2. A different aquarium should contain 15,000 liters of water when filled correctly. It w overflow if it gets to 17,600 liters. One day there is an accident and the tank cracks in 4 places. Water flows out of each crack at a rate of $$ \frac{1}{2} $$ liter per hour. An emergency pump can re-fill the tank at a rate o liters per minute. How many minutes must the pump run to replace the water lost hour?

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To determine how long the emergency pump needs to run to replace the water lost in one hour, let's break down the problem step by step.

### Given:
1. **Water Loss:**
   - The tank has **4 cracks**.
   - **Each crack leaks at a rate of** $$ \frac{1}{2} $$ **liter per hour**.
   - **Total water lost per hour**:
     $$
     4 \text{ cracks} \times \frac{1}{2} \text{ liter/hour per crack} = 2 \text{ liters/hour}
     $$

2. **Pump Refill Rate:**
   - The emergency pump can refill the tank at a rate of **$$ o $$ liters per minute**.

### Objective:
Find the **time ($$ t $$) in minutes** that the pump must run to replace the **2 liters** lost in one hour.

### Calculation:
To replace **2 liters** of water:

$$
\text{Time} \times \text{Pump Rate} = \text{Water Lost}
$$

$$
t \times o = 2 \text{ liters}
$$

Solving for $$ t $$:

$$
t = \frac{2}{o} \text{ minutes}
$$

### Conclusion:
The emergency pump must run for $$ \frac{2}{o} $$ minutes to replace the water lost in one hour.

**Final Answer:**
The pump must run for 2 divided by o minutes. Thus, the required time is 2 ⁄ o minutes.
The pump must run for $$ \frac{2}{o} $$ minutes to replace the water lost in one hour.

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