Question

points a and b are shown on the number line.
a teacher finds the weighted - average, w1, of points a and b by considering their weights as 1 and 2 respectively. a student, however, finds the weighted average, w2, of points a and b by considering their weights as 2 and 1 respectively. select all statements that are true.
□ w1 is less than w2.
□ w1 is greater than w2.
□ w2 is closer to a than b.
□ w1 is closer to b than a.
□ w2 is closer to a than b.
□ w2 is closer to b than a.

Explanation:

Step1: Determine the values of points A and B

From the number - line, \(A = 3\) and \(B = 6\).

Step2: Calculate \(w_1\)

The formula for weighted - average is \(w=\frac{w_1x_1 + w_2x_2}{w_1 + w_2}\). For \(w_1\), where the weights are \(w_1 = 2\) and \(w_2 = 1\), \(x_1=3\) and \(x_2 = 6\). Then \(w_1=\frac{2\times3+1\times6}{2 + 1}=\frac{6 + 6}{3}=\frac{12}{3}=4\).

Step3: Calculate \(w_2\)

For \(w_2\), where the weights are \(w_1 = 1\) and \(w_2 = 2\), \(x_1 = 3\) and \(x_2 = 6\). Then \(w_2=\frac{1\times3+2\times6}{1 + 2}=\frac{3+12}{3}=\frac{15}{3}=5\).

Step4: Analyze the relationships

Since \(4<5\), \(w_1\) is less than \(w_2\). The distance between \(w_1 = 4\) and \(A = 3\) is \(|4 - 3|=1\), and the distance between \(w_1 = 4\) and \(B = 6\) is \(|4 - 6| = 2\), so \(w_1\) is closer to \(A\) than \(B\). The distance between \(w_2 = 5\) and \(A = 3\) is \(|5 - 3|=2\), and the distance between \(w_2 = 5\) and \(B = 6\) is \(|5 - 6| = 1\), so \(w_2\) is closer to \(B\) than \(A\).

Answer:

• \(w_1\) is less than \(w_2\)
• \(w_1\) is closer to \(A\) than \(B\)
• \(w_2\) is closer to \(B\) than \(A\)