Answer
About a 8.91% decrease
Procedure
The initial temperature is 15 °C, raising the temperature by 5 °C results in 20 °C.
Therefore we read 2 values on the table the dissolved oxygen at 15 °C and the dissolved oxygen at 20 °C.
From the table, we see that at 15 °C we have about 10.1 mg/L of dissolved oxygen.
From the table, we see that at 20 °C we have about 9.2 mg/L of dissolved oxygen.
The percent decrease is calculated as follows
![\%\text{ }decrease=(final-initial)/(initial)*100\%](https://img.qammunity.org/2023/formulas/chemistry/college/fl08oq8kxrpzvk452w531etbfr2zbte29t.png)
Substituting the values we have that
![\%\text{ }decrease=(9.2-10.1)/(10.1)*100\%=8.91\%](https://img.qammunity.org/2023/formulas/chemistry/college/h2h53xlcbc9g29va0zkurwfugkgxu4szn3.png)
![\%\text{ }decrease=(9.2-10.1)/(10.1)*100\%=-8.91\%](https://img.qammunity.org/2023/formulas/chemistry/college/arcwnro7n2th7c1iulse12aou0yeyww8xu.png)
The result can be negative or preceded by the word a decrease of.