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gogolik [260]
2 years ago
15

In the context of the components of a typical expert system, _____ is a software package with manual or automated methods for ac

quiring and incorporating new rules and facts so that the expertsystem is capable of growth.A. knowledge baseB. monitoring and surveillance agentC. knowledge acquisition facilityD. personal agent
Computers and Technology
1 answer:
Pani-rosa [81]2 years ago
6 0

Answer:

knowledge acquisition facility

Explanation:

In the context of the components of a typical expert system, Knowledge acquisition facility is defined as that component of an expert system that is responsible for providing an effective and efficient medium for collecting and incorporating relevant information such as data, new rules, relationships and facts used by the expert system.

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Into which of these files would you paste copied information to create an integrated document?
Oksanka [162]
D cause you will need to keep up with data also
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2 years ago
Read 2 more answers
Write a recursive, string-valued method, replace, that accepts a string and returns a new string consisting of the original stri
ch4aika [34]

Answer:

Check the explanation

Explanation:

public String replace(String sentence){

  if(sentence.isEmpty()) return sentence;

  if(sentence.charAt(0) == ' ')

     return '*' + replace(sentence.substring(1,sentence.length()));

  else

     return sentence.charAt(0) +            replace(sentence.substring(1,sentence.length()));

3 0
2 years ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
SVETLANKA909090 [29]

Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

6 0
2 years ago
Write the definition of a function named quadratic that receives three double parameters a, b, c. If the value of a is 0 then th
VMariaS [17]

Answer:

#include <iostream>

#include <cmath>

using namespace std;

//initialize function quadratic

void quadratic(double, double, double);

int main() {

   //declare double variables a, b and c

   double a,b,c;

   //take input from user

   cin>>a>>b>>c;

   //call function quadratic

   quadratic(a,b,c);

return 0;

}

void quadratic(double a, double b, double  c){

   double root,n;

   //check if variable a is equal to zero

   if(a==0){

       cout<<"no solution for a=0"<<endl;

       return;

   }

   //check if b squared - 4ac is less than zero

   else

   if(((b*b)-(4*a*c))<0){

       cout<<"no real solutions"<<endl;

       return;

   }

   //print the largest root if the above conditions are not satisfied

   else{

       n=((b*b)-(4*a*c));

       root=(-b + sqrt(n)) / (2*a);

       cout<<"Largest root is:"<<root<<endl;

   }

   return ;

}

Explanation:

Read three double variables a, b and c from the user and pass it to the function quadratic. Check if the value of variable a is equal to zero. If true, print "no solution for a=0". If this condition is false, check if b squared - 4ac is less than zero. If true, print "no real solutions". If this condition is also false, calculate the largest solution using the following formula:

largest root = (-b + square root of (b squared - 4ac)) / (2*a)

Input 1:

2 5 3

Output 2:

Largest root is:-1

Input 2:

5 6 1

Output 2:

Largest root is:-0.2

5 0
2 years ago
For the (pseudo) assembly code below, replace X, Y, P, and Q with the smallest set of instructions to save/restore values on the
Dimas [21]

Answer:

Explanation:

Let us first consider the procedure procA; the caller in the example given.

  Some results: $s0,$s1,$s2, $t0,$t1 and $t2 are being stored by procA. Out of these registers, few registers are accessing by procA after a call to procB. But, procB might over-write these registers.

       Thus, procA need to save some registers into stack first before calling procB, .

      only $s1,$t0 and $t1 are being used after return from procB in the given example,

       Caller saves and restores only values in $t0-$t9, according to MIPS guidelines for caller-saved and callee-saved registers, .

       Thus, procA needs to save only $t0 and $t1.

    jal instruction overwrites the register $ra by writing the address, to which the control should jump back, after completing the instructions of procB, when procB is called,.

       Therefore, procA also need to save $ra into stack.

 ProcA is writing new values into $a0,$a2, procA must save $a0 and $a1 first before calling procB, .

     In the given example, after return from procB, only $a0 is being used. It is therefore enough to save $a0.

   Also, procA needs to save frame pointer, which points the start of the stack space for each procedure.

       Generally, as soon as the procedure begins, frame pointer is set to the current value of the stack pointer,.

Let us consider the procedure procB; the callee in the given example.

 The callee is responsible for saving values in $s0-$s7 and restoring them before returning to caller, this is according to MIPS guidelines for caller-saved and callee-saved registers,

   procB is expected to over-write the registers $s2 and $t0. Nonetheless, in the first two lines, procB might over-write the registers $s0 and $s1.

   Thus, procB is responsible for saving and restoring $s0,$s1 and $s2.

X:

We need to create space for 5 values on the stack since procA needs to save $a0,$ra,$t0,$t1 and $fp(frame pointer), . Each value(word) takes 4 bytes.

$sp = $sp – 20 # on the stack, create space for 5 values

sw $a0, 16($sp) # store the result in $a0 into the memory address

               # indicated by $sp+20

sw $ra, 12($sp) # save the second value on stack

sw $t0, 8($sp) # save the third value on stack

sw $t1, 4($sp) # save the fourth value on stack

sw $fp, 0($sp) #  To the stack pointer, save the frame pointer

$fp = $sp      #  To the stack pointer, set the frame pointer

Y:

lw $fp, 0($sp) #  from stack, start restoring values

lw $t1, 4($sp)

lw $t0, 8($sp)

lw $ra, 12($sp)

lw $a0, 16($sp)

$sp = $sp + 20 # decrease the size of the stack

P:

$sp = $sp – 12 #  for three values, create space on the stack

sw $s0, 0($sp) # save the value in $s0

sw $s1, 0($sp) # save the value in $s1

sw $s2, 0($sp) # save the value in $s2

Q:

lw $s0, 0($sp) #  from the stack, restore the value of $s0

lw $s1, 0($sp) #  from the stack, restore the value of $s1

lw $s2, 0($sp) #  from the stack, restore the value of $s2

$sp = $sp + 12 # decrease the stack size

8 0
2 years ago
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