Which of the following is a factor that could cause an individual to have excited delirium syndrome?
Choose only ONE best answer.
PMS
Ulcer
Nose bleed
D
Respiratory problems​

Answers

Answer 1
Excited delirium occurs most commonly in males with a history of serious mental illness or acute or chronic substance use disorder, particularly stimulant drugs such as cocaine and MDPV. Alcohol withdrawal or head trauma may also contribute to the condition
Answer 2

Respiratory problems​ is a factor could cause an individual to have excited

delirium syndrome.

People with excited delirium syndrome exhibit unique characteristics such as

Paranoia HallucinationViolenceDistress

This is usually as a result of  the use of some drugs such as cocaine and

methamphetamine which decreases the effect of dopamine modulating

respiratory functions thereby resulting to asphyxia and cardiac arrest.

Read more about Syndromes here https://brainly.com/question/24825576


Related Questions

PLEASE HELP ME ASAP

Answers

The answer is B; network security. If there is an issue with the software corruption, there is an issue with the security and validity of the website. A website encryption could prevent this from happening or even a simple virus cleaner.

The Internet is written in a "language" called
O Usenet
O package switching
O HTTP
O ARPANET​

Answers

Answer:

HTTP

Explanation:

I hope it helps u.

Mark Me as brainlist answer

HTTP

Have a great day! I hope this helped!

Create a game that rolls two dies (number from 1 to 6 on the side) sequentially for 10 times (use loop). If at least once out of 10 times the sum of two random numbers is equal to 10, you win, else you loose. You will need to get your own method getRandom(int n) that generates a random number, see below (you can modify it as needed), a loop, and IF statement in the loop that checks if a sum of two random numbers is 10 or not.
Please start your program as follows, similar to what we did in class:
import java.util.Random;
public class UseRandom{
//your method getRandom() is here below, n is a range for a random number from 0 to n
public int getRandom(int n)
{
Random r=new Random();
int rand=r.nextInt(n);
return rand;
}
//code continues here, don't forget your main() method inside the class, and making your own object in main() using "new" keyword.

Answers

Answer:

Explanation:

The following code is written in Java and loops through 10 times. Each time generating 2 random dice rolls. If the sum is 10 it breaks the loop and outputs a "You Win" statement. Otherwise, it outputs "You Lose"

import java.util.Random;

class Brainly {

   public static void main(String[] args) {

       UseRandom useRandom = new UseRandom();

       boolean youWin = false;

       for (int x = 0; x<10; x++) {

           int num1 = useRandom.getRandom(6);

           int num2 = useRandom.getRandom(6);

           if ((num1 + num2) == 10) {

               System.out.println("Number 1: " + num1);

               System.out.println("Number 2: " + num2);

               System.out.println("You Win");

               youWin = true;

               break;

           }

       }

       if (youWin == false) {

           System.out.println("You Lose");

       }

   }

}

class UseRandom{

   public int getRandom(int n)

   {

       Random r=new Random();

       int rand=r.nextInt(n);

       return rand;

   }}

Write a Java program named Problem 3 that prompts the user to enter two integers, a start value and end value ( you may assume that the start value is less than the end value). As output, the program is to display the odd values from the start value to the end value. For example, if the user enters 2 and 14, the output would be 3, 5, 7, 9, 11, 13 and if the user enters 14 and 3, the output would be 3, 5, 7, 9, 11, 13.

Answers

Answer:

hope this helps

Explanation:

import java.util.Scanner;

public class Problem3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.print("Enter start value: ");

       int start = in.nextInt();

       System.out.print("Enter end value: ");

       int end = in.nextInt();

       if (start > end) {

           int temp = start;

           start = end;

           end = temp;

       }

       for (int i = start; i <= end; i++) {

           if (i % 2 == 1) {

               System.out.print(i);

               if (i == end || i + 1 == end) {

                   System.out.println();

               } else {

                   System.out.print(", ");

               }

           }

       }

   }

}

What is the problem with paying only your minimum credit card balance each month?
A. It lowers your credit score
B. You have to pay interest
C. The bank will cancel your credit card
D. All of the above

Answers

Answer:b

Explanation:

Answer: The answer is B.

Explanation: While yes it is important to make at least the minimum payment, its not ideal to carry a balance from month to month (You'll rack up interest charges). And risk falling into debt. Therefore, the answer is B.  

Hope this helps!

In numerical methods, one source of error occurs when we use an approximation for a mathematical expression that would otherwise be too costly to compute in terms of run-time or memory resources. One routine example is the approximation of infinite series by a finite series that mostly captures the important behavior of the infinite series.
In this problem you will implement an approximation to the exp(x) as represented by the following infinite series,
exp(x) = Infinity n= 0 xn/xn!
Your approximation will be a truncated finite series with N + 1 terms,
exp(x, N) = Infinityn n = 0 xn/xn!
For the first part of this problem, you are given a random real number x and will investigate how well a finite series expansion for exp(x) approximates the infinite series.
Compute exp(x) using a finite series approximation with N E [0, 9] N (i.e. is an integer).
Save the 10 floating point values from your approximation in a numpy array named exp_approx. exp_approx should be of shape (10,) and should be ordered with increasing N (i.e. the first entry of exp_approx should correspond to exp(x, N) when N = 0 and the last entry when N = 9).

Answers

Answer:

The function in Python is as follows:

import math

import numpy as np    

def exp(x):

   mylist = []

   for n in range(10):

       num = (x**n)/(math.factorial(n))

       mylist.append([num])

   exp_approx = np.asarray(mylist)

   sum = 0

   for num in exp_approx:

       sum+=num

   return sum

Explanation:

The imports the python math module

import math

The imports the python numpy module

import numpy as np    

The function begins here

def exp(x):

This creates an empty list

   mylist = []

This iterates from 0 to 9

   for n in range(10):

This calculates each term of the series

       num = (x**n)/(math.factorial(n))

This appends the term to list, mylist

       mylist.append([num])

This appends all elements of mylist to numpy array, exp_approx

   exp_approx = np.asarray(mylist)

This initializes the sum of the series to 0

   sum = 0

This iterates through exp_approx

   for num in exp_approx:

This adds all terms of the series

       sum+=num

This returns the calculated sum

   return sum

Which of the following is NOT an example of a metasearch engine?

Answers

Answer:

Where is the choices mate?

c Write a program that simulates a magic square using 3 one dimensional parallel arrays of integer type. Each one the arrays corresponds to a row of the magic square. The program asks the user to enter the values of the magic square row by row and informs the user if the grid is a magic square or not. flowchart

Answers

Answer:

hope this helps and do consider giving a brainliest to the ans if it helped.

Explanation:

//program to check if the entered grid is magic square or not

/**c++ standard libraries

*/

#include<bits/stdc++.h>

using namespace std;

/**function to check whether the entered grid is magic square or not

*/

int isMagicSquare(int arr[3][3]){

  int i,j,sum=0,sum1=0,rsum,csum;

  for(i=0;i<3;i++){

      sum+=arr[i][i];

      sum1+=arr[i][2-i];

  }

  if(sum!=sum1){

      return 0;

  }

 

  for(i=0;i<3;i++){

      rsum=0;

      csum=0;

      for(j=0;j<3;j++){

          rsum+=arr[i][j];

          csum+=arr[j][i];

      }

      if(sum!=rsum){

          return 0;

      }

      if(sum!=csum){

          return 0;

      }

  }

  return 1;

}

/** main function to get user entries and

* call function

* and print output

*/

int main(){

  int i,j,arr[3][3]={0};

  for(i=0;i<3;i++){

      for(j=0;j<3;j++){

          cout<<"Enter the number for row "<<i<<" and column "<<j<<" : ";

          cin>>arr[i][j];

      }

  }

  int ret = isMagicSquare(arr);

  if(ret==1){

      cout<<"This is a Lo Shu magic square"<<endl;

  }

  else{

      cout<<"This is not a Lo Shu magic square"<<endl;

  }

  return 0;

}

How I did it. Part 4

Answers

Answer:

What are you asking for in this problem?

Explanation:

Your assignment is to write an assembly language program which read a string and print it in uppercase. This program asks the user to enter a string which is subsequently displayed in uppercase. It is important to first ensure that string to be converted is in the a-z range. The program does not recognize any space, symbols or any kind of punctuation marks. Any time the user enters any of this character the program is going to repeatedly ask for the valid string.

An example execution of your program could be:
Please enter your string: 1winter
Invalid Entry!
Please enter your string: summer
Your capitalized string:SUMMER

Answers

Answer:

try this

Explanation:

MSG1 DB 10,13,'enter any string:-$'

MSG2 DB 10,13,'converted string is:-$'

P1 LABEL BYTE

M1 DB 0FFH

L1 DB?

P11 DB 0FFH DUP<'$'>

DATA ENDS

DISPLAY MARCO MSG

MOV AH,9

LEA DX,MSG

INT 21H

ENDM

CODE SEGMENT

ASSUME CS: CODE,DS:DATA

START:

MOV AX,DATA

MOV DS,AX

DISPLAY MSG1

LEA DX,P1

MOV AH,0AH

INT 21H

DISPLAY MSG2

LEA SI,P1

MOV CL,L1

MOV CH,0

CHECK:

CMP[SI],61H

JB DONE

CMP[SI],5BH

UPR: SUB[SI],20H

DONE: INC SI

LOOP CHECK

DISPLAY P11

MOV AH,4CH

INT 21H

CODE ENDS

END START

In c please
Counting the character occurrences in a file
For this task you are asked to write a program that will open a file called “story.txt
and count the number of occurrences of each letter from the alphabet in this file.
At the end your program will output the following report:
Number of occurrences for the alphabets:
a was used-times.
b was used - times.
c was used - times .... ...and so, on
Assume the file contains only lower-case letters and for simplicity just a single
paragraph. Your program should keep a counter associated with each letter of the
alphabet (26 counters) [Hint: Use array|
| Your program should also print a histogram of characters count by adding
a new function print Histogram (int counters []). This function receives the
counters from the previous task and instead of printing the number of times each
character was used, prints a histogram of the counters. An example histogram for
three letters is shown below) [Hint: Use the extended asci character 254]:

Answers

Answer:

#include <stdio.h>

#include <ctype.h>

void printHistogram(int counters[]) {

   int largest = 0;

   int row,i;

   for (i = 0; i < 26; i++) {

       if (counters[i] > largest) {

           largest = counters[i];

       }

   }

   for (row = largest; row > 0; row--) {

       for (i = 0; i < 26; i++) {

           if (counters[i] >= row) {

               putchar(254);

           }

           else {

               putchar(32);

           }

           putchar(32);

       }

       putchar('\n');

   }

   for (i = 0; i < 26; i++) {

       putchar('a' + i);

       putchar(32);

   }

}

int main() {

   int counters[26] = { 0 };

   int i;

   char c;

   FILE* f;

   fopen_s(&f, "story.txt", "r");

   while (!feof(f)) {

       c = tolower(fgetc(f));

       if (c >= 'a' && c <= 'z') {

           counters[c-'a']++;

       }

   }

   for (i = 0; i < 26; i++) {

       printf("%c was used %d times.\n", 'a'+i, counters[i]);

   }

   printf("\nHere is a histogram:\n");

   printHistogram(counters);

}

Plz hurry it’s timed

Answers

Micah and adayln hope this helps u
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