What is a typical use for a MAN?

A.
to connect devices in five offices adjacent to each other
B.
to connect computers across a university campus
C.
to connect corporate offices across three continents
D.
to connect the computers in a private library

Answer:

Following are the responses to these question:

Explanation:

The cache size is 2n words whenever the address bit number is n then So, because cache size is 216 words, its number of address bits required for that  cache is 16 because the recollection is relational 2, there is 2 type for each set. Its cache has 32 blocks, so overall sets are as follows:

[tex]\text{Total Number of sets raluired}= \frac{\text{Number of blocks}}{Associativity}[/tex]

                                               [tex]=\frac{32}{2}\\\\ =16\\\\= 2^4 \ sets[/tex]

The set bits required also are 4. Therefore.

Every other block has 8 words, 23 words, so the field of the word requires 3 bits.

For both the tag field, the remaining portion bits are essential. The bytes in the tag field are calculated as follows:

Bits number in the field tag =Address Bits Total number-Set bits number number-Number of bits of words

=16-4-3

= 9 bit

The number of bits inside the individual fields is therefore as follows:

Tag field: 8 bits Tag field

Fieldset: 4 bits

Field Word:3 bits

Explanation:

Start-up companies can store, manage, process data and use programs through a web-based interface – something that greatly reduces costs. Data stored in the cloud can be accessed from anywhere through various devices with web connectivity, which is an advantage for small companies that don't have a huge IT budget.

Cloud provides storage access and management for files and information using a the cloud computing provider. This could help reduce the cost expended on physical content storage devices.

Hence, the cloud could in the cost expended on physical storage.

Learn more : https://brainly.com/question/22841107

Answer:

10.b

11.c

12.c

13.a

14.d

15.b

16.c

Explanation:

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Answer:

In Java:

import java.util.*;  

public class Main{

   public static void main(String[] args) {

       Map<String, String> students = new HashMap<>();

       students.put("STUD1", "Student 1 Name");

       students.put("STUD2", "Student 2 Name");

       students.put("STUD3", "Student 3 Name");

       students.put("STUD4", "Student 4 Name");

       for(Map.Entry m:students.entrySet()){  

           System.out.println(m.getKey()+" - "+m.getValue());    }  

}}

Explanation:

This creates the map named students

       Map<String, String> students = new HashMap<>();

The next four lines populates the map with the ID and name of the 4 students

       students.put("STUD1", "Student 1 Name");

       students.put("STUD2", "Student 2 Name");

       students.put("STUD3", "Student 3 Name");

       students.put("STUD4", "Student 4 Name");

This iterates through the map

       for(Map.Entry m:students.entrySet()){  

This prints the details of each student

           System.out.println(m.getKey()+" - "+m.getValue());    }  

Please see the following file for your required answer, thank you.

Answer:

nah

Explanation:

cuz ion wanna

Answer:

A. 9

B. 16

C. Number of addresses = 12 number of data lines = 16

Explanation:

The capacity of this chip is 512 x 8

That is 512 words and these words have 8 bits

A. The number of the address lines would be

log2(512) = 2⁹

So the answer is 9

B. The total number of chips required

= 512 x 8 = 4096

(4096 x 2)/ 512 = 16

C. Number of address = log4092 = 2¹²

= 12

The number of data lines = 8x2 = 16

Thank you!

Answer:

http://www.hackshop.org/levels/basic-arduino/hack-the-chromebook

Explanation:

yes

yes

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yes

yes

yes