Which part of the cryosphere comes directly from the atmosphere?

Regarding the Internet of Things (IoT), a business involved in utilities, critical infrastructure, or environmental services can

jeka94

Answer:

A. true

Explanation:

Based on the information provided within the question it can be said that the statement is completely true. This is because, traffic-monitoring applications have the main function of monitoring and analyzing the flow of traffic of one or various different items. Which can be applied to the business involving utility, infrastructure, and environmental services by monitoring the specific items regarding each business. Therefore making sure that enough product is entering or leaving each business as it is suppose to. As well as providing stock analysis and defective product alerts.

6 0

3 years ago

Write a C++ programthat returns the type of a triangle (scalene, equilateral,or

bixtya [17]

Answer:

#include<iostream>

using namespace std;

int main(){

//initialize

int a, b,c;

//print the message

cout<<"Enter the three sides of the triangle: "<<endl;

//store in the variables

cin>>a>>b>>c;

//if-else statement for checking the conditions

if(a == b && b == c && a == c){

cout<<"\nThe triangle is equilateral";

}else if(a != b && b != c && a != c ){

cout<<"\nThe triangle is scalene";

}else{

cout<<"\nThe triangle is isosceles";

}

return 0;

}

Explanation:

Create the main function and declare the three variables for the length of the sides of the triangle.

print the message on the screen for the user. Then the user enters the values and its store in the variables a, b, and c.

use the if-else statement for checking the conditions.

Equilateral triangle: all sides of the triangle are equal.

if condition true, print the equilateral triangle.

Scalene triangle: all sides of the triangle are not equal.

if condition true, print the Scalene triangle.

if both conditions is not true, then, the program moves to else part and print isosceles.

8 0

3 years ago

A contiguous subsequence of a list S is a subsequence made up of consecutive elements of S. For example, if S is 5,15,-30,10,-5,

Sindrei [870]

Answer:

We made use of the dynamic programming method to solve a linear-time algorithm which is given below in the explanation section.

Explanation:

Solution

(1) By making use of the dynamic programming, we can solve the problem in linear time.

Now,

We consider a linear number of sub-problems, each of which can be solved using previously solved sub-problems in constant time, this giving a running time of O(n)

Let G[t] represent the sum of a maximum sum contiguous sub-sequence ending exactly at index t

Thus, given that:

G[t+1] = max{G[t] + A[t+1] ,A[t+1] } (for all 1<=t<= n-1)

Then,

G[0] = A[0].

Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.

However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.

Now the algorithm would be:

Create arrays G and V each of size n.

G[0] = A[0];

V[0] = 0;

max = G[0];

max_start = 0, max_end = 0;

For i going from 1 to n-1:

// We know that G[i] = max { G[i-1] + A[i], A[i] .

If ( G[i-1] > 0)

G[i] = G[i-1] + A[i];

V[i] = V[i-1];

Else

G[i] = A[i];

V[i] = i;

If ( G[i] > max)

max_start = V[i];

max_end = i;

max = G[i];

EndFor.

Output max_start and max_end.

The above algorithm takes O(n) time .

4 0

2 years ago

Write a program that simulates flipping a coin repeatedly and continues until three consecutive heads. are tossed. At that point

quester [9]

Answer and Explanation:

#include <iostream>

#include <string>

#include <time.h>

#include <vector>

using namespace std;

//Takes user info

int user_info(string *name,char *gender){

char G;

cout<<"What is your name?: ";

getline(cin, *name);

cout<<"What is your gender(m/f):";

cin>>G;

cout<<endl;

*gender = tolower(G);

return 0;

}

//Toss coin(part 2)

bool coin_toss(double seed){

double r;

r = (double)seed/(double)RAND_MAX;

//cout<<r;

if (r<0.5) return true;

return false;

}

//check results

float toss_result(vector<char> v,int *h){

int num_heads=0;

int num_toss = v.size();

*h = 0;

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

if(v[i]=='h'){num_heads++;(*h)++;}

}

return (float)num_heads/(float)num_toss;

}

void show_result(int total_num,int head_num,float ratio){

cout<<"it took you "<<total_num<<" tosses to get 3 heads in a row"<<endl;

cout<<"on average you flipped heads "<<(ratio*100)<<"% of the time"<<endl;

}

int main()

{

string name;

char gender;

string K; //Mr or Mrs

cout<<"Welcome to coin toss! Get 3 heads in row to win!"<<endl;

//part 1

user_info(&name,&gender);

if(gender == 'f') K = "Mrs.";

else K = "Mr.";

cout<<"Try to get 3 heads in a row. Good luck "<<K<<name<<"!"<<endl;

char dummy = getchar();

//part 2

int num_toss;

vector<char> toss_results;//store toss results

bool result;//result of toss

srand(time(0));

while(true){

//cout<<rand()<<endl;

result = coin_toss(rand());

cout<<"Press <enter> to flip"<<endl;

while (1)

{

if ('\n' == getchar())

break;

}

if(result) {toss_results.push_back('h');cout<<"HEAD"<<endl;}

else {toss_results.push_back('t');cout<<"TAIL"<<endl;}

num_toss = toss_results.size();

if(num_toss>=3){

if ((toss_results[num_toss-1] == 'h')&&(toss_results[num_toss-2] == 'h')&&(toss_results[num_toss-3] == 'h')){

break;

}

//part 3

float ratio_head;

int num_of_heads;

ratio_head = toss_result(toss_results,&num_of_heads);

//part 4

show_result(toss_results.size(),num_of_heads,ratio_head);

return 0;

}

8 0

3 years ago

Write a program that helps a young student learn to make change. Use the random number generator to generate change amounts betw

erastovalidia [21]

Answer:

#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

int quarters = coins/25;

coins = coins%25;

int dimes = coins/10;

coins = coins%10;

int nickles = coins/5;

coins = coins%5;

int pennies = coins/1;

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

}

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}#include <iostream>

#include <stdlib.h>

using namespace std;

int main()

{

srand(time(0));

int coins = rand()%99 + 1;

int temp = coins;

cout<<"Enter the coins needed to make $0."<<coins<<"?"<<endl<<endl;

int a, b, c, d;

cout<<"Quarters? ";

cin>>a;

cout<<"Dimes? ";

cin>>b;

cout<<"Nickles? ";

cin>>c;

cout<<"Pennies? ";

cin>>d;

int quarters = coins/25;

coins = coins%25;

int dimes = coins/10;

coins = coins%10;

int nickles = coins/5;

coins = coins%5;

int pennies = coins/1;

if(((a*25) + (b*10) + (c*5) + (d*1))==temp)

{

cout<<"That's correct"<<endl;

if(a!=quarters || b!=dimes || c!=nickles || d!=pennies)

{

cout<<"The optimized solution is "<<endl;

}

else

{

cout<<"Not correct"<<endl;

cout<<"The solution is "<<endl;

}

cout<<"\tQuarters: "<<quarters<<endl;

cout<<"\tDimes: "<<dimes<<endl;

cout<<"\tNickles: "<<nickles<<endl;

cout<<"\tPennies: "<<pennies<<endl;

return 0;

}

Explanation:

Use the built-in rand function to generate a random coin.
Get all the required values for quarters, dimes, nickles and pennies and store them in variables.
Check if the coins give a value of temp, then answer is correct and the values of user will be added.
Optimum solution is displayed If it does not matches with the optimum solution. Display that the answer is wrong along with solution.

6 0

2 years ago