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stiv31 [10]
2 years ago
7

Which of the following solutions enables simultaneous digital transmission of voice, video, data, and other network services ove

r a PSTN?1) WiMAX
2) PoE
3) ISDN
4) HSPA+
Computers and Technology
1 answer:
Annette [7]2 years ago
8 0

Answer:

3. ISDN

Explanation:

ISDN ( Integrated Services Digital Network  ) -

It is the used for the simultaneous digital transmission of network service , data , video and voice , instead of the old circuits of the public switched telephone network ( PSTN ) , is referred to as ISDN .

The characteristic feature of ISDN is that it integrates the data and speech on the same lines , which is not possible in PSTN .

Hence, from the given information of the question,

The correct option is ISDN .

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Glenda operates an Airbnb business in which she rents an apartment for $150US per night . There is a mandatory deposit of 50 dol
Kipish [7]

Answer:

Steps of the technological design process include: identify a problem, research the problem, generate possible solutions, select the best solution, create a model, test the model, refine and retest the model as needed, and communicate the final solution.

Explanation:

3 0
2 years ago
Given that it takes 0.08 ms to travel from one track to the next of a hard drive; that the arm is originally positioned at Track
daser333 [38]

Answer:

Time taken to travel from one track to the next = 0.08ms

Initial track= 15     0

4      (15-4)*(0.08)= 0.88

40     (40-4)*(0.08)= 2.88

35      (40-35)*(0.08)= 0.4

11       (35-11)*(0.08)= 1.92

14       (14-11)*(0.08)= 0.24

7         (14-7)*(0.08)= 0.56

----------------------------------------------

Total seek time=0.88+2.88+0.4+1.92+0.24+0.56=6.88ms

Explanation:

We caculate the seek time for each request, and then add them together to find the total seek time. The final track number for the current request becomes the current track of next request, and this process is repeated till the last request is processed.

4 0
3 years ago
The scheme where you can find the greatest common divisor of two integers by repetitive application of the division algorithm is
gavmur [86]

Answer:

False

Explanation:

The scheme where you can find the greatest common divisor (GCD) of two integers by repetitive application of the division algorithm is known as Euclidean Algorithm.

The Euclidean Algorithm for calculating GCD of two numbers X and Y can be given as follows:

  • If X=0 then GCD(X, Y)=Y since the Greatest Common Divisor of 0 and Y is Y.
  • If Y=0 then GCD(X, Y)=X since the Greates Common Divisor of 0 and X is X.
  • Let R be the remainder of dividing X by Y assuming X > Y. (R = X % Y)
  • Find GCD( Y, R ) because GCD( X, Y ) = GCD(Y, R ).
  • Repeat the above steps again till R = 0.

4 0
3 years ago
___refers to self-acting or self-operating.
lisabon 2012 [21]

Answer:

do it to me watch this

Explanation:

trailing zeros in a whole number with no decimal shown are NOT significan

6 0
3 years ago
Write a program that lets the user enter the total rainfall for each of 12 months (starting with January) into an array of doubl
lianna [129]

Answer:

The C++ code is given below with appropriate comments

Explanation:

#include "stdafx.h"

//Header file section

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

//Function prototypes

double rainfallTotal(double[], int);

double averageMonthlyRainfall(double[], int);

double highestAmountRainfall(double[], int, int &);

double lowestAmountRainfall(double[], int, int &);

int main()

{

//Initialize variables

const int ALL_MONTHS = 12;

int highIndex = 0;

int lowIndex = 0;

//Declare variables

double totalRf;

double averageRf;

double mostRf;

double leastRf;

double monthlyRf[ALL_MONTHS];

string monthName[ALL_MONTHS] = { "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December" };

for (int i = 0; i < ALL_MONTHS; i++)

{

 cout << "Enter rainfall for " << monthName[i] << ": ";

 cin >> monthlyRf[i];

 // Check the input as negative numbers

 //for monthly rainfall

 while (monthlyRf[i] < 0)

 {

  cout << "Invalid data (negative rainfall) " << endl;

  cout << "Retry the rainfall in " << monthName[i] << ": ";

  cin >> monthlyRf[i];

 }

}

//Call the methods

totalRf = rainfallTotal(monthlyRf, ALL_MONTHS);

averageRf = averageMonthlyRainfall(monthlyRf, ALL_MONTHS);

mostRf = highestAmountRainfall(monthlyRf, ALL_MONTHS, highIndex);

leastRf = lowestAmountRainfall(monthlyRf, ALL_MONTHS, lowIndex);

//Display output

cout << fixed << showpoint << setprecision(2) << endl;

cout << "Total rainfall: " << totalRf << endl;

cout << "Average rainfall: " << averageRf << endl;

cout << "Least rainfall in " << monthName[lowIndex] <<endl;

cout << "Most rainfall in " << monthName[highIndex] <<endl;

system("pause");

return 0;

}

//Method definition of rainfallTotal

double rainfallTotal(double totalRainfall[], int n)

{

double total = 0;

for (int i = 0; i < n; i++)

 total += totalRainfall[i];

return total;

}

//Method definition of averageMonthlyRainfall

double averageMonthlyRainfall(double totalRainfall[], int n)

{

double average = 0.0;

double total = 0;

for (int i = 0; i < n; i++)

 total += totalRainfall[i];

average = total / n;

return average;

}

//Method definition of lowestAmountRainfall

double lowestAmountRainfall(double totalRainfall[], int n, int &monthIndex)

{

double least;

int i = 0;

least = totalRainfall[i];

while (i < n)

{

 if (totalRainfall[i] < least)

 {

  least = totalRainfall[i];

  monthIndex = i;

 }

 i++;

}

return least;

}

//Method definition of highestAmountRainfall

double highestAmountRainfall(double totalRainfall[], int n, int &monthIndex)

{

double high;

int i = 0;

high = totalRainfall[i];

while (i < n)

{

 if (totalRainfall[i] > high)

 {

  high = totalRainfall[i];

  monthIndex = i;

 }

 i++;

}

return high;

}

3 0
2 years ago
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