Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
Answer:
Answer is in the provided screenshot!
Explanation:
Steps required to solve this problem:
1 - define what characters are "vowels" by assigning them to an array.
2 - create a variable to record the amount of vowels there are in the string.
3 - convert the input string into a character array and iterate through each character
4 - for each of the character of the string we go through, check if it matches any of the vowels
5 - return true if the vowel count is greater than 1
Alternative methods using the Java Stream API have also been wrote, please respond if you require them.
Answer:
myDoubles.add(75.6);
Explanation:
ArrayList<Double> myDoubles = new ArrayList<Double>();
myDoubles.add(10.8);
myDoubles.add(92.46);
myDoubles.add(75.6);
The above code creates a double ArrayList called myDoubles. We add 10.8 and 92.46 initially. After these numbers, 75.6 is added to the myDoubles using <em>add</em> method (You need to type the list name, ".", and the method name to add the number)
Answer:
creating vector:
code:
to20 <-c(1:20) #creating to20 vector
end <-c(40:50) #creating end vector
to50 <-c(to20,21:30,end) # creating to50 vector
print(to50) # printing to50 vector
Explanation:
