We are given that points $X$, $Y$, and $Z$ lie on a number line at positions $-2$, $0$, and $x_3$, respectively. The weights of these points are as follows:

• Point $X$ has twice the weight of point $Y$.
• Point $Z$ has three times the weight of point $Y$.
• The weighted average of the three points is located at $\frac{4}{3}$.

To find the location of point $Z$, we use the formula for a weighted average:

$\text{Weighted average} = \frac{w_X x_X + w_Y x_Y + w_Z x_Z}{w_X + w_Y + w_Z}$

Where:

• $w_X$, $w_Y$, and $w_Z$ are the weights of points $X$, $Y$, and $Z$, respectively.
• $x_X$, $x_Y$, and $x_Z$ are the positions of points $X$, $Y$, and $Z$.

Step-by-Step Solution:

1. Assign weights:

   • Let the weight of point $Y$ be $w_Y = 1$.
   • The weight of point $X$ is $w_X = 2$, since it weighs twice as much as $Y$.
   • The weight of point $Z$ is $w_Z = 3$, since it weighs three times as much as $Y$.

2. Substitute positions and weights:

   • The position of $X$ is $x_X = -2$.
   • The position of $Y$ is $x_Y = 0$.
   • The position of $Z$ is $x_Z$ (which we need to find).
   • The weighted average is given as $\frac{4}{3}$.

Now, substitute all these into the weighted average formula:

$\frac{4}{3} = \frac{2(-2) + 1(0) + 3(x_Z)}{2 + 1 + 3}$

3. Simplify the equation: $\frac{4}{3} = \frac{-4 + 0 + 3x_Z}{6}$ $\frac{4}{3} = \frac{-4 + 3x_Z}{6}$

4. Multiply both sides by 6 to eliminate the denominator: $8 = -4 + 3x_Z$

5. Solve for $x_Z$: $8 + 4 = 3x_Z$ $12 = 3x_Z$ $x_Z = 4$

Final Answer:

The location of point $Z$ is $x_Z = 4$.

Do you want any further clarification on this solution?

Here are 5 related questions to expand your understanding:

1. What is the general formula for a weighted average in higher dimensions?
2. How would the solution change if point $Z$ had a different weight, say twice the weight of point $Y$?
3. How does the concept of center of mass relate to weighted averages in mathematics?
4. If the points were on a 2D plane, how would the weighted average be calculated?
5. How can we apply weighted averages in probability and statistics?

Tip: Always check the sum of weights when dealing with weighted averages to ensure accuracy in calculations.