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telo118 [61]
4 years ago
11

personal digital assistants are handheld devices that serve as personql information managers. what is another term used to refer

to these devices? ​
Computers and Technology
2 answers:
prohojiy [21]4 years ago
6 0

Handheld devices or handheld PCs

A handheld device or a PDA is a device that combines computing, telephone, and internet features together. These devices were used to provide computing and information storage and retrieval capabilities all way up to the early 2010s when they were largely discontinued after the widespread emergence and adoption of powerful smartphones.

ololo11 [35]4 years ago
5 0

Answer:

Handheld devices are portable and compact type of devices which are easy to carry and its size is user-friendly like a tablet, smartphone, etc.

Explanation:

Handheld devices are the combination of the computer, telephone, and internet.

These handheld devices are used to store personal information which refers to personal digital assistants that are used to connect the internet (which is global access) and intranet ( which is restricted within an organization or any specific location or a range of distance ).

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ladessa [460]

Answer:

Abstraction

Explanation: Abstraction is a software engineer terminology that has been found to be very effective and efficient especially as it relates to object-oriented computer programming. It has helped software engineers and other computer experts to effectively manage account objects without having a knowledge of the details with regards to the Implementation process of that account object.

6 0
3 years ago
There are 55 registers, and total 60 instructions available in a general-purpose computer. The computer allows only 2-address in
Elenna [48]

Answer:

total bits = 6 + 6 + 19 = 31 bits

Explanation:

given data

total registers = 55

memory  size = 64 KB

total instructions = 60

solution

here we have given 55 register so we get greater or equal power of 2 that is here 64

2^6 = 64

so here for register operand 6 bit is required

and

when instruction 60 we get here greater or equal power of 2 that is here 64  

2^6 = 64

so here also  for represent instruction 6 bit is required

and

for size 64 kb

2^6 * 2^{10} * 2^3 = 2^{19}

so 19 bits is required for memory location

and

as instruction in 2 parts are opcode and operand

and here given as 2 address instruction

they are memory operand and the register operand

so here

total bits will be = opcode + register operand + memory operand

total bits = 6 + 6 + 19 = 31 bits

total bits =  31 bits

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Explanation:

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