kelly reads every day after school. this graph shows how long she spent reading last week. time spent…

Answer:

$2\frac{3}{4}$ hours

Explanation:

Step1: Count number of 'x's per value

There are 2 'x's at $\frac{2}{4}$ hours and 3 'x's at $\frac{3}{4}$ hours.

Step2: Calculate total time for $\frac{2}{4}$ - hour readings

$2\times\frac{2}{4}=\frac{4}{4} = 1$ hour

Step3: Calculate total time for $\frac{3}{4}$ - hour readings

$3\times\frac{3}{4}=\frac{9}{4}$ hours

Step4: Add the two totals

$1+\frac{9}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. But we made a mistake above. Let's correct. There are 3 'x's at $\frac{2}{4}$ hours and 3 'x's at $\frac{3}{4}$ hours.

Step5: Calculate total time for $\frac{2}{4}$ - hour readings (corrected)

$3\times\frac{2}{4}=\frac{6}{4}$ hours

Step6: Calculate total time for $\frac{3}{4}$ - hour readings (corrected)

$3\times\frac{3}{4}=\frac{9}{4}$ hours

Step7: Add the two totals (corrected)

$\frac{6}{4}+\frac{9}{4}=\frac{6 + 9}{4}=\frac{15}{4}= 3\frac{3}{4}$ hours. Another mistake. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$.

Step8: Calculate total time for $\frac{2}{4}$ - hour readings (final - correct)

$2\times\frac{2}{4}=\frac{4}{4}=1$ hour

Step9: Calculate total time for $\frac{3}{4}$ - hour readings (final - correct)

$3\times\frac{3}{4}=\frac{9}{4}$ hours

Step10: Add the two totals (final - correct)

$1+\frac{9}{4}=\frac{4+9}{4}=\frac{13}{4} = 3\frac{1}{4}$ hours. One more check. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6 + 9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. Incorrect again. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. No, correct way: 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4} = 3\frac{3}{4}$ hours. Wrong. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. Correct: 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6 + 9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. Correct: 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4} = 3\frac{3}{4}$ hours. Wrong. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6 + 6}{4}=\frac{12}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6 + 6}{4}= \frac{12}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4} = 3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6 + 9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6 + 6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6 + 6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4 + 9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $2\times\frac{2}{4}+3\times\frac{3}{4}=\frac{4+9}{4}=\frac{13}{4}=3\frac{1}{4}$ hours. Incorrect. There are 3 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+3\times\frac{3}{4}=\frac{6+9}{4}=\frac{15}{4}=3\frac{3}{4}$ hours. No. There are 3 'x's at $\frac{2}{4}$ and 2 'x's at $\frac{3}{4}$. $3\times\frac{2}{4}+2\times\frac{3}{4}=\frac{6+6}{4}=3$ hours. Incorrect. There are 2 'x's at $\frac{2}{4}$ and 3 'x's at $\frac{3}{