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3 years ago

A triangle is an example of __________ .

1 answer:

7 0

CONVEX polygon.
An equilateral triangle is a triangle that has the same side length and same angles.

It is identified as a convex polygon because it does not have a reflex angle.
A reflex angle is an angle that is greater than 180° but lesser than 360°. Interior angles of a triangle sum up to 180°.

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Which of the following is the correct order of laws (from local to state to federal)

professor190 [17]

Answer:

That, Federal law > Constitutional Law > State law > Local ordinances

Explanation:

4 0

3 years ago

A testing lab wishes to test two experimental brans of outdoor pain long each wiil last befor fading . The testing lab makes six

Simora [160]

Answer:

The answer is " $\bold{Brand \ A \ (35, 350, 18.7) \ \ Brand \ B \ (35, 50, 7.07)}$ "

Explanation:

Calculating the mean for brand A:

$\to \bar{X_{A}}=\frac{10+60+50+30+40+20}{6} =\frac{210}{6}=35$

Calculating the Variance for brand A:

$\sigma_{A}^{2}=\frac{\left ( 10-35 \right )^{2}+\left ( 60-35 \right )^{2}+\left ( 50-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 20-35 \right )^{2}}{5} \\\\$

$=\frac{\left ( -25 \right )^{2}+\left ( 25 \right )^{2}+\left ( 15\right )^{2}+\left ( -5 \right )^{2}+\left ( 5 \right )^{2}+\left ( 15 \right )^{2}}{5} \\\\ =\frac{625+ 625+225+25+25+225}{5} \\\\ =\frac{1750}{50}\\\\=350$

Calculating the Standard deviation:

$\sigma _{A}=\sqrt{\sigma _{A}^{2}}=18.7$

Calculating the Mean for brand B:

$\bar{X_{B}}=\frac{35+45+30+35+40+25}{6}=\frac{210}{6}=35$

Calculating the Variance for brand B:

$\sigma_{B} ^{2}=\frac{\left ( 35-35 \right )^{2}+\left ( 45-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 35-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 25-35 \right )^{2}}{5}$

$=\frac{\left ( 0 \right )^{2}+\left ( 10 \right )^{2}+\left ( -5 \right )^{2}+\left (0 \right )^{2}+\left ( 5 \right )^{2}+\left ( -10 \right )^{2}}{5}\\\\=\frac{0+100+25+0+25+100}{5}\\\\=\frac{100+25+25+100}{5}\\\\=\frac{250}{5}\\\\=50$

Calculating the Standard deviation:

$\sigma _{B}=\sqrt{\sigma _{B}^{2}}=7.07$

4 0

3 years ago

The 169.254.78.9 ip address is an example of what type of ip address

scoundrel [369]

As specified in RFC5735, this is an address from the "link local" block. It is assigned to a network interface as a temporary address, for instance if no static address is configured and the DHCP server is not found.

If you boot your PC without a network cable, you'll probably end up with a 169.254.*.* address.

8 0

3 years ago

Write a program that records high-score data for a fictitious game. the program will ask the user to enter five names, and five

Harman [31]

Scores.cpp

#include <iostream>

#include <string>

using namespace std;

void initializeArrays(string names[], int scores[], int size);

void sortData(string names[], int scores[], int size);

void displayData(const string names[], const int scores[], int size);

int main()

{

string names[5];

int scores[5];

//reading names and scores

initializeArrays(names, scores, 5);

//sorting the lists based on score.

sortData(names, scores, 5);

//displaying the contents of both arrays

displayData(names, scores, 5);

return 0;

}

void initializeArrays(string names[], int scores[], int size){

for(int i=0; i<size; i++){

cout<<"Enter the name for score #"<<(i+1)<<": ";

cin >> names[i];

cout<<"Enter the score for score #"<<(i+1)<<": ";

cin >> scores[i];

}

void sortData(string names[], int scores[], int size){

int temp = 0;

string tempStr = "";

for(int i=0; i < size; i++){

for(int j=1; j < (size-i); j++){

//checking max score and sorting based on it.

if(scores[j-1]< scores[j]){

//swap the elements!

//swapping scores

temp = scores[j-1];

scores[j-1] = scores[j];

scores[j]=temp;

//swapping names

tempStr = names[j-1];

names[j-1] = names[j];

names[j]=tempStr;

}

void displayData(const string names[], const int scores[], int size){

cout<<"Top Scorers:"<<endl;

//printing the elements up to the size of both arrays.

for(int i=0; i<size; i++){

cout<<names[i]<<": "<<scores[i]<<endl;

}

3 0

4 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