A sewage treatment plant discharges a pollutant with a concentration of $$ 10 \mathrm{mg} / \mathrm{L} $$. The plant's total discharge is 0.2 MGD (millions of gallons per day). Determine the rate of pollutant released in pounds pe day. Give your answer accurate to at least two decimal places.

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To determine the rate of pollutant released in pounds per day, we can follow these steps:

1. **Convert the discharge from MGD to liters per day**:
   $$
   1 \text{ MGD} = 3.78541 \text{ m}^3/\text{day} = 3,785,410 \text{ L/day}
   $$
   Therefore, for 0.2 MGD:
   $$
   0.2 \text{ MGD} = 0.2 \times 3,785,410 \text{ L/day} = 757,082 \text{ L/day}
   $$

2. **Calculate the total mass of the pollutant in milligrams per day**:
   The concentration of the pollutant is $$ 10 \text{ mg/L} $$. Thus, the total mass of the pollutant is:
   $$
   \text{Total mass (mg)} = \text{Concentration (mg/L)} \times \text{Volume (L/day)}
   $$
   $$
   \text{Total mass (mg)} = 10 \text{ mg/L} \times 757,082 \text{ L/day} = 7,570,820 \text{ mg/day}
   $$

3. **Convert milligrams to pounds**:
   We know that:
   $$
   1 \text{ pound} = 453,592.37 \text{ mg}
   $$
   Therefore, to convert milligrams to pounds:
   $$
   \text{Total mass (pounds)} = \frac{7,570,820 \text{ mg}}{453,592.37 \text{ mg/pound}} \approx 16.67 \text{ pounds/day}
   $$

Thus, the rate of pollutant released is approximately $$ \boxed{16.67} $$ pounds per day.
The rate of pollutant released is approximately 16.67 pounds per day.

A sewage treatment plant discharges a pollutant with a concentration of \( 10 \mathrm{mg} / \mathrm{L} \). The plant's total discharge is 0.2 MGD (millions of gallons per day). Determine the rate of pollutant released in pounds pe day. Give your answer accurate to at least two decimal places.

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