Question
Use the least squares regression line of this data set to predict a value.
For a science project, Jill wants to see if a larger body of water has more heat energy than a smaller body of water at the same temperature. She prepared a number of buckets filled with various amounts of water at a fixed temperature and dropped an ice cube of the same size into each one.
Jill then recorded the volume of water in each bucket (in milliliters), x, and the amount of time it took for each ice cube to melt (in minutes), y.
Volume of water Melting time (in minutes)
13 16
40 16
69 6
73 9
75 12
The least squares regression line of this data set is:
y=
–
0.126x+18.624
How much time does this line predict it takes for the ice cube to melt in 54 milliliters of water at the fixed temperature?
Round your answer to the nearest thousandth.
minutes
Answers
Answered by
GPT-4o mini
To predict the melting time for 54 milliliters of water using the least squares regression line \( y = -0.126x + 18.624 \), we can substitute \( x = 54 \) into the equation.
\[
y = -0.126(54) + 18.624
\]
Now, calculate \( -0.126 \times 54 \):
\[
-0.126 \times 54 = -6.804
\]
Next, add \( 18.624 \) to \( -6.804 \):
\[
y = -6.804 + 18.624 = 11.82
\]
Thus, the predicted melting time for an ice cube in 54 milliliters of water is:
\[
\boxed{11.820} \text{ minutes}
\]
This is rounded to the nearest thousandth.
\[
y = -0.126(54) + 18.624
\]
Now, calculate \( -0.126 \times 54 \):
\[
-0.126 \times 54 = -6.804
\]
Next, add \( 18.624 \) to \( -6.804 \):
\[
y = -6.804 + 18.624 = 11.82
\]
Thus, the predicted melting time for an ice cube in 54 milliliters of water is:
\[
\boxed{11.820} \text{ minutes}
\]
This is rounded to the nearest thousandth.
There are no human answers yet. A form for humans to post answers is coming very soon!