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3 years ago

Why is it important to not get distracted while driving?

2 answers:

3 0

It is important because it someone is distracted while driving their car, they could get into a car accident, possibly hurting themselves and others.

QveST [7]3 years ago

3 0

Because you can crash, and not only harm or kill yourself , but others too.

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RAID level ________ refers to disk arrays with striping at the level of blocks, but without any redundancy. A. 0 B. 1 C. 2 D. 3

Mamont248 [21]

Answer:

A. 0

Explanation:

RAID level zero refers to disk arrays with striping at the level of blocks, but without any redundancy.

RAID ZERO (disk striping) : This is the process of dividing a body of data into blocks and also spreading the data blocks across multiple storage devices, such as hard disks or solid-state drives (SSDs), in a redundant array of independent disks (RAID) group.

3 0

3 years ago

T in your computer will you use to plug in your camera?

Gnom [1K]

A tool in your computer that you will use to plug in your camera is the Universal Serial Bus (USB). If you camera has this chord and it is an Universal Serial Bus type. Then what you need is your software installer of the camera and only the USB.

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3 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

2 years ago

Why computer is known as versatile and diligent device ?explain

pashok25 [27]

They are versatile because they can be used for all sorts of tasks. They can also do many of the same tasks in different ways. They are diligent because they will do a task thoroughly until it is finished.

They work at a constant speed to do the task. Unlike a human, they will not slow down or get bored or start making mistakes that they were not doing earlier. So once they are programmed correctly to do a task, they will do it diligently.

6 0

3 years ago

The ease of travel in the twentieth century played a role in the advancement of medicine. Based on this lesson, which statement

Shtirlitz [24]

During the twentieth century, transportation had been very significant and played a great influence in the advancement of medicine. This is because this made health care become more accessible to patients no matter where they are located. The answer for this would be option C.

8 0

3 years ago

Read 2 more answers