Question

the line plot shows the distance lena walks each day for 10 days. each x represents 1 day.
distances walked

distance (miles)
how many miles does lena walk in 10 days?
a. 10 miles
b. 11\frac{1}{2} miles
c. 19 miles
d. 23\frac{3}{4} miles

Explanation:

Step1: Count X for each distance

• \( 1\frac{1}{2} \) miles: 1 X
• \( 1\frac{3}{4} \) miles: 2 Xs
• \( 2\frac{1}{2} \) miles: 4 Xs
• \( 2\frac{3}{4} \) miles: 1 X
• 3 miles: 2 Xs

Step2: Calculate total distance for each group

• \( 1\frac{1}{2} \times 1 = \frac{3}{2} \)
• \( 1\frac{3}{4} \times 2 = \frac{7}{4} \times 2 = \frac{7}{2} \)
• \( 2\frac{1}{2} \times 4 = \frac{5}{2} \times 4 = 10 \)
• \( 2\frac{3}{4} \times 1 = \frac{11}{4} \)
• \( 3 \times 2 = 6 \)

Step3: Sum all distances

Convert to quarters:
\( \frac{3}{2} = \frac{6}{4} \), \( \frac{7}{2} = \frac{14}{4} \), \( 10 = \frac{40}{4} \), \( \frac{11}{4} \), \( 6 = \frac{24}{4} \)

Total: \( \frac{6 + 14 + 40 + 11 + 24}{4} = \frac{95}{4} = 23\frac{3}{4} \)? Wait, no, wait—wait, I miscounted Xs. Wait, let's re - check the line plot:

Wait, the distances:

• \( 1\frac{1}{2} \): 1 X
• \( 1\frac{3}{4} \): 2 Xs
• \( 2\frac{1}{2} \): 4 Xs? Wait, no, looking at the plot:

Wait the plot has:

• \( 1\frac{1}{2} \): 1 X
• \( 1\frac{3}{4} \): 2 Xs
• \( 2\frac{1}{2} \): 4 Xs? Wait, no, the Xs:

Wait the labels: \( 1\frac{1}{2} \) (1 X), \( 1\frac{3}{4} \) (2 Xs), \( 2\frac{1}{2} \) (4 Xs? Wait, no, the Xs at \( 2\frac{1}{2} \) are 4? Wait, the original plot:

Wait the user's plot:

• \( 1\frac{1}{2} \): 1 X
• \( 1\frac{3}{4} \): 2 Xs
• \( 2\frac{1}{2} \): 4 Xs? Wait, no, let's count again:

Wait the X positions:

• \( 1\frac{1}{2} \): 1 X
• \( 1\frac{3}{4} \): 2 Xs
• \( 2\frac{1}{2} \): 4 Xs? Wait, no, the Xs:

Wait the problem says 10 days. Let's count the Xs:

1 (at \( 1\frac{1}{2} \)) + 2 (at \( 1\frac{3}{4} \)) + 4 (at \( 2\frac{1}{2} \)) + 1 (at \( 2\frac{3}{4} \)) + 2 (at 3) = 1 + 2 + 4 + 1 + 2 = 10. Correct.

Now recalculate:

• \( 1\frac{1}{2} \times 1 = \frac{3}{2} = 1.5 \)
• \( 1\frac{3}{4} \times 2 = \frac{7}{4} \times 2 = 3.5 \)
• \( 2\frac{1}{2} \times 4 = \frac{5}{2} \times 4 = 10 \)
• \( 2\frac{3}{4} \times 1 = 2.75 \)
• \( 3 \times 2 = 6 \)

Now sum: 1.5 + 3.5 + 10 + 2.75 + 6 = (1.5 + 3.5) + 10 + (2.75 + 6) = 5 + 10 + 8.75 = 23.75 = \( 23\frac{3}{4} \). Wait, but that's option D. But wait, maybe I made a mistake. Wait, no—wait, let's check again.

Wait, maybe the distances are:

Wait, \( 1\frac{1}{2} \) (1), \( 1\frac{3}{4} \) (2), \( 2\frac{1}{2} \) (4), \( 2\frac{3}{4} \) (1), 3 (2).

Calculating:

\( 1\frac{1}{2} = \frac{3}{2} \), \( 1\frac{3}{4} = \frac{7}{4} \), \( 2\frac{1}{2} = \frac{5}{2} \), \( 2\frac{3}{4} = \frac{11}{4} \), 3 = \( \frac{12}{4} \)

Total distance:

\( \frac{3}{2} \times 1 + \frac{7}{4} \times 2 + \frac{5}{2} \times 4 + \frac{11}{4} \times 1 + 3 \times 2 \)

= \( \frac{3}{2} + \frac{7}{2} + 10 + \frac{11}{4} + 6 \)

Convert to quarters:

\( \frac{6}{4} + \frac{14}{4} + \frac{40}{4} + \frac{11}{4} + \frac{24}{4} \)

Sum numerators: 6 + 14 + 40 + 11 + 24 = 95

\( \frac{95}{4} = 23\frac{3}{4} \). Wait, but that's option D. But wait, maybe I misread the X counts. Wait, let's check the options. Option D is \( 23\frac{3}{4} \) miles.

Wait, but let's do it again:

• \( 1\frac{1}{2} \) (1 day): \( 1\frac{1}{2} \times 1 = 1.5 \)
• \( 1\frac{3}{4} \) (2 days): \( 1\frac{3}{4} \times 2 = 3.5 \)
• \( 2\frac{1}{2} \) (4 days): \( 2\frac{1}{2} \times 4 = 10 \)
• \( 2\frac{3}{4} \) (1 day): \( 2\frac{3}{4} \times 1 = 2.75 \)
• 3 (2 days): \( 3 \times 2 = 6 \)

Now sum: 1.5 + 3.5 = 5; 5 + 10 = 15; 15 + 2.75 = 17.75; 17.75 + 6 = 23.75 = \( 23\frac{3}{4} \). So the total is \( 23\frac{3}{4} \) miles.

Answer:

D. \( 23\frac{3}{4} \) miles