We know that a linear equation is given generally as:
y = mx + c
where m = gradient
and c = y intercept
We can see from the scatter plot that the y intercept is 1.12
So, already we have limited our options to B and C.
Now let us find our slope with the second and third data points:
(1, 1.15) and (2, 1.18)
Slope is given as:
![m\text{ = }\frac{P_{2\text{ }}-P_1}{t_{2\text{ }}-t_1}](https://img.qammunity.org/2023/formulas/mathematics/college/r564g6isidsb7pepdwo27bf56uo3rh0fz2.png)
![m\text{ = }\frac{1.18\text{ - 1.15}}{2\text{ - 1}}](https://img.qammunity.org/2023/formulas/mathematics/college/hy8lnsnifyykkkyhslkcbo5hm330pncscn.png)
![m\text{ = }(0.03)/(1)](https://img.qammunity.org/2023/formulas/mathematics/college/898mbbmmvjdvvoz6aw3lmm14np23sszdoq.png)
m = 0.03
Which means that if we were to write a linear equation of this, it would be:
P(t) = 0.03t + 1.12
The only option that is close to this is B which is:
P(t) = 0.02t + 1.12
Hence, this is our answer.