Answer:try redoing everything step by step and see if that solves it
Explanation:
Answer:
Please see the attached file for the complete answer.
Explanation:
C. A misconfiguredaccess control. The fact that Travis is able to access Craig's folder indicates that the access control has been misconfigured, allowing Travis to gain access to datahe is not authorized to view.
This situation indicates that a misconfigured access control has occurred. It appears that Travis has been granted access to Craig's folder, which he should not have access to. This misconfiguration of the access controlsettings may have been an accidental or intentional action taken by someone with the necessary permissions.
Regardless of the cause, this misconfiguration has enabled Travis to access Craig's data, which is a serious security breach and needs to be addressed immediately. The proper security protocols need to be put in place to ensure that only authorized users can access the necessary data and to prevent any further breaches of security.
Since the question isn't complete, here's the full task:
Travis and Craig are both standard users on the network. Each user has a folder on the network server that only they can access. Recently, Travis has been able to access Craig's folder.
This situation indicates which of the following has occurred?
Choose the right option:
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Answer:
a) 0.32%
b) 32%
c) Mbps WAN link is of more benefit
Explanation:
Average SNMP response message = 100 bytes
Avere message for each second when manager sends 400 SNMP
: 400 SNMP per second * 100 bytes = 40000 bytes per second = 40000Bps
To convert byte to bits, We have 1 byte = 8 bits
Therefore, 40000Bps = 40000 * 8 = 320 Kbps
a) Calculating the percentage of a 100 Mbps LAN link’s capacity
100 Mbps = 100000Kbps
320 Kbps is what percent of 100000Kbps = * 100 = 0.32%
So the resulting response traffic would represent 0.32% of a 100 Mbps LAN link’s capacity.
b) Calculating the percentage of a 1Mbps LAN link’s capacity
1Mbps = 1000Kbps
320 Kbps is what percent of 1000Kbps = * 100 = 32%
So the resulting response messages would represent 32% of a 1Mbps LAN link’s capacity.
c) When we are using the 1 Mbps WAN link, we use 32% of its speed to response the message as opposed to 100 Mbps LAN link that uses just 0.32%. This means the 1 Mbps WAN link uses more bandwith than the 100 Mbps LAN link. Therefore the management implication is that it is better to use the 1 Mbps WAN link has it has more benefits.
Answer:
The execution resumes in the finally block if one exists or otherwise from the next statement following the try...catch block.
Explanation:
Once an exception has been thrown and caught in the code, the execution continues with the statements in the finally block if one exists. If there is no finally block defined then execution resumes from the next statement following the try... catch block. For example:
try{
//An exception is raised
}
catch (Exception e){
//Exception is handled
}
System.out.println("After try...catch");
In this code segment, the next statement to be executed after catch is the System.out.println();
instructions:
You need to write code that will print two bricks of numbers, one with integers, one with decimals.
import random
i = 1
while i <= 100:
print("#"+str(i)+": "+str(random.randint(1,100)), end=", ")
i+=1
print()
i = 1
while i <= 100:
print("#"+str(i)+": "+str(random.uniform(1,100)), end=", ")
i += 1
I hope this helps!
(b) Selection-sort on a sequence of size n.
(c) Merge-sort on a sequence of size n.
(d) Radix sort on a sequence of n integer keys, each in the range of[ 0, (n^3) -1]
(e) Find an element in a red-black tree that has n distinct keys.
Answer:
Answers explained below
Explanation:
(a) Construction of a heap of size n , where the keys are not known in advance.
Worst Case Time complexity - O(n log n)
Two procedures - build heap, heapify
Build_heap takes O(n) time and heapify takes O(log n) time. Every time when an element is inserted into the heap, it calls heapify procedure.
=> O(n log n)
(b) Selection-sort on a sequence of size n.
Worst Case Time complexity - O(n^2)
Selection sort finds smallest element in an array repeatedly. So in every iteration it picks the minimum element by comparing it with the other unsorted elements in the array.
=> O(n^2)
(c) Merge-sort on a sequence of size n.
Worst Case Time complexity - O(n log n)
Merge sort has two parts - divide and conquer. First the array is divided (takes O(1) time) into two halves and recursively each half is sorted (takes O(log n) time). Then both halves are combines (takes O(n) time).
=> O(n log n)
(d) Radix sort on a sequence of n integer keys, each in the range of[ 0 , (n^3) -1]
Worst Case Time complexity - O (n log b (a))
b - base of the number system, a - largest number in that range, n - elements in array
Radix sort is based on the number of digits present in an element of an array A. If it has 'd' digits, then it'll loop d times.
(e) Find an element in a red-black tree that has n distinct keys.
Worst Case Time complexity - O (log n)
Red-black tree is a self-balancing binary tree => The time taken to insert, delete, search an element in this tree will always be with respect to its height.
=> O(log n)