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use the corrected line plot to answer the question below. age differenc…

Question

use the corrected line plot to answer the question below. age difference line plot with x marks at positions labeled 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} (x marks at 1, 1\frac{1}{8}, two at 1\frac{1}{4}, one at 1\frac{3}{4}, one at 1\frac{7}{8}) what was the difference between the largest age difference and the smallest age difference? write your answer as a fraction, mixed number, or whole number. \boxed{} years

Explanation:

Step1: Identify largest and smallest

Largest age difference: $1\frac{7}{8}$ (from the line plot, the rightmost X is at $1\frac{7}{8}$).
Smallest age difference: $1$ (leftmost X is at $1$).

Step2: Subtract smallest from largest

Convert mixed number to improper fraction:
$1\frac{7}{8} = \frac{1\times8 + 7}{8} = \frac{15}{8}$
$1 = \frac{8}{8}$
Subtract: $\frac{15}{8} - \frac{8}{8} = \frac{7}{8}$? Wait, no—wait, the rightmost X: wait, the labels are $1, 1\frac{1}{8}, 1\frac{1}{4}, 1\frac{3}{8}, 1\frac{1}{2}, 1\frac{5}{8}, 1\frac{3}{4}, 1\frac{7}{8}$. Wait, the Xs: leftmost at $1$, then $1\frac{1}{8}$, then $1\frac{1}{4}$ (two Xs), then $1\frac{3}{4}$ (one X), $1\frac{7}{8}$ (one X). Wait, no—wait the line plot: the X positions are at $1$, $1\frac{1}{8}$, $1\frac{1}{4}$ (two Xs), $1\frac{3}{4}$ (one X), $1\frac{7}{8}$ (one X). Wait, so largest is $1\frac{7}{8}$, smallest is $1$.

Wait, $1\frac{7}{8} - 1 = \frac{7}{8}$? No, wait: $1\frac{7}{8}$ is $1 + \frac{7}{8}$, minus $1$ is $\frac{7}{8}$? Wait, no, wait the labels: let's check the number line. The marks are at $1, 1\frac{1}{8}, 1\frac{1}{4}, 1\frac{3}{8}, 1\frac{1}{2}, 1\frac{5}{8}, 1\frac{3}{4}, 1\frac{7}{8}$. So the rightmost X is at $1\frac{7}{8}$, leftmost at $1$. So difference is $1\frac{7}{8} - 1 = \frac{7}{8}$? Wait, no, wait: $1\frac{7}{8}$ is $\frac{15}{8}$, $1$ is $\frac{8}{8}$, $\frac{15}{8} - \frac{8}{8} = \frac{7}{8}$. Wait, but maybe I misread the X positions. Wait the Xs: first at $1$, then $1\frac{1}{8}$, then $1\frac{1}{4}$ (two Xs), then $1\frac{3}{4}$ (one X), $1\frac{7}{8}$ (one X). So largest is $1\frac{7}{8}$, smallest is $1$. So $1\frac{7}{8} - 1 = \frac{7}{8}$? Wait, no, wait: $1\frac{7}{8}$ is $1 + 7/8$, subtract $1$ is $7/8$. But wait, maybe the largest is $1\frac{7}{8}$ and smallest is $1$, so difference is $7/8$? Wait, no, wait the number line: let's check the fractions. $1\frac{7}{8} - 1 = \frac{7}{8}$. Yes.

Wait, but let's confirm: $1\frac{7}{8}$ is equal to $1 + 7/8$, and $1$ is $1 + 0/8$. So subtracting, we get $7/8$. So the difference is $\frac{7}{8}$? Wait, no, wait: $1\frac{7}{8} - 1 = \frac{7}{8}$. Yes.

Wait, maybe I made a mistake. Let's re-express:

Largest age difference: $1\frac{7}{8}$ years.
Smallest age difference: $1$ year.

Difference: $1\frac{7}{8} - 1 = \frac{7}{8}$ years.

Answer:

$\frac{7}{8}$