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hjlf
3 years ago
10

An instruction book or program that takes users through a prescribed series of steps to learn how to use a program is called (a)

________.
Computers and Technology
1 answer:
Trava [24]3 years ago
6 0
Manual

An instruction book or program that takes users through a prescribed series of steps to learn how to use a program is called a manual.
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________ is the process of translating a task into a series of commands that a computer will use to perform the task.
IgorC [24]
Programming 
It can also be described as Reverse Engineering 
7 0
4 years ago
Write a python program to calculate the length of any string recursively​
Solnce55 [7]

Answer:

b

Explanation:

6 0
3 years ago
Write a program that declares a two-dimensional array named myFancyArray of the type double. Initialize the array to the followi
kirill115 [55]

Answer:

This solution is provided in C++

#include <iostream>

using namespace std;

double sumFancyArray(double myarr[][6]){

   double sum = 0.0;

   for(int i = 0;i<2;i++)    {

       for(int j = 0;j<6;j++){

       sum+=myarr[i][j];    

       }}    

   return sum;

}

void printFancyArray(double myarr[][6]){

   for(int i = 0;i<2;i++)    {

       for(int j = 0;j<6;j++){

       cout<<myarr[i][j]<<" ";    

       }

   }

}

int main(){

   double myFancyArray[2][6] = {23, 14.12, 17,85.99, 6.06, 13, 1100, 0, 36.36, 90.09, 3.145, 5.4};

   cout<<sumFancyArray(myFancyArray)<<endl;

   printFancyArray(myFancyArray);

   

   return 0;

}

Explanation:

This function returns the sum of the array elements

double sumFancyArray(double myarr[][6]){

This initializes sum to 0

   double sum = 0.0;

This iterates through the row of the array

   for(int i = 0;i<2;i++)    {

This iterates through the column

       for(int j = 0;j<6;j++){

This adds each element

       sum+=myarr[i][j];    

       }}    

This returns the sum

   return sum;

}

This method prints array elements

void printFancyArray(double myarr[][6]){

This iterates through the row of the array

   for(int i = 0;i<2;i++)    {

This iterates through the column

       for(int j = 0;j<6;j++){

This prints each array element

       cout<<myarr[i][j]<<" ";    

       }

   }

}

The main starts here

int main(){

This declares and initializes the array

   double myFancyArray[2][6] = {23, 14.12, 17,85.99, 6.06, 13, 1100, 0, 36.36, 90.09, 3.145, 5.4};

This calls the function  that returns the sum

 cout<<sumFancyArray(myFancyArray)<<endl;

This calls the function that prints the array elements

   printFancyArray(myFancyArray);

   

   return 0;

}

Download cpp
7 0
3 years ago
What is the trade-offs in time complexity between an ArrayList and a LinkedList?
AURORKA [14]

Answer:

  In the time complexity, the array-list can easily be accessible any type of element in the the given list in the fixed amount of time.

On the other hand, the linked list basically require that the list must be traversed from one position to another end position.

The Array-List can get to any component of the rundown in a similar measure of time if the file value is know, while the Linked-List requires the rundown to be crossed from one end or the other to arrive at a position.

4 0
3 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
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