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fgiga [73]
2 years ago
7

The most important thing that users can do to protect their privacy and identity on social sites is to ________. Select one: a.

Read and rely on the privacy policy of the SNS b. Accept responsibility for protecting their own information c. Mark their profiles as confidential d. Use third-party programs
Computers and Technology
1 answer:
Lesechka [4]2 years ago
6 0

Answer:

The answer is "Option b"

Explanation:

The method of monitoring complex social media data, guard toward safety and economic threats is known as the social media protection. In this process, each industry presents a particular set of cultural threats, and some of which, are at both the core of the discussion in the public and certain alternatives were wrong, which can be defined as follows:

  • In option a, It is a platform, that provides the communication services.
  • Option c, and option d both are incorrect, because it can't use a third-party program to protect the data.
You might be interested in
Of 500 sessions that occurred on a website during a one-week period, 200 of them started on the homepage; 100 of these 200 sessi
Elena L [17]

Answer:

the homepage bounce rate

Explanation:

Bounce rate is defined as the percentage of people that visits a page on a website and then exit or leave. That is, the visitors did not view any other page except the first one and they just leave the page after that visit. In order to get the bounce rate, the total number of visitors to a single page is taken and its then divided by the total number of visits to the website. For example, if the total number of visitors to a website over a period of time is 3000, while those that only visited a page on the website is 500, then the bounce rate is

(500/3000) * 100 = 16%

For this question, the homepage recorded 100 visited only once out of 200 which means homepage bounce rate is 50% : (100/200)*100

While website bounce rate is total number of bounces across the website/total visit to the website

100+100 =200 this is the total bounces across the website

200/500 :500 is the total visit to the website

(200/500)*100 = 40 %

Therefore the homepage bounce rate is higher than the site bounce rate

7 0
2 years ago
Using the college registration example from section 6.7.3 as a starting point, do the following:
Arada [10]

Answer:

Explanation:

NCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

jmp L4

L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

8 0
2 years ago
Ben uses a table that has few columns. He knows about a particular value in the first column and wants to find the corresponding
igor_vitrenko [27]
I'd suggest he uses LOOKUP Function.

You can use the LOOKUP function when you need to look in a single column or row and find a value from the same position in a corresponding column or row. We have a much improved VLOOKUP that can also be used to search one row or column or multiple rows and columns.



4 0
2 years ago
Read 2 more answers
Ten 9600-bps lines are to be multiplexed using TDM. a. Ignoring overhead bits in the TDM frame, what is the total capacity requi
lukranit [14]

Answer:

a.

<u>Given Data :</u>

Data rate of each Line (Channel) = 9600 bps

Total lines need to be multiplexed = 10

<u>Solution:</u>

Total Capacity of Synchronous TDM = Data rate of each line x Total Lines

                                                            =  9600 bps x 10

                                                            = 96,000 bps

                                                            = 96 Kbps

Explanation:

In time division multiplexing, multiple channels (Lines) can transmit data over the medium on different time slots. In synchronous TDM, equal time has been allocated to each channel to transmit their data over the medium. so the total capacity required for 10 lines synchronous TDM is 96Kbps.

Answer:

b.

<u>Given Data:</u>

Data rate for each line (Channel) = 9600 bps

Total lines need to be multiplexed = 10

Average TDM link utilization = 80% = 0.8

Each link busy for the time = 50% = 0.5

<u>Solution:</u>

Capacity of Statistical TDM = data rate x number of Lines x each channel duration/Average utilization of TDM

Capacity of Statistical TDM = 9600 bps x 10 x 0.5/0.8

                                               =60,000 bps

                                               = 60 kbps

Explanation:

In TDM, if we want to limit the capacity of TDM and each channel that need to be multiplexed. we can limit the utilization of the TDM by limiting its transmission time of each channel and also the capacity of TDM.

3 0
2 years ago
01000001 01101110 01110011 01110111 01100101 01110010 00100000 01101001 01110011 00100000 00110100 00110010 00101110 There's a w
BARSIC [14]

Answer:

Binary Code

Explanation:

I believe this is Binary code, so if you enter this in Google, click a translator from binary code to text, and it should tell you.

when I did it, it said "Answer is 42"

Hopefully this helped any.

4 0
2 years ago
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