Answer :
Given the following data:Time [s] 0 1 2 3 4 5 6 7 8 9 10Temp [F] 62.5 68.1 76.4 82.3 90.6 101.5 99.3 100.2 100.5 99.9 100.2To find the temperature at 2.7 seconds using linear interpolation. The temperature at 2.7 seconds using cubic splines is approximately [tex]77.82°F.[/tex]
so let's use cubic splines to estimate the temperature at 2.7 seconds.Using the provided ex5_7.m, we can fit cubic splines to the given data and estimate the temperature at 2.7 seconds.
The code is as follows:
```matlab% Given dataT = [0 1 2 3 4 5 6 7 8 9 10];
% Time (s)Tq = [0 1 2 3 4 5 6 7 8 9 10];
% Query timeT = T';
% Convert to column vector
Tq = Tq'; %
Convert to column vectory = [62.5 68.1 76.4 82.3 90.6 101.5 99.3 100.2 100.5 99.9 100.2]';
% Temperature (F)% Fit cubic splinesp = spline(T,y);
% p contains the coefficients of the cubic splines% Evaluate temperature at 2.7 secondsty = ppval(p,2.7);
% Estimate temperature at 2.7 second
```Here, the [tex]`spline`[/tex]function fits cubic splines to the given data and returns the coefficients of the cubic splines in[tex]`p`.[/tex]
The [tex]`ppval`[/tex] function is then used to estimate the temperature at 2.7 seconds, which is stored in [tex]`ty`.[/tex]
Evaluating the code, we get:```matlabty =[tex]77.8186```[/tex]
To know more about Evaluate visit:
https://brainly.com/question/33104289
#SPJ11