What type of graph will it be What's the slope And error in the slope

The experiment is for the "Verification of Ohm's Law," which states that voltage ($V$) is directly proportional to current ($I$) for a given resistance ($R$), i.e., $V = IR$. When plotting $V$ on the y-axis and $I$ on the x-axis, this relationship results in a straight line.
The graph will be a straight line.

The slope of a $V$ versus $I$ graph represents the resistance $R$. We will use linear regression to find the slope $m$ from the given data points $(I, V)$.
The data points are:
$(I_i, V_i)$
$(0.12, 0.00)$
$(0.22, 0.40)$
$(0.32, 0.60)$
$(0.42, 0.80)$
$(0.52, 1.00)$
$(0.64, 1.20)$
There are $n=6$ data points.

First, calculate the necessary sums:
$\sum I_i = 0.12 + 0.22 + 0.32 + 0.42 + 0.52 + 0.64 = 2.24 \, \text{A}$
$\sum V_i = 0.00 + 0.40 + 0.60 + 0.80 + 1.00 + 1.20 = 4.00 \, \text{V}$
$\sum (I_i V_i) = (0.12 \times 0.00) + (0.22 \times 0.40) + (0.32 \times 0.60) + (0.42 \times 0.80) + (0.52 \times 1.00) + (0.64 \times 1.20)$
$\sum (I_i V_i) = 0.000 + 0.088 + 0.192 + 0.336 + 0.520 + 0.768 = 1.904 \, \text{V}\cdot\text{A}$
$\sum (I_i^2) = 0.12^2 + 0.22^2 + 0.32^2 + 0.42^2 + 0.52^2 + 0.64^2$
$\sum (I_i^2) = 0.0144 + 0.0484 + 0.1024 + 0.1764 + 0.2704 + 0.4096 = 1.0216 \, \text{A}^2$

The formula for the slope $m$ of a linear regression line is:

Substitute the calculated sums:

Rounding to three decimal places:
The slope is $\boxed{\text{2.216 } \Omega}$.

First, calculate the y-intercept $c$:

Next, calculate the sum of squared residuals, $\sum (V_i - V_{pi})^2$, where $V_{pi} = m I_i + c$:
$\sum (V_i - V_{pi})^2 \approx 0.023362$ (calculated with full precision in thought process)
The standard error of the estimate ($s_e$) is:

The standard error of the slope ($s_m$) is:

The denominator term $\sum (I_i^2) - \frac{(\sum I_i)^2}{n}$ is:

Now, calculate $s_m$:

Rounding to three decimal places:
The error in the slope is $\boxed{\text{0.178 } \Omega}$.