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steposvetlana [31]
3 years ago
11

In one week, your company received the following quantities of e-mail messages. Monday 240 Tuesday 315 Wednesday 290 Thursday 18

0 Friday 375 What is the average number of e-mail messages? A. 280 B. 285 C. 275 D. 310
Computers and Technology
1 answer:
Vika [28.1K]3 years ago
3 0

Answer:

280

Explanation:

Average = (240 + 315 + 290 + 180 + 375) ÷ 5

= 1400 ÷ 5

= 280

Cheers

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Which IDEs support multiple high-level programming languages? Select all that apply.
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Match the image to its type.
boyakko [2]

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  • <em>Red</em> flowers
  • Oranges

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Reds, oranges, and yellows are considered warm, and greens, teals, blues, and purples are considered cool. Colors such as white, black, brown, or gray are all neutral.

3 0
3 years ago
Need some help writing a simple PYTHON Student registration program:
Y_Kistochka [10]

import random

database = {}

while True:

   try:

       user_choice = int(input("Enter your student id to bring up your information (Press enter if you don't have one) : "))

       if user_choice in database:

           print("Student ID:",database[user_choice][0],"\nFirst name:",database[user_choice][1],"\nLast name:",database[user_choice][2],"\nPhone number:",database[user_choice][3],"\nEmail address:",database[user_choice][4],"\nPassword:",database[user_choice][5],"\nPoints:",database[user_choice][6])

   except ValueError:

       name = input("Enter your first name: ")

       last_name = input("Enter your last name: ")

       phone_number = input("Enter your phone number: ")

       email = input("Enter your email address: ")

       password = input("Enter a password: ")

       points = 100

       student_id = random.randint(1,1000)

       while student_id in database:

           student_id = random.randint(1,1000)

       print("Hello,",name,"your student ID is:",student_id)

       database[student_id] = [student_id,name, last_name, phone_number, email, password, points]

I wrote my code in python 3.8. I hope this helps.

6 0
2 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\rightarrowif( 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

\rightarrowvisited[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"

\rightarrowelse

\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
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