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3 years ago

Which of the following Wi-Fi chalking method refers to drawing symbols in public places to advertise open Wi-Fi networks?

Computers and Technology

2 answers:

valina [46]3 years ago

7 0

Answer:

War Chalking is the correct answer

dezoksy [38]3 years ago

3 0

Answer War Chalking

Explanation: War chalking is the technique for presenting the WiFi network publicly. This signifies about the WiFi is present but not describes the manner of it , that is whether it is open connection or closed connection.

It can be used by the people in general as well as hackers . Hackers usually tend to attack and hack its security that is present.Hackers then use the WiFi network is then used for their own work after being hacked.

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Answer:

Explanation:

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3 years ago

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2 years ago

The computer-like model used to describe the way humans encode, store, and retrieve information is the ________ model.

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2 years ago

Exercises: Solve the following problems on paper and bring your sheet of paper to your section on Thursday: Consider the followi

Volgvan

Answer:

Hi Lzvcorri! Based on the inputs, we can expect the answers:

a. [4, 4]

b. [3, 6, 6]

c. [7, 8, 8, 4]

d. [5, 7, 7, 4, 4]

e. [4, 6, 6, 7, 9, 9]

Explanation:

For a, the input is {2,4} so the array length is 2, and the loop will run for array.length - 1. So the first run with this array will see if array[i] (i=0 as initially set in the for loop with the statement "for (i=0;)") is less than the value in the array after it, array[i+1]. Since 2, is less than 4, it will assign the value of 4 in the first array index as per this condition being met (array[i] = array[i + 1];). Similarly, if you trace the execution of the program with the remaining inputs b, c, d, and e, you should expect the resultant array as in the answer above.

4 0

3 years ago

(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-

I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit memory to add in 2 location:

$\to \frac{0}{1} =2^1 \ (\frac{m}{m} \ location)$

The address of 2-bit memory to add in 4 location:

$\to \frac{\frac{00}{01}}{\frac{10}{11}} =2^2 \ (\frac{m}{m} \ location)$

similarly,

Complete 'n'-bit memory address' location number is = $2^n.$ Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory: