Question 1011308
What is expected as "Perform a linear regression on this data" depends on your course/book/teacher.
Before calculators (yes, I am that old),
we would do a lot of pencil and paper calculations,
using complicated formulas.
Maybe you are expected to use the functions in your calculator.
Those function uses formulas you do not need to remember,
to calculate results such as
correlation coefficient,
slope, and
y-intercept
of the linear regression best fit line.
Since I am sitting in front of my computer,
I would use the spreadsheet program I have installed (Microsoft Excel)
to tabulate, graph, and calculate the needed results.
 
My tabulation, graph and results look like this:
{{{matrix(2,8,
minutes,3,5,9,18,30,37,48,
degrees,93.5,90,84.8,70.2,54.4,42.5,24.9)}}} {{{drawing(300,360,-20,80,-10,110,grid(0),
circle(3,93.5,1),circle(5,90,1),
circle(9,84.8,1),circle(18,70.2,1),
circle(30,54.4,1),circle(37,42.5,1),
circle(48,24.9,1),green(line(0,98.02,68,-4.36))
)}}}
correlation coefficient: {{{r=-0.9995}}} (rounded to the first decimal place that is not a 9),
slope: {{{m=-1.50556}}} (rounded to 5 decimal places),
y-intercept: {{{b=98.0192}}} (rounded to 4 decimal places),
best fit line (y=mx+b): {{{temperature=-1.50556*minutes+98.0192}}} .
 
The absolute value of the correlation coefficient is very close to exactly {{{1}}} ,
and that means that the line fits the data very closely (as we see in the graph),
meaning that the linear model is right (that the temperature varies linearly with time).
The fact that the correlation coefficient is negative
shows that temperature decreases with time. 
 
The slope tells us that the temperature decreases {{{1.50556}}} degrees per minute,
so on average, the temperature decreases {{{5*1.50556=about}}}{{{highlight(7.5)}}} degrees (rounded) every {{{5}}} minutes.
 
The data allows us to predict temperature between 3 and 48 minutes,
but we may feel somewhat confident that the relationship still holds for
values of {{{x}}} (seconds) close to the edges of the [3,48] interval.
That is why I extended the best fit line in the graph to cover [0,68].
The line seems to cross the {{{y=0}}} x-axis at {{{x=65}}} (65 minutes).
We can solve the linear best fit equation for the time when {{{temperature=-7degrees}}} , and the solution freezes.
{{{system(temperature=-7,temperature=-1.50556*minutes+98.0192)}}}--->{{{system(temperature=-7,1.50556*minutes=98.0192+7)}}}--->{{{system(temperature=-7,1.50556*minutes=105.0192)}}}--->{{{system(temperature=-7,temperature=105.0192/1.50556=65.1)}}}(rounded).
Since Natalie was measuring the time in whole minutes,
we can report the time the solution freezes with the same precision as {{{highlight(65minutes)}}} .