During a science lab, students heated water, allowed it to cool, and recorded the temperature over time. They computed the
difference between the water temperature and the room temperature. The results are in the table. $$\begin{array} { | l | l | l | l | l | l | l | }
\hline \text { Time (in minutes) } & 10 & 20 & 30 & 40 & 50 & 60 \
\hline \text { Difference in temp. (degrees F) } & 68 & 36 & 20 & 10 & 6 & 4 \
\hline
\end{array}$$
-Use the equation = 2.057 - 0.025\) time to predict the difference in temperature after 45 minutes.

Question

During a science lab, students heated water, allowed it to cool, and recorded the temperature over time. They computed the
difference between the water temperature and the room temperature. The results are in the table. $$\begin{array} { | l | l | l | l | l | l | l | } 
\hline \text { Time (in minutes) } & 10 & 20 & 30 & 40 & 50 & 60 \
\hline \text { Difference in temp. (degrees F) } & 68 & 36 & 20 & 10 & 6 & 4 \
\hline
\end{array}$$
-Use the equation = 2.057 - 0.025\) time to predict the difference in temperature after 45 minutes.

Quizplus · Accepted Answer

The Answer of During a science lab, students heated water, allowed it to cool, and recorded the temperature over time. They computed the
difference between the water temperature and the room temperature. The results are in the table. $$\begin{array} { | l | l | l | l | l | l | l | } 
\hline \text { Time (in minutes) } & 10 & 20 & 30 & 40 & 50 & 60 \
\hline \text { Difference in temp. (degrees F) } & 68 & 36 & 20 & 10 & 6 & 4 \
\hline
\end{array}$$
-Use the equation = 2.057 - 0.025\) time to predict the difference in temperature after 45 minutes.

During a Science Lab, Students Heated Water, Allowed It to Cool