The table shows experimental values of two variables, x and y, which are connected by an equation of the form $y = \frac{k}{x^{n}}$, where n and k are constants.

1. Calculate the values of $\lg x$ and $\lg y$ for each data point, rounding to 2 decimal places.

For $x = 1.58$, $\lg x \approx 0.20$. For $y = 20.89$, $\lg y \approx 1.32$.
For $x = 2.00$, $\lg x \approx 0.30$. For $y = 15.14$, $\lg y \approx 1.18$.
For $x = 5.01$, $\lg x \approx 0.70$. For $y = 4.37$, $\lg y \approx 0.64$.
For $x = 13.80$, $\lg x \approx 1.14$. For $y = 1.10$, $\lg y \approx 0.04$.

2. Plot the points on a Cartesian plane and draw the line of best fit.

Choose appropriate scales for the axes (e.g., $2\text{ cm}$ to $0.2$ units on both axes).
Plot the points $(0.20, 1.32)$, $(0.30, 1.18)$, $(0.70, 0.64)$, and $(1.14, 0.04)$.
Draw a straight line passing through or as close as possible to all the plotted points.

1. Transform the given non-linear equation into a linear form involving $\lg y$ and $\lg x$.

$y = \frac{k}{x^n}$
$\lg y = \lg \left(\frac{k}{x^n}\right)$
$\lg y = \lg k - \lg x^n$
$\lg y = -n \lg x + \lg k$

2. Calculate the gradient of the line to estimate $n$.

Using points from the table to find the gradient of the line of best fit (or using the theoretical points if the line passes through them perfectly):
Let's use $(x_1, y_1) = (1.58, 20.89)$ and $(x_2, y_2) = (13.80, 1.10)$.
$\lg x_1 = \lg 1.58 \approx 0.1987$
$\lg y_1 = \lg 20.89 \approx 1.3199$
$\lg x_2 = \lg 13.80 \approx 1.1399$
$\lg y_2 = \lg 1.10 \approx 0.0414$

$\text{Gradient} = \frac{0.0414 - 1.3199}{1.1399 - 0.1987} = \frac{-1.2785}{0.9412} \approx -1.358$
Since gradient $= -n$, $n \approx 1.36$ (accept $1.35$ to $1.37$).

3. Calculate the vertical intercept to estimate $k$.

Using the equation $\lg y = -1.358 \lg x + \lg k$ and the point $(0.1987, 1.3199)$:
$1.3199 = -1.358(0.1987) + \lg k$
$1.3199 = -0.2698 + \lg k$
$\lg k = 1.5897$
$k = 10^{1.5897} \approx 38.9$ (accept $38.0$ to $40.0$ depending on the line of best fit).

1. Identify the equation of the straight line to be drawn

The given equation is $\lg y = \lg x^{2}$.
Using the power law of logarithms, this simplifies to $\lg y = 2\lg x$.
Letting $Y = \lg y$ and $X = \lg x$, the line to draw is $Y = 2X$.

2. Find the intersection point and solve for x

Draw the line $Y = 2X$ on the same axes.
From the graph, the intersection of this line with the line of best fit occurs at approximately $X = 0.473$.
Since $X = \lg x$, we have $\lg x = 0.473$.
$x = 10^{0.473} \approx 2.97$