umber of people who eat at least one item =(Sum of individuals)−(Sum of two-set intersections)+(All-three intersection)
∣R∪B∪C∣=152−60+8
∣R∪B∪C∣=92+8
∣R∪B∪C∣=100
The number of people who eat at least one of the three items is 100.

The number of ways to choose the 3 cards one after the other is the product of the number of choices at each step:
First card: There are 42 choices.
Second card: Since the first card is not replaced, there are 42−1=41 choices left.
Third card: Since the first two cards are not replaced, there are 42−2=40 choices left.
Total number of ways =42×41×40
Total ways=68,880

The set of positive integers is S={1,2,3,…,200}. Total Outcomes (N): The total number of integers is 200. Let A be the event that the number is divisible by 2. Let B be the event that the number is divisible by 5. We want to find the probability of the union, P(A∪B). 1. Count of Favorable Outcomes We use the formula for the number of elements in the union of two sets: ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣ Count divisible by 2 (∣A∣): ∣A∣=⌊200/2​⌋=100 Count divisible by 5 (∣B∣): ∣B∣=⌊200/5​⌋=40 Count divisible by both 2 and 5 (∣A∩B∣): A number divisible by both 2 and 5 is divisible by their Least Common Multiple (LCM), which is 10. ∣A∩B∣=⌊200/10​⌋=20 Count divisible by 2 or 5 (∣A∪B∣): ∣A∪B∣=100+40−20=120 2. Probability The probability is the ratio of favorable outcomes to the total outcomes: P(A∪B)=∣A∪B∣​/ N=120​/200 Simplifying the fraction: P(A∪B)=200÷40/120÷40​=3​/5 The probability that the selected number is divisible by either 2 or 5 is 3/5.

The complete sum-of-products (SOP) form of the Boolean expression A(x,y,z)=x(y′z)′ is xyz′+xy′z′+xy′z (after correcting a simplification step) or xyz′+xy′z′

Conversion of (A0F)16​
0 in hex is 0000 in binary.
F in hex is 1111 in binary.
Concatenating the binary groups in order:
A∣0∣F
1010∣0000∣1111 ,br> The binary equivalent of (A0F)16​ is 101000001111.

In the context of standard Input/Output (I/O) operations, the terms Read and Write describe the direction of data flow relative to the computer's main memory (RAM):
Write Operation: Data is taken from the memory and moved to an external device (I/O device), such as saving a file to a hard disk, sending data to a network card, or displaying characters on a screen.
Read Operation: Data is taken from an external device (I/O device) and moved to the memory, such as loading a file from a hard disk, receiving data from a network, or reading input from a keyboard.
The CPU registers are temporary storage locations within the CPU itself and are typically involved in intermediate steps, but the primary data transfer path for an I/O write operation is between memory and the I/O device.

The term Mapping (or Cache Mapping) refers to the method or algorithm used to determine where a block of data from the larger main memory will be placed in the smaller, faster cache memory.
The three primary mapping techniques are:
Direct Mapping: Each block of main memory has only one possible location in the cache.
Associative Mapping: A block of main memory can be placed anywhere in the cache.
Set-Associative Mapping: A block of main memory can be placed in any line within a specific "set" in the cache.

The conversion from a decimal number to Gray code involves two steps:
Step 1: Decimal to Binary
First, convert the decimal number 8 into its binary equivalent. (8)10​=(1000)2​
Step 2: Binary to Gray Code
The Gray code (G) is derived from the binary code (B) using the following rule:
The Most Significant Bit (MSB) of the Gray code is the same as the MSB of the binary code (G3​=B3​).
Any other bit Gi​ is the result of the Exclusive OR (XOR) operation between the corresponding binary bit Bi​ and the next higher bit Bi+1​ (Gi​=Bi+1​⊕Bi​).

The loop terminates when x becomes equal to y. At this point, x=2 and y=2

The Sutherland–Cohen algorithm is a well-known method used in computer graphics to perform line clipping.
Clipping is the process of determining the portion of a geometric object (like a line segment, polygon, or text) that is visible within a rectangular area, called the clipping window. Any part of the object outside this window is discarded.

Any 2-attribute relation is always in 4NF, which implies BCNF. However, the "all prime attributes" rule is a more general and theoretically strong statement often tested against the BCNF definition. Given a choice, the BCNF definition based on superkeys is the primary focus.
But option-2(If all attributes of a relation are prime attributes then the relation is in BCNF.) is also true