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sp2606 [1]
3 years ago
5

A user complains that he cannot access a particular website, although he is able to access other websites. at which layer of the

osi model should you begin troubleshooting this problem
Computers and Technology
1 answer:
Annette [7]3 years ago
5 0
Hi,

If user is unable to access the website, than website requires credentials which he cannot prompt correctly. This often happens when you search for specific IP address on website. For example you search "194.186.0.33" if this is an IP from a device, the website will ask you to enter your username and password. You are unable to troubleshoot this you can only set your own devices and access them through web.

Hope this helps.
r3t40
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What is the output of the code below assuming that global variable x has value 2 and global y has value 3? def f1(): return "ab"
Lostsunrise [7]

Answer:

ababababab

Explanation:

The code above is written in python and python uses indentation .So let me rephrase the code accordingly and explain what the code really do.

Note x and y is a global variable which can be used by any of the function declared.  According to the question x and y are 2 and 3 respectively

The first block of code describes a function f1 without any argument but the code should return the string "ab"

def f1():

      return "ab"

The second block of code defines a function f2 and returns the value of f1 multiply by x. This means you are multiplying the string "ab" by 2 which will be equals to abab

def f2():

           return f1() * x

The third block of code declared a function f3 and returns the sum of  f2 and product of f1 and y. using PEMDAS principle the multiplication aspect will be solved first so, ab × 3 = ababab, then we add it to f2  . ababab + abab = ababababab.

def f3():

        return f2() + f1() * y

Finally, we print the function f3 value to get ababababab

print(f3())

If you run the code on your IDE like below you will get  ababababab

x = 2

y = 3

def f1():

      return "ab"  

def f2():

      return f1() * x  

def f3():  

      return f2() + f1() * y  

print(f3())

     

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