Question

use the corrected line plot to answer the question below. age difference line plot with x’s at 1, (1\frac{1}{8}), (1\frac{1}{4}) (two x’s), (1\frac{3}{4}), (1\frac{7}{8}) years apart. how many students reported an age difference of (1\frac{3}{4}) years or less? students

Explanation:

Step1: Identify values ≤ \(1\frac{3}{4}\)

First, we list the age differences from the line plot and check which are less than or equal to \(1\frac{3}{4}\) (which is \(1.75\) or \(\frac{7}{4}\) or \(1\frac{6}{8}\)). The values on the line plot are \(1\), \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\), \(1\frac{3}{4}\), \(1\frac{7}{8}\). So we consider \(1\), \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\), \(1\frac{3}{4}\).

Step2: Count the X's for each value

- For \(1\): 1 X
- For \(1\frac{1}{8}\): 1 X
- For \(1\frac{1}{4}\): 2 X's (wait, looking at the plot: the first mark (1) has 1 X, second (\(1\frac{1}{8}\)) has 1 X, third (\(1\frac{1}{4}\)) has 2 X's, then the next marks up to \(1\frac{3}{4}\): wait, the plot shows: at 1: 1 X, at \(1\frac{1}{8}\): 1 X, at \(1\frac{1}{4}\): 2 X's, then the next marks ( \(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\)) have 0 X's, and at \(1\frac{3}{4}\): 1 X? Wait no, let's re - examine the plot. The X's are at: 1 (1 X), \(1\frac{1}{8}\) (1 X), \(1\frac{1}{4}\) (2 X's), \(1\frac{3}{4}\) (1 X), \(1\frac{7}{8}\) (1 X). Wait, no, the original plot: the first X at 1, second at \(1\frac{1}{8}\), third at \(1\frac{1}{4}\) (two X's), then later at \(1\frac{3}{4}\) (one X), and \(1\frac{7}{8}\) (one X). Wait, we need to count all X's where the value is ≤ \(1\frac{3}{4}\). So the values are 1, \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{4}\) (and the ones in between with 0 X's). So:

- At 1: 1 X
- At \(1\frac{1}{8}\): 1 X
- At \(1\frac{1}{4}\): 2 X's
- At \(1\frac{3}{4}\): 1 X

Wait, no, maybe I misread. Let's list the positions:

The line plot has marks at \(1\), \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\), \(1\frac{3}{4}\), \(1\frac{7}{8}\).

The X's are:

- \(1\): 1 X
- \(1\frac{1}{8}\): 1 X
- \(1\frac{1}{4}\): 2 X's (so total 1 + 1+ 2)
- Then, the next marks (\(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\)) have 0 X's.
- Then \(1\frac{3}{4}\): 1 X

Wait, but \(1\frac{3}{4}\) is equal to \(1\frac{3}{4}\), so we include it. Then \(1\frac{7}{8}\) is greater than \(1\frac{3}{4}\) (since \(1\frac{7}{8}=1.875\) and \(1\frac{3}{4} = 1.75\)), so we exclude it.

So total X's: 1 (at 1) + 1 (at \(1\frac{1}{8}\)) + 2 (at \(1\frac{1}{4}\)) + 0 (at \(1\frac{3}{8}\)) + 0 (at \(1\frac{1}{2}\)) + 0 (at \(1\frac{5}{8}\)) + 1 (at \(1\frac{3}{4}\)) = 1 + 1+ 2 + 1=5? Wait, no, maybe I made a mistake. Wait the original plot: looking at the image, the X's are:

- At 1: 1 X
- At \(1\frac{1}{8}\): 1 X
- At \(1\frac{1}{4}\): 2 X's (so two X's stacked)
- Then, later at \(1\frac{3}{4}\): 1 X
- At \(1\frac{7}{8}\): 1 X

Wait, but we need age difference of \(1\frac{3}{4}\) or less. So \(1\frac{7}{8}\) is more than \(1\frac{3}{4}\) (since \(\frac{7}{8}=0.875\), \(\frac{3}{4}=0.75\), so \(1\frac{7}{8}=1.875\), \(1\frac{3}{4}=1.75\)), so we exclude \(1\frac{7}{8}\).

So the X's at \(1\) (1), \(1\frac{1}{8}\) (1), \(1\frac{1}{4}\) (2), \(1\frac{3}{4}\) (1). Wait, but what about the marks between \(1\frac{1}{4}\) and \(1\frac{3}{4}\) (i.e., \(1\frac{3}{8}\), \(1\frac{1}{2}\), \(1\frac{5}{8}\))? They have 0 X's. So total is 1 + 1+ 2+ 1 = 5? Wait, no, maybe I miscounted the X's at \(1\frac{1}{4}\). Wait the plot shows at \(1\frac{1}{4}\) there are two X's (so 2), at 1:1, at \(1\frac{1}{8}\):1, at \(1\frac{3}{4}\):1. So 1 + 1+ 2 + 1=5? Wait, but let's check again. Wait the problem is to count all students (X's) where the age difference is \(1\frac{3}{4}\) or less. So we need to sum the number of X's for each data point with value ≤ \(1\frac{3}{4}\).

Data points and their counts:

- \(1\): 1
- \(1\frac{1}{8}\): 1
- \(1\frac{1}{4}\): 2
- \(1\frac{3}{8}\): 0
- \(1\frac{1}{2}\): 0
- \(1\frac{5}{8}\): 0
- \(1\frac{3}{4}\): 1
- \(1\frac{7}{8}\): 0 (since it's more than \(1\frac{3}{4}\))

Now sum these counts: 1 + 1+ 2 + 0+ 0 + 0+ 1=5? Wait, no, wait the X at \(1\frac{3}{4}\) is 1, and the X at \(1\frac{7}{8}\) is 1, but we exclude \(1\frac{7}{8}\). Wait, maybe I made a mistake in the number of X's at \(1\frac{1}{4}\). Let's look at the plot again: the user's plot shows:

- At 1: 1 X
- At \(1\frac{1}{8}\): 1 X
- At \(1\frac{1}{4}\): 2 X's (two green X's)
- Then, later at \(1\frac{3}{4}\): 1 X
- At \(1\frac{7}{8}\): 1 X

So the values ≤ \(1\frac{3}{4}\) are \(1\), \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{4}\) (and the ones in between with 0). So the number of X's: 1 (for 1) + 1 (for \(1\frac{1}{8}\)) + 2 (for \(1\frac{1}{4}\)) + 1 (for \(1\frac{3}{4}\)) = 5? Wait, but wait, is \(1\frac{3}{4}\) included? Yes, because the question is "or less". So \(1\frac{3}{4}\) is equal, so we include it. Then what about the X at \(1\frac{3}{4}\): 1 X. Then the X at \(1\frac{7}{8}\) is 1 X, which is more than \(1\frac{3}{4}\), so we exclude it.

Wait, but maybe I missed some X's. Wait the total X's: 1 (at 1) + 1 (at \(1\frac{1}{8}\)) + 2 (at \(1\frac{1}{4}\)) + 1 (at \(1\frac{3}{4}\)) + 1 (at \(1\frac{7}{8}\)) = 6? But we need to exclude \(1\frac{7}{8}\), so 6 - 1 = 5? Wait, no, let's count the X's:

- 1: 1
- \(1\frac{1}{8}\): 1 (total 2)
- \(1\frac{1}{4}\): 2 (total 4)
- \(1\frac{3}{4}\): 1 (total 5)
- \(1\frac{7}{8}\): 1 (total 6)

So the ones ≤ \(1\frac{3}{4}\) are the first four (1, \(1\frac{1}{8}\), \(1\frac{1}{4}\), \(1\frac{3}{4}\)) with counts 1,1,2,1. So 1 + 1+ 2 + 1 = 5? Wait, no, 1+1=2, 2+2=4, 4 + 1=5. Yes.

Wait, but maybe the correct count is 1 (1) + 1 (\(1\frac{1}{8}\)) + 2 (\(1\frac{1}{4}\)) + 1 (\(1\frac{3}{4}\)) = 5. Wait, but let's check the fractions:

\(1\frac{3}{4}=\frac{7}{4}=\frac{14}{8}\)

\(1=\frac{8}{8}\), \(1\frac{1}{8}=\frac{9}{8}\), \(1\frac{1}{4}=\frac{10}{8}\), \(1\frac{3}{8}=\frac{11}{8}\), \(1\frac{1}{2}=\frac{12}{8}\), \(1\frac{5}{8}=\frac{13}{8}\), \(1\frac{3}{4}=\frac{14}{8}\), \(1\frac{7}{8}=\frac{15}{8}\)

So all values with numerator ≤ 14 (when denominator is 8) are included. So:

- \(\frac{8}{8}\) (1): 1 X
- \(\frac{9}{8}\) (\(1\frac{1}{8}\)): 1 X
- \(\frac{10}{8}\) (\(1\frac{1}{4}\)): 2 X's
- \(\frac{11}{8}\) (\(1\frac{3}{8}\)): 0 X's
- \(\frac{12}{8}\) (\(1\frac{1}{2}\)): 0 X's
- \(\frac{13}{8}\) (\(1\frac{5}{8}\)): 0 X's
- \(\frac{14}{8}\) (\(1\frac{3}{4}\)): 1 X
- \(\frac{15}{8}\) (\(1\frac{7}{8}\)): 0 X's (excluded)

Summing the counts: 1 + 1+ 2 + 0+ 0 + 0+ 1 = 5.

Wait, but maybe I made a mistake. Let's count the X's again from the plot:

- At 1: 1 X
- At \(1\frac{1}{8}\): 1 X (total 2)
- At \(1\frac{1}{4}\): 2 X's (total 4)
- At \(1\frac{3}{4}\): 1 X (total 5)
- At \(1\frac{7}{8}\): 1 X (total 6, but we exclude this one)

So the number of students with age difference \(1\frac{3}{4}\) or less is 5? Wait, no, wait the X at \(1\frac{3}{4}\) is 1, and the X at \(1\frac{7}{8}\) is 1. So when we count "or less", we include \(1\frac{3}{4}\) and all below. So the X's are:

1 (1), 1 (\(1\frac{1}{8}\)), 2 (\(1\frac{1}{4}\)), 1 (\(1\frac{3}{4}\)) → total 1 + 1+ 2 + 1 = 5.

Answer:

5