Question

Difficulty: MediumInterpreting Linear Relationships in Context

A municipal water reservoir contains 250 million gallons of water. During a dry spell, water is released from the reservoir at a constant rate of 3.5 million gallons per day. Additionally, water evaporates from the reservoir at a constant rate of 0.3 million gallons per day. If no water enters the reservoir, the total amount of water WW, in million gallons, remaining in the reservoir after dd days of the dry spell is modeled by the equation W=250rdW = 250 - r d, where rr is a constant. What is the value of rr?

Answer: 3.8 million gallons per day

Answer

The correct value of rr is 3.8.
The constant rr in the linear equation W=250rdW = 250 - r d represents the total rate, in million gallons per day, at which the water volume in the reservoir decreases. Since water is lost through both release (3.53.5 million gallons per day) and evaporation (0.30.3 million gallons per day), the total rate of decrease is the sum of these two rates, which is 3.5+0.3=3.83.5 + 0.3 = 3.8 million gallons per day. Therefore, the value of rr is 3.83.8.

Step-by-Step Solution

1
Identify the factors causing a decrease in the reservoir's water volume.
Water is lost through release at 3.53.5 million gallons per day and evaporation at 0.30.3 million gallons per day.
Both release and evaporation contribute to the total rate of water depletion.
2
Calculate the total daily rate of water loss.
3.5+0.3=3.83.5 + 0.3 = 3.8 million gallons per day.
Adding the individual rates of loss yields the overall rate of decrease.
3
Compare the total daily rate of water loss to the model equation W=250rdW = 250 - r d.
r=3.8r = 3.8.
In the linear model, 250250 represents the initial amount of water, and rr represents the constant rate at which water decreases per day. Thus, rr is the total daily rate of water loss.

Key Concept

Interpreting the slope (rate of change) in a linear equation in context.
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