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sineoko [7]
3 years ago
9

What are the fundamental activities that are common to all software processes?

Computers and Technology
1 answer:
erica [24]3 years ago
7 0

Answer:

There are some fundamental activities that are common to all software processes:

•Software specification. In this activity the functionality of the software and constraints on its operation must be defined.

•Software design and implementation. ...

•Software validation. ...

•Software evolution

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Karo-lina-s [1.5K]
The answer is the first response, by order of elimination, you can eliminate the rest.

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3 0
3 years ago
Which statement best justifies the use of a high-speed cache in a CPU?
crimeas [40]

Cache memory is a high-speed memory that stores the instructions and data that have been frequently accessed.  It decreases the time it takes to decode the instructions stored in the instruction pipeline.

A.  It decreases the time it takes to decode instructions stored in the instruction pipeline.

<u>Explanation:</u>

Whenever an instruction is invoked or some data is accessed, the CPU looks for it in the cache memory before accessing the main memory.

If the content is found in the cache memory, it accessed from there and then and hence the access time and decode time is reduced as there were no main memory lockups.

6 0
3 years ago
Decimal numbers are based on __________.
MaRussiya [10]
They are based on 10 digits.

I’m not sure.
8 0
3 years ago
Read 2 more answers
Why does the definition of fair use remain ambiguous?
bogdanovich [222]
Millions of dollars in legal fees have been spent attempting to define what qualifies as a fair use.

There are no hard-and-fast rules, only general guidelines and varied court decisions, because the judges and lawmakers who created the fair use exception did not want to limit its definition.

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8 0
3 years ago
Given a Fully Associative cache with 4 lines using LRU replacement. The word size is one byte, there is one word per block, and
uranmaximum [27]

Answer:

3 bits

Explanation:

Capacity of main memory=16 Bytes=24

The number of address bits= 4 bits.

The size of the word= 1 Byte=20

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The number of offset bits=20=0

Number of tag bits= total number of address bits - (word bits + offset bits + set bits)

= 4 - 0 -0- 1

= 3 bits

8 0
3 years ago
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