Question

use your line plot to answer the question below. amount of laundry detergent line plot with x - marks at 1, (1\frac{1}{4}), (1\frac{1}{2}), (1\frac{3}{4}), 2 (labeled as tablespoons) what fraction of the loads contain no more than (1\frac{1}{4}) tablespoons of detergent? write your answer as a fraction, mixed number, or whole number. (square) of the laundry loads

Explanation:

Step1: Count loads at or below \(1\frac{1}{4}\)

At 1: 2 X's; At \(1\frac{1}{4}\): 3 X's. Total: \(2 + 3 = 5\).

Step2: Count total loads

At 1: 2; \(1\frac{1}{4}\): 3; \(1\frac{1}{2}\): 2; \(1\frac{3}{4}\): 3; 2: 3. Total: \(2 + 3 + 2 + 3 + 3 = 13\)? Wait, no, re - count:
Wait, let's re - examine the line plot:
- For 1: 2 X's (second row and third row)
- For \(1\frac{1}{4}\): 3 X's (first row:1, second row:1, third row:1? Wait no, looking at the plot:
First row (top row): \(1\frac{1}{4}\) has 1, \(1\frac{3}{4}\) has 1, 2 has 1.
Second row: 1 has 1, \(1\frac{1}{4}\) has 1, \(1\frac{1}{2}\) has 1, \(1\frac{3}{4}\) has 1, 2 has 1.
Third row: 1 has 1, \(1\frac{1}{4}\) has 1, \(1\frac{1}{2}\) has 1, \(1\frac{3}{4}\) has 1, 2 has 1.
Wait, maybe better to count each column:
- Column 1 (1): second row X, third row X → 2
- Column \(1\frac{1}{4}\): first row X, second row X, third row X → 3
- Column \(1\frac{1}{2}\): second row X, third row X → 2
- Column \(1\frac{3}{4}\): first row X, second row X, third row X → 3
- Column 2: first row X, second row X, third row X → 3

Total loads: \(2 + 3+2 + 3+3=13\)? Wait, no, that can't be. Wait, maybe I miscounted the first row. Let's look again:

First row (top): \(1\frac{1}{4}\) (1 X), \(1\frac{3}{4}\) (1 X), 2 (1 X) → 3 X's.

Second row: 1 (1 X), \(1\frac{1}{4}\) (1 X), \(1\frac{1}{2}\) (1 X), \(1\frac{3}{4}\) (1 X), 2 (1 X) → 5 X's.

Third row: 1 (1 X), \(1\frac{1}{4}\) (1 X), \(1\frac{1}{2}\) (1 X), \(1\frac{3}{4}\) (1 X), 2 (1 X) → 5 X's.

Wait, no, the original plot:

Looking at the image:

First row (top): two X's at \(1\frac{3}{4}\) and 2? No, the user's plot:

"First row (top): X at \(1\frac{1}{4}\), X at \(1\frac{3}{4}\), X at 2"

Second row: X at 1, X at \(1\frac{1}{4}\), X at \(1\frac{1}{2}\), X at \(1\frac{3}{4}\), X at 2

Third row: X at 1, X at \(1\frac{1}{4}\), X at \(1\frac{1}{2}\), X at \(1\frac{3}{4}\), X at 2

Wait, maybe the correct count:

- 1: second row X, third row X → 2
- \(1\frac{1}{4}\): first row X, second row X, third row X → 3
- \(1\frac{1}{2}\): second row X, third row X → 2
- \(1\frac{3}{4}\): first row X, second row X, third row X → 3
- 2: first row X, second row X, third row X → 3

Total: \(2 + 3+2 + 3+3 = 13\)? Wait, but let's check the "no more than \(1\frac{1}{4}\)" which is 1 and \(1\frac{1}{4}\). So 1 has 2, \(1\frac{1}{4}\) has 3. So total favorable: \(2 + 3=5\). Total loads: let's sum all X's.

First row: 3 X's

Second row: 5 X's

Third row: 5 X's

Wait, no, first row: 3, second:5, third:5? No, the first row has 3 X's (as per the plot: \(1\frac{1}{4}\), \(1\frac{3}{4}\), 2), second row has 5 (1, \(1\frac{1}{4}\), \(1\frac{1}{2}\), \(1\frac{3}{4}\), 2), third row has 5 (same as second). So total X's: \(3 + 5+5 = 13\)? Wait, but 3 + 5+5 = 13. But the favorable (1 and \(1\frac{1}{4}\)): at 1: 2 (second and third row), at \(1\frac{1}{4}\): 3 (first, second, third). So 2 + 3 = 5. So the fraction is \(\frac{5}{13}\)? Wait, no, maybe I miscounted the total.

Wait, let's do it again:

- 1: number of X's: look at the column for 1. Second row: X, third row: X → 2
- \(1\frac{1}{4}\): first row: X, second row: X, third row: X → 3
- \(1\frac{1}{2}\): second row: X, third row: X → 2
- \(1\frac{3}{4}\): first row: X, second row: X, third row: X → 3
- 2: first row: X, second row: X, third row: X → 3

Now sum all: 2 + 3+2 + 3+3 = 13. Favorable (≤\(1\frac{1}{4}\)): 2 + 3 = 5. So the fraction is \(\frac{5}{13}\)? Wait, but maybe I made a mistake in counting the first row. Let's check the original plot again. The user's plot:

First row (top): X at \(1\frac{1}{4}\), X at \(1\frac{3}{4}\), X at 2 → 3 X's.

Second row: X at 1, X at \(1\frac{1}{4}\), X at \(1\frac{1}{2}\), X at \(1\frac{3}{4}\), X at 2 → 5 X's.

Third row: X at 1, X at \(1\frac{1}{4}\), X at \(1\frac{1}{2}\), X at \(1\frac{3}{4}\), X at 2 → 5 X's.

So total X's: 3 + 5 + 5 = 13.

Number of loads with no more than \(1\frac{1}{4}\) (i.e., 1 or \(1\frac{1}{4}\)):

- At 1: in second and third row → 2
- At \(1\frac{1}{4}\): in first, second, third row → 3

Total: 2 + 3 = 5.

So the fraction is \(\frac{5}{13}\)? Wait, but maybe I miscounted. Wait, maybe the first row has 1 X at \(1\frac{1}{4}\), 1 at \(1\frac{3}{4}\), 1 at 2 (3 X's). Second row: 1 at 1, 1 at \(1\frac{1}{4}\), 1 at \(1\frac{1}{2}\), 1 at \(1\frac{3}{4}\), 1 at 2 (5 X's). Third row: same as second (5 X's). So total 3 + 5 + 5 = 13. Favorable: 1 (1 in second) + 1 (1 in third) + 1 (\(1\frac{1}{4}\) in first) + 1 (\(1\frac{1}{4}\) in second) + 1 (\(1\frac{1}{4}\) in third) → 5. So yes, \(\frac{5}{13}\)? Wait, but maybe the total is different. Wait, maybe the first row has 1 X at \(1\frac{1}{4}\), 2 X's at \(1\frac{3}{4}\) and 2? No, the plot shows:

Looking at the user's image:

First row (top): one X at \(1\frac{1}{4}\), one X at \(1\frac{3}{4}\), one X at 2.

Second row: one X at 1, one X at \(1\frac{1}{4}\), one X at \(1\frac{1}{2}\), one X at \(1\frac{3}{4}\), one X at 2.

Third row: one X at 1, one X at \(1\frac{1}{4}\), one X at \(1\frac{1}{2}\), one X at \(1\frac{3}{4}\), one X at 2.

So:

- 1: 2 (second and third)
- \(1\frac{1}{4}\): 3 (first, second, third)
- \(1\frac{1}{2}\): 2 (second, third)
- \(1\frac{3}{4}\): 3 (first, second, third)
- 2: 3 (first, second, third)

Total: 2 + 3 + 2 + 3 + 3 = 13.

Favorable: 2 + 3 = 5. So fraction is \(\frac{5}{13}\). Wait, but maybe I made a mistake. Let's check again.

Wait, maybe the first row has 2 X's at \(1\frac{3}{4}\) and 2? No, the user's plot shows:

"First row (top): X X (at \(1\frac{3}{4}\) and 2?) No, the original text's plot:

"First row: X (at \(1\frac{1}{4}\)), X (at \(1\frac{3}{4}\)), X (at 2)"

"Second row: X (at 1), X (at \(1\frac{1}{4}\)), X (at \(1\frac{1}{2}\)), X (at \(1\frac{3}{4}\)), X (at 2)"

"Third row: X (at 1), X (at \(1\frac{1}{4}\)), X (at \(1\frac{1}{2}\)), X (at \(1\frac{3}{4}\)), X (at 2)"

So:

- 1: 2 (second, third)
- \(1\frac{1}{4}\): 3 (first, second, third)
- \(1\frac{1}{2}\): 2 (second, third)
- \(1\frac{3}{4}\): 3 (first, second, third)
- 2: 3 (first, second, third)

Total: 2 + 3 + 2 + 3 + 3 = 13.

Favorable (≤\(1\frac{1}{4}\)): 2 + 3 = 5. So the fraction is \(\frac{5}{13}\).

Answer:

\(\frac{5}{13}\)