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To solve the problem, we need to calculate how much Ellis can expect to pay to run the air conditioning (AC) unit for his entire house in July.

### Given Information:
- The AC unit's average active time in July is provided in the table:
  - 1 hour active per day: 4 days
  - 3 hours active per day: 7 days
  - 5 hours active per day: 10 days
  - 7 hours active per day: 6 days
  - 8 hours active per day: 4 days
- It takes 1.4 kWh to cool an area of 200 square feet for one hour.
- Ellis' house measures 700 square feet.
- Ellis' electricity provider charges 14.37p per kWh.

### Step 1: Determine the energy consumption per hour for the entire house.
- Since 1.4 kWh cools 200 square feet in 1 hour, the energy needed to cool 700 square feet (which is \( \frac{700}{200} = 3.5 \) times more area) will be:
  \[
  \text{Energy per hour for the entire house} = 1.4 \times 3.5 = 4.9 \text{ kWh}
  \]

### Step 2: Calculate the total energy consumption for each day type.
We will calculate the energy consumption for each day type given in the table:
- **1 hour active per day for 4 days:**
  \[
  \text{Total energy} = 4.9 \text{ kWh/day} \times 1 \text{ hour/day} \times 4 \text{ days} = 19.6 \text{ kWh}
  \]
- **3 hours active per day for 7 days:**
  \[
  \text{Total energy} = 4.9 \text{ kWh/day} \times 3 \text{ hours/day} \times 7 \text{ days} = 102.9 \text{ kWh}
  \]
- **5 hours active per day for 10 days:**
  \[
  \text{Total energy} = 4.9 \text{ kWh/day} \times 5 \text{ hours/day} \times 10 \text{ days} = 245 \text{ kWh}
  \]
- **7 hours active per day for 6 days:**
  \[
  \text{Total energy} = 4.9 \text{ kWh/day} \times 7 \text{ hours/day} \times 6 \text{ days} = 205.8 \text{ kWh}
  \]
- **8 hours active per day for 4 days:**
  \[
  \text{Total energy} = 4.9 \text{ kWh/day} \times 8 \text{ hours/day} \times 4 \text{ days} = 156.8 \text{ kWh}
  \]

### Step 3: Sum up the total energy consumption for the month of July.
\[
\text{Total energy consumption} = 19.6 + 102.9 + 245 + 205.8 + 156.8 = 730.1 \text{ kWh}
\]

### Step 4: Calculate the total cost.
- The cost of electricity per kWh is 14.37p, so:
  \[
  \text{Total cost} = 730.1 \text{ kWh} \times 14.37 \text{p/kWh} = 10494.337 \text{ p} \approx £104.94
  \]

### Final Answer:
Ellis can expect to pay approximately **£104.94** to run the AC unit in July.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the cost change if the electricity rate increased to 16p per kWh?
2. What would be the total energy consumption if Ellis’ house was 1000 square feet instead of 700?
3. How much would it cost to run the AC if it was active for 10 hours every day in July?
4. If Ellis could improve insulation to reduce the energy needed by 10%, how would that affect the cost?
5. How would the cost compare if the AC was run for the same number of hours but only cooled 500 square feet?

**Tip:** Always check the units of energy consumption (kWh) and cost (pence per kWh) to ensure accuracy in calculations.

Learn how to calculate the approximate cost Ellis can expect to pay for running his air conditioning unit throughout July. Detailed calculations based on energy consumption and electricity cost per kWh are provided.

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