When it comes to the best possible security for your wireless router, be sure to use WEP encryption to ensure that your transmissions are unreadable by hackers. a. True.
b. False.

Answers

Answer 1
Answer:

Answer:

The correct answer to the following question will be "False".

Explanation:

  • Data equivalent protection for Wired Equivalent Privacy (WEP) is an older authentication and data encryption mode that was replaced with better authentication methods for WPA-PSK and WPA2-PSK and data encryption.
  • Only when you select Up to 54 Mbps from the Mode menu will the WEP option appear.
  • WEP can easily be decoded that is why it is not good for secure transmissions.

Therefore, it's the right answer


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COLLEGE

This is a form of load balancing where larger workloads are issued to IT resources with higher processing capacities a. Pay-Per-Use Monitor

b. Asymmetric Distribution

c. SLA Monitor

d. Workload Prioritization

Answers

Answer:

The correct answer is b. Asymmetric Distribution.

Explanation:

Asymmetric Distribution has to do with those larger workloads which are issued to IT resources with higher processing capacities. The Pay-Per-Use Monitor is where the billing system relies on. The Workload Prioritization is where workloads are prioritized according to their level. And the SLA monitor is a contract between a service provider and the customer. So, the most correct answer is b.Asymmetric Distribution.

HIGH SCHOOL

The taskbar can display a maximum of ten items. true, false

Answers

False I know for a fact Windows 10 can display more because the bar is smaller. Could be ten on others tho not sure.

False is your answer.

HIGH SCHOOL

33. Universal containers need to add an additional recipient to a workflow email alert that is fired from the case object. What type of field could be added to the case object to allow the additional desired recipient on the email alert? (A) Text field
(B) Email field
(C) Formula field
(D) Lookup field

Answers

Email field hope I help u
COLLEGE

The code below returns the number of zeros at the end of n! [factorial n]int zeros(int n){int res = 0;while (n!=0){res += n/5;n /= 5;}return res;}Rewrite this method recursively:6. Write a recursive function that returns the product of the digits of its integer input parameter, n. You omay assume that n is non-negative. For example, productDigits(243) should return 24, since 2 x 4 x 3 = 24.int productDigits (int n)

Answers

Answer to 1:

The recursive method is as follows:

int zeros(int n) {

if (n == 0) return 0;

return n/5 + zeros(n/5);  }

Explanation:

In the above code, the zeros() function is a recursive function as it is calling itself. When this function executes, it calls itself. Lets see how this method works.

Suppose n = 10

first it checks if n is equals to 0. As n is 10 so the program flow moves to the next statement:  return n/5 + zeros(n/5); which calls zeros() function again. This recursive function will keep calling itself until the base condition evaluates to true.

Here the base case is when the value of n becomes 0 and the recursive case is the n/5 + zeros(n/5) which keeps calling itself (zeros function) recursively.

Now lets see what  return n/5 + zeros(n/5); statement does.

At first recursive call  

10/5 + zeros(10/5)

2 + zeros(2)

Now here the zeros function called itself again so:

zeros(2) the if condition checks if 2 == 0 which is false as 2>0 So return n/5 + zeros(n/5); statement executes again

2/5 + zeros(2/5)

This basically takes this form:

2+ [2/5 + zeros(2/5)]

2 + [0 + zeros(0)]

The answer to 2/5 is 0.4 but here we are using int for n so the rounded number is 0.

Now zeros(0) will again call zeros() function. Now n = 0 So if condition checks if n==0 which is true. The base condition evaluates to true means the recursion will stop now.

Now what is returned in output finally?

2 + [0] = 2

So this recursive function returns 2.

If you want to see the results of this function in output screen you can write this main() function which takes a number in input and calls zeros() function.

int main()

{int num;

cin>>num;

cout

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