Answer:
D. Low-level components are combined into clusters that perform a specific software sub-function
Explanation:
Normally, regression testing are done on a manual basis by simply re-executing a subset of the entire available test cases or it can be done automatically by utilizing playback tools or automated capture. Hence, from the options, a combination of low-level components into clusters that perform a specific sub-function does not align with how regression testing are done either manually or automatically.
Answer:
/*
Find Largest and Smallest Number in an Array Example
This Java Example shows how to find largest and smallest number in an
array.
*/
public class FindLargestSmallestNumber {
public static void main(String[] args) {
//array of 10 numbers
int numbers[] = new int[]{32,43,53,54,32,65,63,98,43,23};
//assign first element of an array to largest and smallest
int smallest = numbers[0];
int largetst = numbers[0];
for(int i=1; i< numbers.length; i++)
{
if(numbers[i] > largetst)
largetst = numbers[i];
else if (numbers[i] < smallest)
smallest = numbers[i];
}
System.out.println("Largest Number is : " + largetst);
System.out.println("Smallest Number is : " + smallest);
}
}
/*
Output of this program would be
Largest Number is : 98
Smallest Number is : 23
*/
Explanation:
Answer:
function out=circular_primes(no)
prim=primes(no);% find the all prime number till the given number
pr=0;
nos=[];
po=[];
for p=1:length(prim)
n=prim(p); % picking up each prime no one by one
n=num2str(n);% change into string for rotation of no
d=length(n); % length of string
if d>1 % take nos greater than 10 beacuase below 10 no need for rotation
for h=1:d
a=n(1);
for r=1:d % for rotation right to left
if r==d 5 % for the last element of string
n(d)=a;
else
n(r)=n(r+1); %shifting
end
end
s=str2num(n); % string to number
nos=[nos,s]; % store rotated elements in array
end
if nos(end)==no %if given no is also a circular prime we need smaller
break;
end
for gr=1:length(nos) % checking rotated nos are prime or not
p1=isprime(nos(gr));
po=[po,p1]; %storing logical result in array
end
if sum(po(:))==length(nos) %if all are prime the length and sum are must be equal
pr=pr+1;
out=pr;
else
out=pr;
end
po=[];
nos=[];
else
s=str2num(n); %numbers less than 10
f=isprime(s);
if f==1
pr=pr+1;
out=pr;
else
out=pr;
end
end
end
end
Explanation:
Answer:
A cardboard box on the edge of your lane
Explanation:
