Describe, and estimate the costs of, a procedure to insert a new item into an existing binary max-heap.

Resolution=Voltage Range​/2^n

Where:

• Voltage Range = 3.3V (the output voltage range from 0V to 3.3V)
• n = number of bits in the digital input (12 bits in this case)

Now, let's calculate the resolution:

Resolution = 3.3/2^12 
                  ​=0.80586mV

So, the resolution of the analogue output is approximately 0.80586 mV per bit.

To find the sets \( X \) and \( Y \) based on the given information, we need to interpret the provided conditions step by step.

### Given:

1. \( X \cup Y = \{1, 2, 3, 5, 6, 8, 9, 10\} \)
2. \( X \cap Y = \{1, 5\} \)
3. \( Y - X = \{2, 6, 9, 10\} \)

### Step 1: Using \( Y - X \)

From condition 3, we know that:
\[
Y - X = \{2, 6, 9, 10\}
\]
This means that the elements \( 2, 6, 9, \) and \( 10 \) are in \( Y \) but not in \( X \). Therefore, we can express \( Y \) as:
\[
Y = (Y - X) \cup (X \cap Y)
\]
Since \( X \cap Y = \{1, 5\} \), we can write:
\[
Y = \{2, 6, 9, 10\} \cup \{1, 5\} = \{1, 2, 5, 6, 9, 10\}
\]

### Step 2: Using \( X \cup Y \)

Now, we know:
\[
X \cup Y = \{1, 2, 3, 5, 6, 8, 9, 10\}
\]
And we have \( Y = \{1, 2, 5, 6, 9, 10\} \). Now we can find \( X \):
\[
X = (X \cup Y) - Y
\]
This gives us:
\[
X = \{1, 2, 3, 5, 6, 8, 9, 10\} - \{1, 2, 5, 6, 9, 10\}
\]
Calculating this, we get:
\[
X = \{3, 8\}
\]

### Step 3: Verification

Now we need to verify if the derived sets satisfy all the conditions.

- **Checking \( X \cup Y \)**:
\[
X \cup Y = \{3, 8\} \cup \{1, 2, 5, 6, 9, 10\} = \{1, 2, 3, 5, 6, 8, 9, 10\} \quad \text{(True)}
\]

- **Checking \( X \cap Y \)**:
\[
X \cap Y = \{3, 8\} \cap \{1, 2, 5, 6, 9, 10\} = \{1, 5\} \quad \text{(True)}
\]

- **Checking \( Y - X \)**:
\[
Y - X = \{1, 2, 5, 6, 9, 10\} - \{3, 8\} = \{2, 6, 9, 10\} \quad \text{(True)}
\]

### Conclusion

Thus, the sets \( X \) and \( Y \) are:

\[
X = \{3, 8\}
\]
\[
Y = \{1, 2, 5, 6, 9, 10\}
\]