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alexandr1967 [171]
2 years ago
10

Please post detailed answers to the following questions. Please use complete sentences.

Computers and Technology
1 answer:
Aloiza [94]2 years ago
4 0

Please post detailed answers to the following questions. Please use complete sentences t two artists from the following list: Phil Borges,Lee Jeffries.

<h3>What are photography opportunities?</h3>

The taking of constant or nevertheless snap shots furnished a brand new medium with which to seize reality, and modified the manner humans in general, and artists in particular, noticed the arena and created new inventive opportunities. Learning from the technological know-how of photography, artists evolved a selection of the latest portrayal techniques.

Photography modified our imaginative and prescient of the arena with the aid of using presenting greater get entry to to greater snap shots drawn from greater locations and instances withinside the global than ever before. Photography enabled snap shots to be copied and mass-distributed. The media-sphere became burgeoning.

Read more about the internet :

brainly.com/question/2780939

#SPJ1

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Which of the following is a public location that sells Internet access? Internet café Extranet Intranet LogMeIn,
Goshia [24]

Answer:A) Internet cafe

Explanation: Internet cafe is the place that provides the internet service to the public and helps them to access the internet .These cafes can be used by the public who are travelers, people who don't have access at home etc.This internet cafe works on the basis of the hourly rate that it takes from the public.

Rest of the options are incorrect  as intranet and extranet are types of network and LogMeIn is a desktop service .thus, the correct option is option(A).

8 0
3 years ago
Write a statement to create a Google Guava Multimap instance which contains student ids as keys and the associated values are st
Sergeu [11.5K]

Answer: provided in the explanation section

Explanation:

Note: take note for indentation so as to prevent error.

So we import com.google.common.collect.Multimap;

import java.util.Collection;

import java.util.Map;

class Main {

   public static void main(String[] args) {

       // key as studentId and List of Profile as value.

       Multimap<String,Profile> multimap = ArrayListMultimap.create();

       multimap.put("id1", new ExamProfile("examStudentId1"));

       multimap.put("id1", new ELearningProfile("userId1"));

       multimap.put("id2", new ExamProfile("examStudentId2"));

       multimap.put("id2", new ELearningProfile("userId2"));

       // print the data now.

       for (Map.Entry<String, Collection<Profile>> entry : multimap.entrySet()) {

           String key = entry.getKey();

           Collection<String> value =  multimap.get(key);

           System.out.println(key + ":" + value);

       }

   }

}

// assuming String as the studentId.

// assuming Profile as the interface. And Profile can multiple implementations that would be

// specific to student.

interface Profile {

}

class ExamProfile implements Profile {

   private String examStudentId;

   public ExamProfile(String examStudentId) {

       this.examStudentId = examStudentId;

   }

}

class ELearningProfile implements Profile {

   private String userId;

   public ELearningProfile(String userId) {

       this.userId = userId;

   }

}

This code is able to iterate through all entries in the Google Guava multimap and display all the students name in the associated students profile.

8 0
3 years ago
Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a
kirill115 [55]

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = \frac{CPI*Instruction Count}{ClockRate}

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is =\frac{2.5*100,000}{750*10^6}

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =\frac{2.5*Instruction}{750*10^6}

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

3 0
3 years ago
Why it is important for everyday people to have access to credible online information?
Citrus2011 [14]

Answer:

so you know what is true

Explanation:

5 0
3 years ago
Read 2 more answers
How does a typewriter involve math and science?
patriot [66]
You have a specific number of words typed a minute and you need to know the science involved in making it
8 0
3 years ago
Read 2 more answers
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