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Roman55 [17]
3 years ago
14

Choose the response that best completes the following statement.

Computers and Technology
2 answers:
lisabon 2012 [21]3 years ago
7 0
I am guessing. My guess is code.
dem82 [27]3 years ago
6 0

Answer:

The answer to this is actually A) Binary

Explanation:

At the lowest level, a computer can only understand patterns of bits - which is to say, binary numbers. These lowest-level instructions are usually called “opcodes,” short for operation codes, or “machine language.”

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A Pool charges $4 each visit,or you can buy a membership. Write and solve an inequality to find how many times a person should u
MAXImum [283]
$4 x number of visits (V)> price of membership (M)
V > M/4
8 0
4 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
4 years ago
What is ‘verification’?
aev [14]

Answer: Verification in the terms of the computer field is defined as the process which is for testing of the consistency and the accuracy of the information, algorithm, data, etc. Verification is carried out by the approval or disapproval method after the checking process.

This techniques helps in verifying whether the function,algorithm ,data that has been accessed in the system is complete and precise or not so that it can properly function in the operating system.

5 0
4 years ago
Question 2 of 10
Roman55 [17]

Answer:

C

Explanation:

correct answer

3 0
3 years ago
Considering the following algorithm, which of the following requirements are satisfied?
Alisiya [41]

Answer:

b) Bounded Waiting

Explanation:

int currentThread = 1;

bool thread1Access = true;

bool thread2Access = true;

thread1 { thread2 {

While (true) {

                   While (true)

                                   {

                     while(thread2Access == true)

                                       {

                                      while(thread1Access == true)

                                       {

                                            If (currentThread == 2) {

                                              If (currentThread == 1)

                                                {        

                                                  thread1Access = false; thread2Access = false;

                                                  While (currentThread == 2);

                                                 While (currentThread == 1);

                                                  thread1Access = true; thread2Access = true;

} }

/* start of critical section */ /* start of critical section */

currentThread = 2 currentThread = 1

… ...

/* end of critical section */ /* end of critical section */

thread1Access = false; thread2Access = false;

… ...

} }

} }

} }

It can be seen that in all the instances, both threads are programmed to share same resource at the same time, and hence this is the bounded waiting. For Mutual exclusion, two threads cannot share one resource at one time. They must share simultaneously. Also there should be no deadlock. For Progress each thread should have exclusive access to all the resources. Thus its definitely the not the Progress. And hence its Bounded waiting.

4 0
4 years ago
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