১ কিলোবাইট ১০২৪ বাইটের সমান

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.

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.

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.

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

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

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]

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.

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 }

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

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.