Answer: "Natural barriers that can separate species group members are volcanos, and mountain ranges. Other examples of natural barriers include rivers, lakes, and other bodies of water; cliffs and other types of terrain that are difficult to traverse; and areas dense with certain types of plant life (e.g., blackberry bushes)."
Hope it helps <3
The Internet is written in a "language" called
O Usenet
O package switching
O HTTP
O ARPANET
Answer:
HTTP
Explanation:
I hope it helps u.
Mark Me as brainlist answer
How I did it. Part 4
Answer:
What are you asking for in this problem?
Explanation:
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]:
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);
}
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.
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;
}}
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
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;
}
Which of the following is NOT an example of a metasearch engine?
Answer:
Where is the choices mate?
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).
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
PLEASE HELP ME ASAP
Does anybody have the full answer sheet for Cisco Packet Tracer 11.10.1 (Design and implement a vlsm addressing scheme)? Will give brainliest!
Answer:
Red font color or Gray highlights indicate text that appears ... In this lab, you will design a VLSM addressing scheme given a network ... You have been asked to design, implement, and test addressing ...
Explanation: