There are three more important concept, inheritance, which makes the OOP code more modular, easier to reuse and build a relationship between classes. Encapsulation can hide some of the private details of a class from other objects, while polymorphism can allow us to use a common operation in different ways.

Singly linked lists and doubly linked lists are two types of Linked lists. A singly-linked list consists of data and a link to the next element, while a doubly-linked list node contains a link to the previous node

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 ∪ Y = { 1 , 2 , 3 , 5 , 6 , 8 , 9 , 10 } 

2.  X ∩ 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 ) ∪ ( X ∩ Y ) 

Since  X ∩ Y = { 1 , 5 } , we can write:

Y = { 2 , 6 , 9 , 10 } ∪ { 1 , 5 } = { 1 , 2 , 5 , 6 , 9 , 10 }

### Step 2: Using  X ∪ Y

Now, we know:

X ∪ 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 ∪ 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 ∪ Y **:

X ∪ Y = { 3 , 8 } ∪ { 1 , 2 , 5 , 6 , 9 , 10 } = { 1 , 2 , 3 , 5 , 6 , 8 , 9 , 10 } (True)

- **Checking  X ∩ Y **:

X ∩ Y = { 3 , 8 } ∩ { 1 , 2 , 5 , 6 , 9 , 10 } = { 1 , 5 } (True)

- **Checking  Y − X **:

Y − X = { 1 , 2 , 5 , 6 , 9 , 10 } − { 3 , 8 } = { 2 , 6 , 9 , 10 } (True)

### Conclusion 

Thus, the sets  X  and  Y  are: 

X = { 3 , 8 } 

Y = { 1 , 2 , 5 , 6 , 9 , 10 }

Find the total cache size: Multiply the number of blocks by the bits per block: 2^10 blocks x 147 bits = 147 KibiBits. Therefore, for a direct mapped cache with 16KB of data, 4-word blocks, and a 32-bit address, the total bits required would be 147 KibiBits.

Each dice has 6 faces only one face at time is counted so there are 6 x 6 =36 possible combinations. Like this

[1][1], [1][2], [1][3], [1][4], [1][5], [1][6],
[2][1], [2][2], [2][3], [2][4], [2][5], [2][6],
[3][1], [3][2], [3][3], [3][4], [3][5], [3][6],
[4][1], [4][2], [4][3], [4][4], [4][5], [4][6],
[5][1], [5][2], [5][3], [5][4], [5][5], [5][6],
[6][1], [6][2], [6][3], [6][4], [6][5], [6][6]

Reading is an excellent habit. Here, the word 'reading' is a Gerund

The main difference between flow control and congestion control is that, In flow control, rate of traffic received from a sender can be controlled by a receiver. On the other hand, In congestion control, rate of traffic from sender to the network is controlled

The average-case time complexity of Quicksort is O(n*log(n)), which is quicker than Merge Sort, Bubble Sort, and other sorting algorithms. However, the worst-case time complexity is O(n^2) when the pivot choice consistently results in unbalanced partitions. To mitigate this, randomized pivot selection is commonly used.

Input : arr[] = {40, 50, 90}
Output : Yes
We can construct a number which is
divisible by 3, for example 945000. 
So the answer is Yes. 

Input : arr[] = {1, 4}
Output : No
The only possible numbers are 14 and 41,
but both of them are not divisible by 3, 
so the answer is No.

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.

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.

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