, 7.82) with the line of best fit \( y = 0.23x + 9.9 \):
- Calculate the predicted y-value when \( x = 6 \):
\[
y_{\text{pred}} = 0.23(6) + 9.9 = 1.38 + 9.9 = 11.28
\]
- The actual y-value is 7.82. Now we calculate the residual:
\[
\text{Residual} = \text{Actual} - \text{Predicted} = 7.82 - 11.28 = -3.46
\]
So, the residual of the data point (6, 7.82) is **-3.46**.
In summary:
- The residual for (1, 16.88) is **-0.22**.
- The residual for (6, 7.82) is **-3.46**.
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