All categories

3 years ago

The user cannot use a computer system without.................... software

Computers and Technology

2 answers:

Nat2105 [25]3 years ago

6 0

Answer:

the user cannot use a computer system without.........hardware.....and...... software

zheka24 [161]3 years ago

4 0

Answer:

the user cannot use a computer system without hardware and software

You might be interested in

Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

6 0

3 years ago

What is the duty of business to contribute to the well-being of society

NikAS [45]

Answer:

To be able to explain what’s wrong and what needs to be done

Explanation:

So if your looking at a worksheet and the costumer needs help they are gonna ask you about it so your gonna need to be able to explain what’s going on and what needs to be/get done

5 0

3 years ago

A _____ cloud allows an organization to take advantage of the scalability and cost-effectiveness that a public cloud computing e

balu736 [363]

Answer:

Hybrid

Explanation:

Hybrid cloud is a solution that combines a private cloud with one or more public cloud services, with proprietary software enabling communication between each distinct service.

5 0

3 years ago

You place a new disc in an optical drive, and then double-click the drive in the operating system's file browser interface. An e

lidiya [134]

Answer:

Wait. Then you try it again

Explanation:

You wait and try again later and if this still persists, you may have to check the optical drive from the drive manager peradventure there is an issue there probably a software issue, also you may have to clean the optical drive using the cleaner disc, likewise you may need to test an old disc on the optical drive to be sure the fault is not from the disc you inserted.

7 0

3 years ago

Two electronics technicians are looking at the piece of testing equipment shown in the figure above. Technician A says that this

Ilya [14]

B is your correct answer.

8 0

3 years ago

Read 2 more answers

The user cannot use a computer system without.................... software​

The user cannot use a computer system without.................... software