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cricket20 [7]
3 years ago
9

Which term describes the process of training a machine to do simple, repetitive tasks, and adapt or correct its performance base

d on changing conditions at speed and scale?​
Computers and Technology
1 answer:
yKpoI14uk [10]3 years ago
8 0
Automation. ... It involves taking a machine or software that was taught to do simple repetitive tasks (traditional automation) and teaching it to intuitively adapt or correct its performance based on changing conditions, at speed and scale.
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What is the MINIMUM number of paths that would need to be broken to prevent Computing Device A from connecting with Computing De
Sveta_85 [38]

Answer:

The answer is "Option c".

Explanation:

In the given question the device A is connected by 3 wires, contributing all of them, which also includes several connector paths. When all the wires of A are broken down, and if all of this leaves no routes that can be used. Even so, if it is done to E, it's also linked to four different routes. Its solution would've been C because its value will be the MINIMUM.

4 0
3 years ago
Read 2 more answers
In traditional programming, probably the most often used error-handling outcome was to ____.Group of answer choicesterminate the
iVinArrow [24]

In traditional programming, probably the most often used error-handling outcome was to terminate the program in which the offending statement occurred.

<h3>What is a traditional programming?</h3>

Traditional programming is known to be a form of manual way that one or a user makes a program.

Note that in the case above, In traditional programming, probably the most often used error-handling outcome was to terminate the program in which the offending statement occurred.

Learn more about programming from

brainly.com/question/23275071

#SPJ1

4 0
2 years ago
Which of the following are the lines defining the borders of a shape? (1 point)
shtirl [24]

Answer:

no I is edges

2 is gaming

3 is creativity

3 0
3 years ago
A circuit contains three resistors connected in parallel. The value of R1 is 2 K , the value or R2 is 6 K , and the value of R3
mezya [45]
In a parallel connection, the voltage is same in every branch.
Now, three <span>three resistors connected in parallel.
R1 = 2k ohm
</span>R2 = 6k ohm
R3 = 10k ohm
in parallel, net resisitance  = \frac{ R_{1}R_{2}R_{3} }{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}

Now, putting the values, we get, R net = 1.30 k ohm.
Ans, voltage = 100 VDC
Thus, power = \frac{ V^{2} }{R} = \frac{ 100^{2} }{1.3 k} \\ \\ where \\ k = 10^{3} 
                      = 7.69 Watt
6 0
3 years ago
Write a program that accepts an integer value called multiplier as user input. Create an array of integers with ARRAY_SIZE eleme
Tasya [4]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

void PrintForward(int myarray[], int size){

   for(int i = 0; i<size;i++){        cout<<myarray[i]<<" ";    }

}

void PrintBackward(int myarray[], int size){

   for(int i = size-1; i>=0;i--){        cout<<myarray[i]<<" ";    }

}

int main(){

   const int ARRAY_SIZE = 12;

   int multiplier;

   cout<<"Multiplier: ";

   cin>>multiplier;

   int myarray [ARRAY_SIZE];

   for(int i = 0; i<ARRAY_SIZE;i++){        myarray[i] = i * multiplier;    }

   PrintForward(myarray,ARRAY_SIZE);

   PrintBackward(myarray,ARRAY_SIZE);

   return 0;}

Explanation:

The PrintForward function begins here

void PrintForward(int myarray[], int size){

This iterates through the array in ascending order and print each array element

<em>    for(int i = 0; i<size;i++){        cout<<myarray[i]<<" ";    }</em>

}

The PrintBackward function begins here

void PrintBackward(int myarray[], int size){

This iterates through the array in descending order and print each array element

<em>    for(int i = size-1; i>=0;i--){        cout<<myarray[i]<<" ";    }</em>

}

The main begins here

int main(){

This declares and initializes the array size

   const int ARRAY_SIZE = 12;

This declares the multiplier as an integer

   int multiplier;

This gets input for the multiplier

   cout<<"Multiplier: ";    cin>>multiplier;

This declares the array

   int myarray [ARRAY_SIZE];

This iterates through the array and populate the array by i * multiplier

<em>    for(int i = 0; i<ARRAY_SIZE;i++){        myarray[i] = i * multiplier;    }</em>

This calls the PrintForward method

   PrintForward(myarray,ARRAY_SIZE);

This calls the PrintBackward method

   PrintBackward(myarray,ARRAY_SIZE);

   return 0;}

6 0
3 years ago
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