A nibble is equal to _______ A. 4 bits B. 8 bits C. 16 bits D. 32 bits

A nibble is equal to _______ A. 4 bits B. 8 bits C. 16 bits D. 32 bits
by

1 Answer

Answer :

A. 4 bits
by
selected

Related questions

A nibble is equal to_____ bits. (1) 32 (2) 4 (3) 8 (4) 16

Description : A nibble is equal to_____ bits. (1) 32 (2) 4 (3) 8 (4) 16

Last Answer : 4

Because we wish to allow each ASCII code to occupy one location in memory, most memories are _____ addressable. a. BYTE b. NIBBLE c. WORD (16 bits) d. DOUBLEWORD (32 bits)

Description : Because we wish to allow each ASCII code to occupy one location in memory, most memories are _____ addressable. a. BYTE b. NIBBLE c. WORD (16 bits) d. DOUBLEWORD (32 bits)

Last Answer : a. BYTE

A nibble is equal to – (1) 32 (2) 4 (3) 8 (4) 16

Description : A nibble is equal to – (1) 32 (2) 4 (3) 8 (4) 16

Last Answer : (2) 4 Explanation: In computing, a nibble is a fourbit aggregation, or half an octet. A nibble contains 4 bits.

1 nibble equals to A) 1 bits B) 2 bits C) 4 bits D) 8 bits

Description : 1 nibble equals to A) 1 bits B) 2 bits C) 4 bits D) 8 bits

Last Answer : Answer : C

One nibble is equivalent to how many bits? a) 2 b) 4 c) 8 d) 1

Description : One nibble is equivalent to how many bits? a) 2 b) 4 c) 8 d) 1

Last Answer : Answer: b Explanation: A nibble is defined as a unit of data which comprises of 4 binary digits or half of 8-bit byte. Therefore, 1 nibble = 4 bits. A bit is the smallest unit of data in a computer system. A byte = 8 bits, therefore, half of a byte=4 bits=a nibble

A binary number with 8 bits is called asa___ a. Bytes b. Bits c. Nibble d. __ Allof these

Description : A binary number with 8 bits is called asa___ a. Bytes b. Bits c. Nibble d. __ Allof these

Last Answer : a. Bytes

The Width of a processor’s data path is measured in bits. Which of the following are common data paths? A) 8 bits B) 12 bits C) 16 bits D) 32 bits

Description : The Width of a processor’s data path is measured in bits. Which of the following are common data paths? A) 8 bits B) 12 bits C) 16 bits D) 32 bits

Last Answer : Answer : A

The word length of Micro computers lies in the range between ________ (A) 8 and 16 bits (B) 8 and 21 bits (C) 8 and 24 bits (D) 8 and 32 bits

Description : The word length of Micro computers lies in the range between ________ (A) 8 and 16 bits (B) 8 and 21 bits (C) 8 and 24 bits (D) 8 and 32 bits

Last Answer : (D) 8 and 32 bits

A group of 4 binary bits is called. a. Nibble b. Byte c. Decimal d. Digit

Description : A group of 4 binary bits is called. a. Nibble b. Byte c. Decimal d. Digit

Last Answer : a. Nibble

The message in pub/sub is an opaque blob of ________ a) Bits b) Bytes c) Word d) Nibble

Description : The message in pub/sub is an opaque blob of ________ a) Bits b) Bytes c) Word d) Nibble

Last Answer : Bytes

How many bits are required to represent one nibble ?

Description : How many bits are required to represent one nibble ?

Last Answer : 4-bits are required to represent one nibble.

An IPv4 address consistsof _______ bits. A) 4 B) 8 C) 32 D) 64

Description : An IPv4 address consistsof _______ bits. A) 4 B) 8 C) 32 D) 64

Last Answer : 32

The Width of a processor’s data path is measured in bits. Which of the following are common data paths? a. 8 bits b. 12 bits c. 16 bits d. 32 bits

Description : The Width of a processor’s data path is measured in bits. Which of the following are common data paths? a. 8 bits b. 12 bits c. 16 bits d. 32 bits

Last Answer : a. 8 bits

The Width of a processor’s data path is measured in bits. Which of the following are common data paths? a. 8 bits b. 12 bits c. 16 bits d. 32 bits

Description : The Width of a processor’s data path is measured in bits. Which of the following are common data paths? a. 8 bits b. 12 bits c. 16 bits d. 32 bits

Last Answer : a. 8 bits

What is the size of the 'total length' field in IPv 4 datagram ? (A) 4 bits (B) 8 bits (C) 16 bits (D) 32 bits

Description : What is the size of the 'total length' field in IPv 4 datagram ? (A) 4 bits (B) 8 bits (C) 16 bits (D) 32 bits

Last Answer : (C) 16 bits 

One Byte is equal to (1) 8 bits (2) 12 bits (3) 16 bits (4) 20 bits

Description : One Byte is equal to (1) 8 bits (2) 12 bits (3) 16 bits (4) 20 bits

Last Answer : 8 bits

The VPIof an NNI is _______ bits in length. A) 8 B) 12 C) 16 D) 24

Description : The VPIof an NNI is _______ bits in length. A) 8 B) 12 C) 16 D) 24

Last Answer : 12

The VPIof a UNI is _______ bits in length. A) 8 B) 12 C) 16 D) 24

Description : The VPIof a UNI is _______ bits in length. A) 8 B) 12 C) 16 D) 24

Last Answer : 8

A memory unit stores 2^16 bytes in 32-bit words. How many address bits are necessary in order to retrieve a single word from this memory unit?

Description : A memory unit stores 2^16 bytes in 32-bit words. How many address bits are necessary in order to retrieve a single word from this memory unit?

Last Answer : Each word is 4 bytes, which occupies 4 consecutive addresses at 16 bits each. The first 14 bits of these addresses will be the same, differing only in the two least significant bits. You only need to specify the first 14 bits, because there are only 2^14 words occupying those 2^16 bytes.

Which instruction are 32 bits long , with extra 16 bits. a. Memory reference instruction b. Memory reference format c. Both d. None of these

Description : Which instruction are 32 bits long , with extra 16 bits. a. Memory reference instruction b. Memory reference format c. Both d. None of these

Last Answer : a. Memory reference instruction

How many bits internet address isassigned to eachhost on a TCP/IP internet which is used in all communications withthehost? A. 16 - bits B. 32 - bits C. 48 - bits D. 64 - bits E. None of the above

Description : How many bits internet address isassigned to eachhost on a TCP/IP internet which is used in all communications withthehost? A. 16 - bits B. 32 - bits C. 48 - bits D. 64 - bits E. None of the above

Last Answer : 32 - bits

The Internet addresses are ______ bits in length in IPV4 addressing scheme A. 16 B. 64 C. 32 D. 48

Description : The Internet addresses are ______ bits in length in IPV4 addressing scheme A. 16 B. 64 C. 32 D. 48

Last Answer : C. 32

The Internet addresses are ……….. bits in length in IPV4 addressing scheme a. 16 b. 64 c. 48 d. 32

Description : The Internet addresses are ……….. bits in length in IPV4 addressing scheme a. 16 b. 64 c. 48 d. 32

Last Answer : d. 32

The processor 80386/80486 and the Pentium processor uses _____ bits address bus: a. 16 b. 32 c. 36 d. 64

Description : The processor 80386/80486 and the Pentium processor uses _____ bits address bus: a. 16 b. 32 c. 36 d. 64

Last Answer : b. 32

How many address bits are required to represent a 32 K memory (A) 10 bits. (B) 12 bits. (C) 14 bits. (D) 16 bits.

Description : How many address bits are required to represent a 32 K memory (A) 10 bits. (B) 12 bits. (C) 14 bits. (D) 16 bits.

Last Answer : (D) 16 bits.

The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits

Description : The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits

Last Answer : (C) 16 bits Explanation: VirtualAddress = 32 bits PageSize = 4KB = 12 bits therefore : VPNTag = 20 bits, OffsetTag = 12 bits TLBEntryLength = VPNTag = 20 bits TotalTLBEntries = 64, 4-way implies ... therefore : TLBIndex = 4 bits TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits

The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64

Description : The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64

Last Answer : (D) 64

A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32

Description : A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32

Last Answer : (3) 15 and 14

A byte is a collection of _______ A. 4 bits B. 12 bits C. 6 bits D. 8 bits

Description : A byte is a collection of _______ A. 4 bits B. 12 bits C. 6 bits D. 8 bits

Last Answer : D. 8 bits

Number of bits in ALU is _________ a) 4 b) 8 c) 16 d) 2

Description : Number of bits in ALU is _________ a) 4 b) 8 c) 16 d) 2

Last Answer : Answer: c Explanation: Arithmetic and Logic Unit consists of 16bits. They perform certain Arithmetic and bitwise operations (add, subtract, AND, OR, XOR, Increment, decrement, shift)

Word Length of a Personal Computer is ______: a) 4 Bits b) 8 Bits c) 16 Bits d) None of The Above

Description : Word Length of a Personal Computer is ______: a) 4 Bits b) 8 Bits c) 16 Bits d) None of The Above

Last Answer : b) 8 Bits

Word length if a personal computer ia _________ 1) 4 bits 2) 8 bits 3) 16 bits 4) 64 bits

Description : Word length if a personal computer ia _________ 1) 4 bits 2) 8 bits 3) 16 bits 4) 64 bits

Last Answer : 1) 4 bits

Word length if a personal computer ia _________ 1 4 bits 2 8 bits 3 16 bits 4 64 bits

Description : Word length if a personal computer ia _________ 1 4 bits 2 8 bits 3 16 bits 4 64 bits

Last Answer : 1 4 bits

The intel 8086 microprocessor is a _______ processor A. 8 bit B. 16 bit C. 32 bit D. 4 bit

Description : The intel 8086 microprocessor is a _______ processor A. 8 bit B. 16 bit C. 32 bit D. 4 bit

Last Answer : The intel 8086 microprocessor is a 16 bit processor 

. In addresses for _______ networks, the first 16 bits specify a particular network, and the last 16 bits specify a particular host. A. class A B. class C C. class B D. class D

Description : . In addresses for _______ networks, the first 16 bits specify a particular network, and the last 16 bits specify a particular host. A. class A B. class C C. class B D. class D

Last Answer : C. class B

The smallest unit of data in computer is ________________ a) Byte b) Nibble c) Bit d) KB

Description : The smallest unit of data in computer is ________________ a) Byte b) Nibble c) Bit d) KB

Last Answer : Answer: c Explanation: A bit is defined as the smallest unit of data in a computer system. It is used as a short form of Binary Digit. A bit can have only two values 0 or 1. A nibble comprises of 4 bits, a byte is a collection of 8 bits whereas KB (Kilobyte) is equal to 1024 bytes

If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______. a. 64 Kbps b. 32 Mbps c. 32 Kbps d. 64 MbpS

Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______. a. 64 Kbps b. 32 Mbps c. 32 Kbps d. 64 MbpS

Last Answer : c. 32 Kbps

Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is _____. a. 2 sec b. 3 sec c. 4 sec d. 1 sec

Description : Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is _____. a. 2 sec b. 3 sec c. 4 sec d. 1 sec

Last Answer : a. 2 sec

Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it is assumed that each sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps

Description : Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it ... sample requires 8 bits? (A) 32 kbps (B) 64 kbps (C) 128 kbps (D) 256 kbps

Last Answer : (C) 128 kbps

If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ............... (A) 64 Kbps (B) 32 Mbps (C) 32 Kbps (D) 64 Mbps

Description : If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ............... (A) 64 Kbps (B) 32 Mbps (C) 32 Kbps (D) 64 Mbps

Last Answer : (C) 32 Kbps

One Byte is equal to: a) 10 Bits b) 7 Bits c) 8 Bits d) None of These

Description : One Byte is equal to: a) 10 Bits b) 7 Bits c) 8 Bits d) None of These

Last Answer : c) 8 Bits

How many bits are equal to one byte? (1) 8 (2) 10 (3) 12 (4) 14

Description : How many bits are equal to one byte? (1) 8 (2) 10 (3) 12 (4) 14

Last Answer : 8

_______ is a method of storing data bits using magnetic charges instead of the electrical charges used by DRAM. A. VRAM B. WRAM C. MRAM D. None of the above

Description : _______ is a method of storing data bits using magnetic charges instead of the electrical charges used by DRAM. A. VRAM B. WRAM C. MRAM D. None of the above

Last Answer : C. MRAM Magneto resistive RAM

1 Kilobyte (KB) is _______: a) 1000 bits b) 1 Kilo KB c) Both (a) and (b) d) 1024 Bytes e) None of The Above

Description : 1 Kilobyte (KB) is _______: a) 1000 bits b) 1 Kilo KB c) Both (a) and (b) d) 1024 Bytes e) None of The Above

Last Answer : d) 1024 Bytes

_______ can detect burst error of length less than or equal to degree of the polynomial and detects burst errors that affect odd number of bits. a. Hamming Code b. CRC c. VRC d. None of the above

Description : _______ can detect burst error of length less than or equal to degree of the polynomial and detects burst errors that affect odd number of bits. a. Hamming Code b. CRC c. VRC d. None of the above

Last Answer : b. CRC

What should be minimum requirement of random access memory (RAM) for internet access: a) 8 MB b) 16 MB c) 32 MB d) 64 MB

Description : What should be minimum requirement of random access memory (RAM) for internet access: a) 8 MB b) 16 MB c) 32 MB d) 64 MB

Last Answer : c) 32 MB

If in a computer, 16 bits are used to specify address in a RAM, the number of addresses will be A) 216 B) 65,536 C) 64K D) Any of the above

Description : If in a computer, 16 bits are used to specify address in a RAM, the number of addresses will be A) 216 B) 65,536 C) 64K D) Any of the above

Last Answer : Answer : B

When a voltage of 442.7 VDC is applied to the illustrated circuit with a resistance of 1.25 ohms the current will be _______. EL-0018 A. 28.25 amps B. 35.32 amps C. 354.16 amps D. 443.62 amps

Description : When a voltage of 442.7 VDC is applied to the illustrated circuit with a resistance of 1.25 ohms the current will be _______. EL-0018 A. 28.25 amps B. 35.32 amps C. 354.16 amps D. 443.62 amps

Last Answer : Answer: C

What is the typical number of bits in a static memory location? A. 1 B. 2 C. 16 D. 8

Description : What is the typical number of bits in a static memory location? A. 1 B. 2 C. 16 D. 8

Last Answer : D. 8