A student has these readings for a spring balance experiment: the load x in N is 1, 2, 3, 4, 5, and…

Step 1. The sums. $\sum x = 15$ N, $\sum y$ $= 2.1 + 3.9 + 6.2 + 7.9 + 10.1$ $= 30.2$ cm, $\sum xy$ $= 2.1 + 7.8 + 18.6 + 31.6$ $+ 50.5$ $= 110.6$ N cm and $\sum x^2 = 55\ \text{N}^2$.

Step 2. The slope. $m = \dfrac{5 \times 110.6 - 15 \times 30.2}{5 \times 55 - 15^2}$ $= \dfrac{553 - 453}{50}\ \text{cm/N}$ $= 2.0\ \text{cm/N}$.

Step 3. The intercept. $c = \dfrac{30.2 - 2.0 \times 15}{5}\ \text{cm}$ $= 0.04$ cm.

Step 4. The prediction. For a load of $6$ N, $y$ $= 2.0\ \text{cm/N} \times 6\ \text{N} + 0.04\ \text{cm}$ $= 12.04$ cm.