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3 years ago

Which kind of device is a printer? A. an output device B. an input device

2 answers:

7 0

It's an output device, think about it, it's outputting something - aka your image onto paper.

Input is the opposite, it inputs something onto into the computer, for example a scanner would be an input device.

Andrew [12]3 years ago

5 0

Well a printer takes data and puts it on a piece of paper so it would be an output device

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Given the code that reads a list of integers, complete the number_guess() function, which should choose a random number between

Cerrena [4.2K]

Answer:

see explaination

Explanation:

import random

def number_guess(num):

n = random.randint(1, 100)

if num < n:

print(num, "is too low. Random number was " + str(n) + ".")

elif num > n:

print(num, "is too high. Random number was " + str(n) + ".")

else:

print(num, "is correct!")

if __name__ == '__main__':

# Use the seed 900 to get the same pseudo random numbers every time

random.seed(900)

# Convert the string tokens into integers

user_input = input()

tokens = user_input.split()

for token in tokens:

num = int(token)

number_guess(num)

5 0

3 years ago

Alguien me prestta una cuenta de osu!, porfa

Mariana [72]

Answer:

english please........

7 0

3 years ago

The ____ algorithm tries to extend a partial solution toward completion

Fittoniya [83]

Answer: Backtracking

Explanation:

Backtracking algorithm approaches a solution in a recursive fashion whereby it tries to build answers and modify them in time intervals as we progress through the solution. One such backtracking algorithm is the N Queen problem whereby we place N Queen in a chessboard of size NxN such that no two queens attack each other. So we place a queen and backtrack if there is a possibility that the queen is under attack from other queen. This process continues with time and thereby it tends to extend a partial solution towards the completion.

8 0

3 years ago

There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou

Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:- n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex u ∈ V

$\rightarrow$ for each vertex v ∈ V

$\rightarrow\rightarrow$ if( d(pu, pj) > k )

$\rightarrow\rightarrow\rightarrow$ add vertex u to Adj[v] // Adj[v] represents adjacency list of v

$\rightarrow\rightarrow\rightarrow$ add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3. bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

$\rightarrow$ visited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph

6. for each vertex u ∈ V

$\rightarrow$ if ( visited[u] == false)

$\rightarrow\rightarrow$ color[u] = 0;

$\rightarrow\rightarrow$ isbipartite = BFS(G,u,color,visited) // if the vertices reachable from u form a bipartite graph, it returns true

$\rightarrow\rightarrow$ if (isbipartite == false)

$\rightarrow\rightarrow\rightarrow$ print " No solution exists "

$\rightarrow\rightarrow\rightarrow$ exit(0)

7. for each vertex u ∈V

$\rightarrow$ if (color[u] == 0 )

$\rightarrow\rightarrow$ print " Student u is assigned Bus 1"

$\rightarrow$ else

$\rightarrow\rightarrow$ print " Student v is assigned Bus 2"

BFS(G,s,color,visited)

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

$\rightarrow$ u = Q.pop()

$\rightarrow$ for each vertex v ∈ Adj[u]

$\rightarrow\rightarrow$ if (visited[v] == false)

$\rightarrow\rightarrow\rightarrow$ color[v] = (color[u] + 1) % 2

$\rightarrow\rightarrow\rightarrow$ visited[v] = true

$\rightarrow\rightarrow\rightarrow$ Q.push(v)

$\rightarrow\rightarrow$ else

$\rightarrow\rightarrow\rightarrow$ if (color[u] == color[v])

$\rightarrow\rightarrow\rightarrow\rightarrow$ return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0

3 years ago

In what ways are computers being used to improve our quality of life

serg [7]

Explanation:

makes info easy to get

improves communication

enhances education

etc

plsss mark me brainliesttt

3 0

3 years ago