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aleksandr82 [10.1K]
3 years ago
7

In comparing a multi-boot environment and an environment using a type II hypervisor which offers the best performance? Which wou

ld offer the most convenience in switching from one OS to another? Explain your answer.
Computers and Technology
1 answer:
ElenaW [278]3 years ago
4 0

Answer / Explanation

To answer this question, let us first understand what a multi boot environment is and what an environment using a hypervisor is:

Multiple-booting can be defined or simply refers to the process of installing different multiple operating system on a single computer and also to be able to determine which of the operating systems boot per time.

While:

Hypervisor or a type 2 hypervisor can be described as a form of virtual machine in form of a software that is installed on in an existing operating system of a computer. One of the main benefit of an hypervisor is it enables easy navigation of guest on your computer without actually navigating into your original or primary operating system plus it aid in fast system recovery.

If we now refer back to the question asked in respect to which i would offer as the most convenience, i would prefer the type 2 hypervisor as convenient because i am able to run an operating system inside another operating system so navigating between two operating system would be very easy as compared to the Multi-booting where i have to always shut down an operating system to navigate between the operating system because i can only use one operating system per time.

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Write a program that records high-score data for a fictitious game. the program will ask the user to enter five names, and five
Harman [31]

Scores.cpp

#include <iostream>

#include <string>

using namespace std;

void initializeArrays(string names[], int scores[], int size);

void sortData(string names[], int scores[], int size);

void displayData(const string names[], const int scores[], int size);

int main()

{

   string names[5];

   int scores[5];

   //reading names and scores

   initializeArrays(names, scores, 5);

   //sorting the lists based on score.

   sortData(names, scores, 5);

   //displaying the contents of both arrays

   displayData(names, scores, 5);

 

  return 0;

}

void initializeArrays(string names[], int scores[], int size){

   for(int i=0; i<size; i++){

       cout<<"Enter the name for score #"<<(i+1)<<": ";

       cin >> names[i];

       cout<<"Enter the score for score #"<<(i+1)<<": ";

       cin >> scores[i];

       }

}

void sortData(string names[], int scores[], int size){

   

       int temp = 0;

       string tempStr = "";

       

       for(int i=0; i < size; i++){

               for(int j=1; j < (size-i); j++){

                      //checking max score and sorting based on it.

                       if(scores[j-1]< scores[j]){

                               //swap the elements!

                           //swapping scores

                               temp = scores[j-1];

                               scores[j-1] = scores[j];

                               scores[j]=temp;

                               

                               //swapping names

                               tempStr = names[j-1];

                               names[j-1] = names[j];

                               names[j]=tempStr;

                       }

                       

               }

       }

}

void displayData(const string names[], const int scores[], int size){

   cout<<"Top Scorers:"<<endl;

       //printing the elements up to the size of both arrays.

       for(int i=0; i<size; i++){

           cout<<names[i]<<": "<<scores[i]<<endl;

       }

}

3 0
4 years ago
Supposethatyoubet$5oneachofasequenceof50independentfairgames. Usethecentrallimittheorem to approximate the probability that you
lions [1.4K]

Answer:

0.119

Explanation:

50 independent fair games

Mean will be 50/2 = 25

Standard deviation of the loosing probability is :

\sqrt{50*0.5*0.5} * 10 = 33.536

we have to loose 33 times out of 50 games.

To find the approximate probability we use z- value table.

z value =  \frac{33-25}{3.35}

z - value = 2.2659

Looking at the z-value table we get 0.119.

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Answer:

step 1:- start

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step 3:- calculate interest using formula sI=((amount*rate*time)/100)

step 4:- print simple interest

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Traceroute ensures each hop on the way to a destination device drops a packet and sends back an ICMP error message.

This means traceroute can measure the duration of time between when the data is sent and when the ICMP message is received back for each hop—giving you the RTT value for each hop.

To learn more about IP address, refer

https://brainly.ph/question/4632300

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2 years ago
By default, after how much time has elapsed in a client's DHCP lease will the client attempt to renew the lease?
maw [93]
The client will attempt to renew halfway through the time of the lease.
3 0
3 years ago
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