What is the IP address of the second Ethernet adapter?

Answer:

A half note is a note that lasts two quarter notes. For most songs, it does last half a measure, but this is not always the case. An 8th note is a note that lasts half of a quarter note, and is usually (but again not always) an 8th of a measure. A quarter note is the most common type of note in a measure.

If you want to find out what exactly that stuff means, well, you're out of luck here, but it's probably better to just look that up anyways.

Answer:

It could be that an ssh virtual terminal has not been configured for the port connected to PC-A

Explanation:

The computer network is an interconnection and intercommunication of computer devices to share resources. Intermediate devices like the Router and Switches are used to create physical and virtual connections between networks and nodes.

virtual connections like the default Telnet and the and more secure SSH are used to remotely communicate with intermediate devices for management purposes. For example, to use the ssh, the line vty and the specified virtual port range must be configured with a username, a password, an encryption key and other ssh features before a connection can be made.

Answer:

Here is the JAVA program that prompts the user to enter positive integers from standard input and prints out the maximum and minimum values:

import java.util.*;

public class Ex0603 {

   public static void main(String[] args) { //start of main function

   Scanner in = new Scanner(System.in); //creates Scanner class object to take input from user

   int minimum = 0; //stores the minimum of the integers

   int maximum = 0; //stores the minimum of the integers

   int num; //stores the positive integer values

   System.out.println("enter a positive integer: ");//prompts user to enter a positive integer

   num = in.nextInt();//scans and reads the input integer value from user

   minimum = num;  //assigns value of num to minimum

   while(true) {//while loop keeps prompting user to enter a positive integer

       System.out.println("enter a positive integer: ");//prompts user to enter a positive integer

       num = in.nextInt();//scans and reads the input integer value from user

       if(num <=0)//if the user inputs a negative integer or 0

           break; //breaks the loop

       if (num > maximum) {//if the value of positive integer is greater than the maximum of value

           maximum = num; } //assigns that maximum value of num to maximum variable

       if (num < minimum) {//if the value of positive integer is less than the minimum of value

           minimum = num; //assigns that minimum value of num to minimum       }    }            

   System.out.println("The maximum integer is : " + maximum); //displays the maximum of the positive integers

   System.out.println("The minimum integer is : " + minimum); }}   //displays the minimum of the positive integers

Explanation:

Here is the JAVA program that that prompts the user to enter an integer N, then reads N double values, and prints their mean (average value) and sample standard deviation:

import java.util.*;  

public class Ex0603 {  

public static void main(String[] args) { //start of main function

Scanner in = new Scanner(System.in);  //creates Scanner class object to take input from user

double integer= 0;  //to store the number of input values

double sum = 0;  //to store the sum of input numbers

double mean = 0;  //to store the average of numbers

double sd = 0;  //to store the standard deviation result

double variance = 0;  //to store the variance result

double sumSquare = 0;  //to store the sum of (num-mean)^2

int n = 0;  //to store the sample size N

   System.out.println("Enter an integer: ");  //prompts user to enter an integer that is how many elements user wants to input

   integer = in.nextInt();  //reads the value of integer from user

   double size=integer; // declares a double type variable and assigns value of int type variable integer to this double type variable size

   int i=0;  //used to point each element of the num array

   double num[] = new double[(int)size];  //declares an array to hold the numbers input by user

   while(integer>0){  //loop executes until value of integer is greater than 0

   System.out.println("Enter a number: "); // prompts to enter a number

   num[i] = in.nextDouble(); //reads double values

   sum += num[i];  //adds the values stored in num array

   i++;  //increments i to point to the next element in num array

   n++;  //increments n to count the total number of elements

  integer--; //decrements integer value at each iteration

   mean = sum / n;    }  //compute the average of values

   i=0;  

    while(size>0){  //loop executes until value of size exceeds 0

       sumSquare += Math.pow(num[i]-mean,2); //takes the sum of square difference between each element of num array and value of mean

       i++; //increments i

      size--;   }  //decrements size

variance = sumSquare / (n-1);  //compute variance by dividing the result of sum of the squares of their differences from the mean by n-1

   sd = Math.sqrt(variance);  //computes standard deviation by using sqrt function which takes the sqrt of the result of variance computed above

System.out.println("Average value is: " + mean+ " and the standard deviation is " + String. format("%.2f", sd));   }} //displays the average of numbers stored in mean variable and value of standard deviation is displayed up to 2 decimal places

Screenshots of both programs and their outputs is attached.

Answer:

true

Explanation:

Answer:

C) Normalization

Explanation:

Normalization is an approach that is to be done in a systematic manner in order to remove the data duplicity and there are non-desirable attributes such as insert, update, delete, etc. In this the process should be followed in multi step that contains the data in a tabular form so that it could eliminate the repetition of data to maintain the efficiency of the data

Therefore according to the given situation, the option C is correct

The process of ensuring that data values are ordered such that the values in the columns confirms to a certain range and expunged duplicate values is called Normalization.

Hence, the process is known as Normalization.

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Answer:

false

Explanation:

If the condition is false at the start, the contents of the loop should never be executed. If the CMP is at the end, you cannot prevent the code to run at least once. So this approach will work for a DO { ... } WHILE() loop, but not for a regular WHILE() { ... }

Answer:

I believe its confidentiality

Explanation:

confidentiality means the state of being kept secret or private

A loss of confidentiality is the unauthorized disclosure of information. The correct option is c).

Confidentiality is a set of rules or a promise that limits access to specific information. There aren't enough boundaries in confidential information. One shouldn't give your information to random people, but you also shouldn't give it to people close to you because not everyone is trustworthy.

One should only give out one's information when it is absolutely necessary, which is if you are. Unless there is a compelling reason not to, request consent to share information.

Unless it is going to hurt you or another person, a counselor will keep all of the information you give them confidential. When a detective keeps all crime-related clues and information confidential and hidden from civilians.

Therefore, the correct option is c) confidentiality.

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Answer:

Assignment is the operation of a variable to a value

The difference between an attribute and a value set is:

In programming, there are some terms that are useful for a person that wants to write a code that works.

Some of these terms include:

As a result of this, an attribute is the name that  is given to the property of a class.  On the other hand, a value set are those things that are used to represent an object in a given class.

For example, an attribute can have the name "Apples" and can hold the value of "abc" while the value set are those things that can be attributed to a class.

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Answer:

The RTP (real time protocol) which uses the UDP ( user datagram protocol) and Real-time control protocol.

Explanation:

The RTP of a voice over IP protocol (VoIP) which is used to prioritise the transfer of video and audio data over text data. It is used for video conferencing and telephony services in an enterprise.

It uses the user-datagram protocol to transfer or access videos and audio data for its real time effect and the real-time control protocol for quality of service analysis and maintenance.

An example of shareware a compression program like Stuffit for Macs or Windows. An example of shareware is a budgeting software program that only allows you to use three categories instead of having the ability to create a full budget

Explanation:

Blackbaud's software products are specially designed to support the unique needs of nonprofit and social good organizations.

Explore our software solutions that can help you advance your organization's mission.

Answer:

Moore's law is currently dead

Explanation:

But it was relevant to give a metric of the computational power that will exist, allowing programmers to make software that lasts longer, Moore's law said that a chip's transistors would be doubled every 18 months for the same price

Answer:

1.Shows roads

2.shows the contours and physical features of an area

Explanation:

Answer:

A.shows roads

D.shows the contours and physical features of an area

Explanation:

Answer:

import java.util.Scanner;

public class Main

{

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

 System.out.print("Enter a number: ");

 int userVal = input.nextInt();

 String aString;

 if(userVal < 0)

     aString = "negative";

 else

     aString = "non-negative";

 System.out.println(aString);

}

}

Explanation:

Ask the user to enter a number and set it to userVal

Check the value of userVal. If it is smaller than 0, set the string as "negative". If it is not, set it as "non-negative"

Print the string

Answer:

B. Scheme is a fully curried language that does not do delayed evaluation.

Explanation:

Scheme is a functional programming language that accepts first or higher-order functions and is a family of the LISP programming language. The syntax in scheme supports both curring and uncurring.

It also supports delayed evaluation and lazy evaluation of its coding statements

I've had this question before.

The answers are 2 & 4.

Leadership & career specific skills.

Answer:

2. Leadership

4. Career-specific skills

Explanation:

Just took the assignment

Hope this helps! :)

Answer:

I think secure digital is needed

Answer:

Film

Explanation:

I chose Film and got it correct

Answer:

   See explanation

Explanation:

Given:

Processors:

P1  

P2

P3

Clock Rate of Processors:

Clock rate of P1 = 3 GHz

Clock rate of P2 = 2.5 GHz  

Clock rate of P3 = 4.0 GHz  

CPI of Processors:

Cycles per instruction = CPI of P1 = 1.5  

Cycles per instruction = CPI of P2 = 1.0

Cycles per instruction = CPI of P3 = 2.2

To find:

a) Which processor has the highest performance expressed in instructions per second

Solution:

Performance = clock rate / CPI  

The performance of processor depends on instructions count and CPU time.  

As we know that  

CPU time = Instructions * Cycles Per Instruction / clock rate  

CPU time = Instructions * CPI / clock rate

Instructions per second = Instruction count / CPU time  

As:    

CPU time = Instructions * CPI/ clock rate  

Instructions/CPU time = clock rate/CPI  

Instructions per second = clock rate / CPI

IPS = clock rate/ CPI  

Hence

Performance = Clock rate / CPI = clock rate/ CPI

Compute Performance of P1:

Performance for P1 = IPS of P1 = clock rate of P1 / CPI of P1 = 3 GHz / 1.5 = 2

As we know that 1 GHz = 10⁹ Hz. So:  

Performance of P1  expressed in instructions per second is  2 x 10⁹

Compute Performance of P2:

Performance for P2 = IPS of P2= clock rate of P2 / CPI of P2 = 2.5 GHz / 1.0 = 2 .5

As we know that 1 GHz = 10⁹ Hz. So:  

Performance of P1  expressed in instructions per second is  2.5 x 10⁹

Compute Performance of P3:

Performance for P3 = IPS of P3= clock rate of P3 / CPI of P3 = 4.0 GHz / 2.2 = 1.82

As we know that 1 GHz = 10⁹ Hz. So:  

Performance of P1  expressed in instructions per second is 1.82 x 10⁹

From the above computed performances of each processor it can be seen that Processor 2 (P2) has the highest performance expressed in instructions per second i.e. 2.5 x 10⁹

b) find the number of cycles and the number of instructions.

Given:

processors each execute a program in 10 seconds, So,

CPU time = 10 sec

Solution:

Compute number of cycles:

As we know that:

CPU time = cycles count / clock rate   = clock cycles/clock rate

So

clock cycles = CPU time x clock rate

Compute number of cycles of P1:

clock cycles  = 10 x 3 GHz

                     = 30

As we know that 1 GHz = 10⁹ Hz. So:  

clock cycles of P1 = 3 x 10¹⁰

Compute number of cycles of P2:

clock cycles  = 10 x 2.5 GHz

                     = 25

As we know that 1 GHz = 10⁹ Hz. So:  

clock cycles of P2 = 2.5 x 10¹⁰

Compute number of cycles of P3:

clock cycles  = 10 x 4.0 GHz

                     = 40

As we know that 1 GHz = 10⁹ Hz. So:  

clock cycles of P3 = 4 x 10¹⁰

Now as we know that:

Instructions per second = Instruction count / CPU time  

IPS = IC + CPU time

So to find number of instructions:

instruction count = Instructions per second x CPU time

Compute number of instructions of P1:

instructions of P1 = Instructions per second of P1 x CPU time

                            = 2 x 10⁹  x 10

                            = 2 x 10¹⁰

Compute number of instructions of P2:

instructions of P2 = Instructions per second of P2 x CPU time

                            = 2.5 x 10⁹  x 10

                            = 2.5 x 10¹⁰

Compute number of instructions of P3:

instructions of P3 = Instructions per second of P3 x CPU time

                            = 1.82 x 10⁹  x 10

                            = 1.82 x 10¹⁰

c) What clock rate should we have to reduce the execution time by 30%

As we know

CPU time = Execution time = instructions x CPI / clock rate

We have to find new clock rate to reduce execution time by 30%

This means we have to find:

New Execution Time = 70% of Old Execution Time

According to formula of Execution time:

instructions(new) x CPI(new) / clock rate(new) = 0.7 [instructions(old) x CPI(old) / clock rate(old)]

As the instructions(new)  = instructions(old)

So,

CPI(new) / clock rate(new) = 0.7 [CPI(old) / clock rate(old)]

When trying to reduce the execution time by 30%, this leads to an increase of 20% in the CPI.

CPI(new) = 1.2 CPI(old)

New CPI of P1:

CPI(new P1) = 1.2 CPI(old P1)

                    = 1.2 x 1.5

CPI(new P1) = 1.8

New CPI of P2:

CPI(new P2) = 1.2 CPI(old P2)

                    = 1.2 x 1.0

CPI(new P2) = 1.2

New CPI of P3:

CPI(new P3) = 1.2 CPI(old P3)

                    = 1.2 x 2.2

CPI(new P3) = 2.6

1.2 / clock rate (new) = 0.7 / clock rate(old)

So new clock rate is computed as:

clock rate (new)  = (1.2 / 0.7 ) x clock rate(old)

clock rate (new)  = 1.71 x clock rate(old)

clock rate (new)  = 1.71 x clock rate(old)

Hence the clock rate should be increased by 71% approx.

Now new clock rate for each processor is:

clock rate (new) for P1 = 3 GHz x 1.71     = 5.13 GHz

clock rate (new) for P2 = 2.5 GHz x 1.71 = 4.27 GHz

clock rate (new) for P3 = 4.0 GHz x 1.71 = 6.84 GHz  

In this exercise we have to use CPU and processor knowledge, so we can say that:

A)Processador 2

B) Processador 3

C) Processador 3

From the information given in the text we have:

a) With the formula bellow, we find:

[tex]Performance = clock \ rate / CPI[/tex]

The performance of processor depends on instructions count and CPU time.  

[tex]P= 3 GHz / 1.5 = 2*10^9[/tex]

[tex]P = 2.5 GHz / 1.0 = 2 .5*10^9[/tex]

[tex]P = 4.0 GHz / 2.2 = 1.82 *10^9[/tex]

b) Processors each execute a program in 10 seconds, So:

[tex]10 * 3 GHz = 30*10^{10}[/tex]

[tex]10 * 2.5 GHz= 25*10^{10[/tex]

[tex]10 * 4.0 GHz = 40*10^{10[/tex]

c) So for reduce we have:

[tex]CPI = 1.2 CPI = 1.2 * 1.5CPI = 1.8[/tex]

[tex]CPI= 1.2 CPI = 1.2 * 1.0\\CPI= 1.2[/tex]

[tex]CPI= 1.2 CPI = 1.2 *2.2\\CPI = 2.6[/tex]

Now new clock rate for each processor is:

[tex]clock rate (new) for P1 = 3 GHz x 1.71 = 5.13 GHz\\clock rate (new) for P2 = 2.5 GHz x 1.71 = 4.27 GHz\\clock rate (new) for P3 = 4.0 GHz x 1.71 = 6.84 GHz[/tex]

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Answer:

A5:G12

Hope it helps yuh

# Using the input function

# Prompt the user to input some number

# Convert the input to a float - this will cater for both floating

# numbers and integers

# Store the result in a variable called my_number

my_number = float(input('Please enter some number '))

#Create a variable named calculation which equals my_number plus 9

calculation = my_number + 9

#Multiply calculation by 2

calculation *= 2

#Subtract 4 from calculation

calculation -= 4

#Divide calculation by 2

calculation /= 2

#Subtract my_number from calculation

calculation -= my_number

#Print calculation

print('calculation = ' + str(calculation))

>> Please enter some number 8

calculation = 7.0

The above program has been written in Python and it contains comments explaining each line of the code. Please go through the comments. A sample output has also been provided.

Answer:

The Following are the answer to the given statement:

Explanation:

Given value:

[tex]a) \ Y= \bar{A}BC+\bar{B\bar{C}} +BC \\\\b) \ Y= \bar{(A+B+C)}D+AD+B \\\\c) \ Y= ABCD+\bar{A} B \bar{C} D + \bar (\bar{B} + D) \ \ E[/tex]

In latex code, in typing there is some error so, please find the question.

Solve point a:

[tex]a) \ Y= \bar{A}BC+\bar{B\bar{C}} +BC \\\\[/tex]

        [tex]= (\bar{A} + 1 ) BC+\bar{B\bar{C}} \\\\= (\bar{B} + C ) +BC\\\\= \bar{B} + C +BC\\\\= \bar{B} + C(B+1) \\\\= \bar{B} + C \\\\[/tex]

Solve point b:

[tex]b) \ Y= \bar{(A+B+C)}D+AD+B \\\\[/tex]

[tex]Y= \bar{A}+\bar{B}+\bar{C}D+AD+B \\\\[/tex]

   [tex]= \bar{C}D+AD+B \\\\[/tex]

Solve point c:

[tex]c) \ Y= ABCD+\bar{A} B \bar{C} D + \bar (\bar{B} + D) \ \ E[/tex]

        [tex]=ABCD+\bar{A}B\bar{C}D+ B\bar{D}E\\\\=(AC+\bar{A}\bar{C} )BD+ B\bar{D}E\\\\=(A \circ C)BD+ B\bar{D}E\\\\[/tex]

 please find the circuits attachment for the above point.

A. digital encryption

Answer:

the answer is a for all u edg users

Explanation:

Answer:

b. Wolfram Alpha

Explanation:

A specialty search engine can be defined as a type of database query engine which searches for informations and indexes them based on a particular content, subject or focus such as location, topic, government, industry etc. A specialty search engine is different from the general-purpose search engines such as Bing, Google or Yahoo because it gives results that are tailored or more relevant to a particular user.

Wolfram Alpha is an example of a specialty search engine. A Wolfram Alpha is an answer search engine which gives answers to specific questions such as computational, arithmetic, geometric mathematics or calculus.

Also, note that a specialty search engine is commonly referred to as a topical or vertical search engine.

Wolfram Alpha is an example of a specialty search engine. Thus, option B is correct.

A specialty search engine can be defined as a type of database query engine which searches for informations and indexes them based on a particular content, subject or focus such as location, topic, government, industry etc. A specialty search engine is different from the general-purpose search engines such as Bing, Yahoo because it gives results that are tailored or more relevant to a particular user.

Wolfram Alpha is an example of a specialty search engine. A Wolfram Alpha is an answer search engine which gives answers to specific questions such as computational, arithmetic, geometric mathematics or calculus.

Also, note that a specialty search engine is commonly referred to as a topical or vertical search engine.

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Answer:C

Explanation:

Answer:

See explanation

Explanation:

First let us find what this algorithm compute:

Algorithm Mystery(n) {

        S = 0;

        for i = 1 to n do

             { S = S + i * i; }

         return S }

1)

Let suppose n = 3

S = 0

The algorithm has a for loop that has a loop variable i initialized to 1

At first iteration:

i = 1

S = S + i * i;

   = 0 + 1 * 1

S = 1

At second iteration:

i = 2

S = 1

S = S + 2 * 2;

   = 1 + 2 * 2

   = 1 + 4

S = 5

At third iteration:

i = 3

S = 5

S = S + 3 * 3;

   = 5 + 3 * 3

   = 5 + 9

S = 14

Now the loop breaks at i=4 because loop iterates up to n and n =3

So from above example it is clear that the algorithm computes the sum of squares of numbers from 1 to n. Or you can say it compute the sum of first n squares.  Let us represent this statement in mathematical form:

∑[tex]\left \ {{n} \atop {i=1}} \right.[/tex] i²

2)

From above example we can see that the basic operation is multiplication. At every iteration loop variable i is multiplied by itself from 1 to n or till n times. However we can also say that addition is the basic operation because at each iteration the value of S is added to the square of i. So it takes the same amount of time.

3)

In the for loop, the basic operation executes once. We can say that at each iteration of the loop, multiplication is performed once. Suppose A(n) represents the number of times basic operation executes then,

A(n) = ∑[tex]\left \ {{n} \atop {i=1}} \right.[/tex] 1 = n

4)

Since for loop executes once for each of the numbers from 1 to n, so this shows that for loop executes for n times. The basic operation in best, worst or average case runs n times. Hence the running time of Θ(n) . We can say that A(n) = n ∈ Θ(n)  

If b is the number of bits needed to  represent n then

b = log₂n + 1

b = log₂n

So

n ≈ [tex]2^{b}[/tex]

A(n) ≈ [tex]2^{b}[/tex] ≈ Θ( [tex]2^{b}[/tex] )  

5)

One solution is to calculate sum of squares without using Θ(n) algorithm.  So the efficient algorithm that takes less time than the previous algorithm is:

Algorithm Mystery(n) {

        S = 0;

        S = (n * ( n + 1 ) (2n + 1) ) / 6 ;

        return S}

Now the sum can be calculated in Θ(1)  times. This is because regardless of the size and number of operands/operations, the time of arithmetic operation stays the same. Now lets find out how this algorithm works for

n = 3

S = (n * ( n + 1 ) (2n + 1) ) / 6 ;

 = (3 * ( 3 + 1 ) * (2(3) + 1) ) / 6

 = (3 * (4) * (6+1)) / 6

= (3 * (4) * (7)) / 6

= (3 * 4 * 7) / 6

= 84 / 6

S = 14

Answer:

C: 6

Explanation:

The values are separated by ` ,`  with this it is enough to count all the numbers that are separated by ` , ` making it have 6 elements

Following are the program to calculate the length of the array:

Program Explanation:

OR

In the given array all the element is separated by the "," symbol, and when the user counts the array value that is equal to 6.

Program:

#include <iostream>//header file

using namespace std;

int main()//main method

{

   int data[] = { 12, 34, 9, 0, -62, 88 };//defining an integer array that hold given value

   cout<<sizeof(data)/sizeof(data[0])<<endl;//using print method calculate and print the size of array

   return 0;

}

Output:

Please find the attached file.

Therefore, the final answer is "Option c".

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Answer:

total = 0

for i in range(5):

   score = int(input("Enter a score: "))

   total += score

average = total / 5

print("The average is " + str(average))

Explanation:

*The code is in Python.

Initialize the total as 0

Create a for loop that iterates five times. Inside the loop, ask the user to enter a score. Add the score to the total (cumulative sum)

After the loop, calculate the average, divide the total by 5

Print the average