what privacy risks do new technologies present,
and how do we decide if they're worth it?

Answer:

AZAX is key sentence for without reloading page.

Explanation:

AJAX is also known as XHR. If some one wants to send the file without reloading then look ahead at the search function and prompt the users by auto saving the documents. This request is sent to the server. The server return the data to your network. Whenever you have seen the file form, some action on the page. You will see some request from AJAX server request on your functional area. AJAX is called a reloading file without page. AZAX is stands for Asynchronous java scripts And XML. AJAX is used to send an email. AZAX is not used in front end, it is used from back end process.

Answer:

An operating system (OS) is a software which is responsible for the management of computer hardware, software, and also provides common services for computer programs.

Operating System is pivotal in teaching and learning because:

1. It enables computer programs to run smoothly on various computer devices.

2. The Operating System gives teachers the opportunity to install learning apps on their devices for ease of teaching.

3. It enables students to download and install learning applications, download and upload assignments, etc.

4. The Operating System makes video conferencing for online tuition easy and smooth.

5. It makes computer users to interact with other applications and softwares within a device.

Operating systems are found on many computer devices e.g: mobile phones, video games, PCs, supercomputers, etc.

Answer:

Option A

Availability

Explanation:

The implementation of too many security protocols will lead to a reduction of the ease at which a piece of information is accessible.  Accessing the piece of information will become hard even for legitimate users.

The security protocols used should not be few, however, they should be just adequate to maintain the necessary level of confidentiality and integrity that the piece of information should have, While ensuring that the legitimate users can still access it without much difficulty.

Answer:

Using the terminal in linux OS to configure a static ip address, gateway and subnet mask.

Explanation:

-Enter the terminal in the linux environment and use the ifconfig eth0 or -a to bring up the main network interface and other network interfaces.

- for static network configuration, use;

  - ifconfig eth0 10.254.1.42  

  - ifconfig eth0 netmask 255.255.0.0

  - ifconfig eth0 broadcast 10.254.255.255

- and add a default gateway with;

 -  route add default gw 10.254.0.1  eth0.

- Now verify the settings with the ifconfig eth0 command.

Answer:

c. Cartography.

Explanation:

Cartography is the science and art of making an illustrated map or chart. Geographic information system (GIS) allows users to interpret, analyze, and visualize data in different ways that reveal patterns and trends in the form of reports, charts, and maps.

Basically, cartography is the science and art of depicting a geographical area graphically, mostly on flat surfaces or media like maps or charts. It is an ancient art that was peculiar to the fishing and hunting geographical regions. Geographic information system is an improved and technological form of cartography used for performing a whole lot of activities or functions on data generated from different locations of the Earth.

Answer:

C. Cartography

Explanation:

Answer:

You'll need dual monitor cables and an adapter.

Explanation:

First Step

Position your monitors on your desk or workspace. Make sure the systems are off.

Second Step

Make sure your power strip is close by. Then plug your power strip and connect the first monitor to your PC via your HDMI

Use an adapter to do the same for the second monitor.

Turn the entire system on

Third Step

On your PC, right click on a blank place in your home screen and click on Display Settings.

If you want to have two separate displays showing the same thing, select Duplicate, but if you want to have the two displays independent of each other, select Extend Display.

Apply the settings and select Done.

Answer:

Here is the Python program:

import keyword  #module that contains list of keywords of python

filename = input("Enter Python source code filename: ") # prompts user to enter the filename of a source code

code = open(filename, "r") # opens the file in read mode

keywords = keyword.kwlist #extract list of all keywords of Python and stored it into keywords

dictionary = dict() #creates a dictionary to store each keyword and its number of occurrence in source code

for statement in code: # iterates through each line of the source code in the file

   statement = statement.strip() # removes the spaces in the statement of source code  

   words = statement.split(" ") #break each statement of the source code into a list of words by empty space separator

   for word in words:# iterates through each word/item of the words list  

       if word in keywords:#checks if word in the code is present in the keywords list of Python  

           if word in dictionary: #checks if word is already present in the dictionary

               dictionary[word] = dictionary[word] + 1 #if word is present in dictionary add one to the count of the existing word

           else: #if word is not already present in the dictionary  

               dictionary[word] = 1 #add the word to the dictionary and set the count of word to 1

for key in list(dictionary.keys()): #iterates through each word in the list of all keys in dictionary  

   print(key, ":", dictionary[key])# prints keyword: occurrences in key:value format of dict

Explanation:

The program is well explained in the comments attached with each line of the program.  

The program prompts the user to enter the name of the file that contains the Python source code. Then the file is opened in read mode using open() method.

Then the keyword.kwlist statement contains the list of all keywords of Python. These are stored in keywords.

Then a dictionary is created which is used to store the words from the source code that are the keywords along with their number of occurrences in the file.

Then source code is split into the lines (statements) and the first for loop iterates through each line and removes the spaces in the statement of source code .

Then the lines are split into a list of words using split() method. The second for loop is used to iterate through each word in the list of words of the source code. Now each word is matched with the list of keywords of Python that is stored in keywords. If a word in the source code of the file is present in the keywords then that word is added to the dictionary and the count of that word is set to 1. If the word is already present in the dictionary. For example if there are 3 "import" keywords in the source code and if 1 of the import keywords is already in the dictionary. So when the second import keyword is found, then the count of that keyword is increased by 1 so that becomes 2.

Then the last loop is used to print each word of the Python that is a keyword along with its number of occurrences in the file.

The program and its output is attached in a screenshot. I have used this program as source code file.

Answer:

In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.

Explanation:

In main() function three integer type variables are declared: a,b,c

In fun1() three int type variables are declared/defined: b,c,d

In fun2() three int type variables are declared/defined: c,d,e

In fun3() three int type variables are declared/defined: d,e,f

a. Main calls fun1; fun1 calls fun2; fun2 calls fun3

Here the main() calls fun1() which calls fun2() and fun2() calls func3() . This means first the func3() executes, then fun2(), then fun1() and last main()

Visible Variable:  d, e, f        Defined in: fun3

Visible Variable: c                 Defined in: fun2 (the variables d and e of fun2  

                                                                                                     are not visible)

Visible Variable: b                  Defined in: fun1 ( c and d of func1 are hidden)

Visible Variable: a                 Defined in: main (b,c are hidden)

b. Main calls fun1; fun1 calls fun3

Here the main() calls fun1, fun1 calls fun3. This means the body of fun3 executes first, then of fun1 and then in last, of main()

Visible Variable: d, e, f           Defined in: fun3

Visible Variable:  b, c              Defined in: fun1 (d not visible)

Visible Variable:  a                  Defined in: main ( b and c not visible)

c. Main calls fun2; fun2 calls fun3; fun3 calls fun1

Here the main() calls fun2, fun2 calls fun3 and fun3 calls fun1. This means the body of fun1 executes first, then of fun3, then fun2 and in last, of main()

Visible Variable:  b, c, d        Defined in: fun1

Visible Variable:  e, f             Defined in: fun3 ( d not visible)

Visible Variable:  a                Defined in: main ( b and c not visible)

Here variables c, d and e of fun2 are not visible

d. Main calls fun3; fun3 calls fun1

Here the main() calls fun3, fun3 calls fun1. This means the body of fun1 executes first, then of fun3 and then in last, of main()

Visible Variable: b, c, d     Defined in: fun1  

Visible Variable:   e, f        Defined in:  fun3   ( d not visible )

Visible Variable:    a          Defined in: main (b and c not visible)

e. Main calls fun1; fun1 calls fun3; fun3 calls fun2

Here the main() calls fun1, fun1 calls fun3 and fun3 calls fun2. This means the body of fun2 executes first, then of fun3, then of fun1 and then in last, of main()

Visible Variable: c, d, e        Defined in: fun2

Visible Variable:  f               Defined in: fun3 ( d and e not visible)

Visible Variable:  b               Defined in:  fun1 ( c and d not visible)

Visible Variable: a                Defined in: main ( b and c not visible)

f. Main calls fun3; fun3 calls fun2; fun2 calls fun1

Here the main() calls fun3, fun3 calls fun2 and fun2 calls fun1. This means the body of fun1 executes first, then of fun2, then of fun3 and then in last, of main()

Visible Variable: b, c, d       Defined in: fun1  

Visible Variable: e               Defined in: fun2  

Visible Variable: f                Defined in: fun3  

Visible Variable: a               Defined in: main

Answer:

A. UC ONE Self-Service Center

Explanation:

The UC ONE Self-Service Center is an online platform where one can get detailed information about registration, classes, finances, and other personal details. This is also the portal where one can check class schedule, bill payment, viewing available courses, checking final grades, book ordering, etc.

it gives students all the convenience required for effective learning experience.

The UC ONE platform is a platform found in the portal of University of the Cumberland.

Answer:

B. UDP 67

C. UDP 68

Explanation:

In this scenario, after a new firewall is installed, users report that they do not have connectivity to the Internet. The output of the ipconfig command shows an IP address of 169.254.0.101. The ports that would need to be opened on the firewall to allow the users to obtain an IP address are both the UDP 67 and UDP 68. UDP is an acronym for user datagram protocol in computer networking and it is part of the transmission control protocol/internet protocol (TCP/IP) suite.

Generally, the standard Internet communications protocols which allow digital computers to transfer (prepare and forward) data over long distances is the TCP/IP suite.

Also, the UDP 67 and 68 ports is responsible for the connectionless framework which is typically being used by the dynamic host configuration protocol (DHCP) that is used to assign IP addresses to various computer users and manage their leases.

UDP 67 is the destination port for a DHCP server while the UDP 68 is the port number for the computer user (client).

Hence, for the users to have access to the internet or internet connectivity both UDP 67 and 68 must be opened on the firewall.

Answer:

The technician should either provide a route from the user's IP scheme to the printer or change the user's IP scheme to be within the same scheme as the printer.

Explanation:

The most likely reason the user is unable to print to the printer is that the user's computer cannot path the current address to the address of the printer.  Without the printer being on the same subnet as the user's computer, the computer has no way of being able to tell where the printer is.

The two solutions to this are to either add a route to the user's computer explicitly telling the user's computer where to trace to find the printer or to change the user's computer to match the IP scheme of the printer.

Cheers.

Answer:

2 correct answer out of 3

Answer:

When students have completed the assignment, they simply click the Mark As Done button to let the teacher know they have finished.

Explanation: Note: The teacher does NOT receive an alert or email notification when work has been turned in, or marked as done.

Answer:

a. There are improvements in security and high availability during outage

Explanation:

Virtualization occurs when data that could be in several formats (for example, storage devices) are made to appear real through a software. There are virtualization providers who act as third-party between users and the original manufacturers of the software. While virtualization has received wide popularity in recent times because of the several advantages it offers of which cost-saving is included, it also has some disadvantages. One of them from the options provided is the possibility of a security breach and its uncertain availability.

The security breach arises because of the proliferation of information on the virtual space. This information can be accessed by unauthorized hackers. Also since third-party providers are the usual providers of virtualization services, the availability depends on them because if they shut down, users can no longer access the software.

Answer:

Are characterized by radical novelty

Explanation:

Example, intelligent enterprises are reimaging and reinventing the way they do bussines

Answer:

1.024x10^-5

Explanation:

To calculate the transmission delay bytes for the token,

we have the token to be = 64bytes and 10mbps rate.

The transmission delay = 64bytes/10mbps

= 51.2 microseconds

A microsecond is a millionth of a second.

= 5.12 x 10^-5

The question says the average number of stations that the token will pass through is 20. Remember this value is gotten from 40/2

20 x 5.12 x 10^-5

= 0.001024

= 1.024x10^-5

Therefore on an idle network with 40 stations, the average wait is

= 1.024x10^-5

Answer:

SELECT COUNT (DISTICT constraint_name)

FROM apd_schema.constraint_column_usage

RUN

Explanation:

General syntax to return a catalog view of information about foreign keys.

SELECT DISTINCT PARENT_TABLE =

           RIGHT(Replace(DC.constraint_name, 'fkeys_', ''),

           Len(Replace(DC.constraint_name, 'fkeys_', '')) - Charindex('_', Replace(DC.constraint_name, 'fkeys_', ''))),

           CHILD_TABLE = DC.table_name,

           CCU.column_name,

           DC.constraint_name,

           DC.constraint_type

FROM apd_schema.table_constraints DC

INNER JOIN apd_schema.constraint_column_usage CCU

           ON DC.constraint_name = CCU.constraint_name

WHERE  DC.constraint_type LIKE '%foreign'

           OR DC.constraint_type LIKE '%foreign%'

           OR DC.constraint_type LIKE 'foreign%'

RUN

Answer:

The programming language is not stated;

However, the program written in Python is as follows

def solveEquation(x,y,z):

     result = z - y + 2 * x

     print(result)

x = float(input("x = "))

y = float(input("y = "))

z = float(input("z = "))

print(solveEquation(x,y,z))

Explanation:

This line defines the function solveEquation

def solveEquation(x,y,z):

This line calculates the expression in the question

     result = z - y + 2 * x

This line returns the result of the above expression

     print(result)

The next three lines prompts user for x, y and z

x = float(input("x = "))

y = float(input("y = "))

z = float(input("z = "))

This line prints the result of the expression

print(solveEquation(x,y,z))

Answer:

The answer is "check the worksheet is not read only"

Explanation:

The read only mode is used for read the file data, and it doesn't allows the user to update the file, and for updating the worksheet we should check iut does not open in the read-only mode.

If it is open, then we close it and for close we goto the office button and click on the tools option after that goto general setting, in this there is a check box for turn off the read-only mode.

Answer:

Following are the algorithm to this question:

Explanation:

In the RSA algorithm can be defined as follows:  

In this algorithm, we select two separate prime numbers that are the "P and Q", To protection purposes, both p and q combines are supposed to become dynamically chosen but must be similar in scale but 'unique in length' so render it easier to influence. Its value can be found by the main analysis effectively.  

Computing N = PQ.  

In this, N can be used for key pair, that is public and private together as the unit and the Length was its key length, normally is spoken bits. Measure,

[tex]\lambda (N) = \ lcm( \lambda (P), \lambda (Q)) = \ lcm(P- 1, Q - 1)[/tex]  where [tex]\lambda[/tex] is the total function of Carmichaels. It is a privately held value. Selecting the integer E to be relatively prime from [tex]1<E < \lambda (N)[/tex]and [tex]gcd(E, \lambda (N) ) = 1;[/tex] that is [tex]E \ \ and \ \ \lambda (N)[/tex].  D was its complex number equivalent to E (modulo [tex]\lambda (N)[/tex] ); that is d was its design multiplicative equivalent of E-1.  

It's more evident as a fix for d provided of DE ≡ 1 (modulo [tex]\lambda (N)[/tex] ).E with an automatic warning latitude or little mass of bigging contribute most frequently to 216 + 1 = 65,537 more qualified encrypted data.

In some situations it's was shown that far lower E values (such as 3) are less stable.  

E is eligible as a supporter of the public key.  

D is retained as the personal supporter of its key.  

Its digital signature was its module N and the assistance for the community (or authentication). Its secret key includes that modulus N and coded (or decoding) sponsor D, that must be kept private. P, Q, and [tex]\lambda (N)[/tex] will also be confined as they can be used in measuring D. The Euler totient operates [tex]\varphi (N) = (P-1)(Q - 1)[/tex] however, could even, as mentioned throughout the initial RSA paper, have been used to compute the private exponent D rather than λ(N).

It applies because [tex]\varphi (N)[/tex], which can always be split into  [tex]\lambda (N)[/tex], and thus any D satisfying DE ≡ 1, it can also satisfy (mod  [tex]\lambda (N)[/tex]). It works because [tex]\varphi (N)[/tex], will always be divided by [tex]\varphi (N)[/tex],. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >   [tex]\lambda (N)[/tex] ) for time - to - time). Many RSA frameworks assume notation are generated either by methodology, however, some concepts like fips, 186-4, may demand that D<   [tex]\lambda (N)[/tex]. if they use a private follower D, rather than by streamlined decoding method mostly based on a china rest theorem. Every sensitive "over-sized" exponential which does not cooperate may always be reduced to a shorter corresponding exponential by modulo  [tex]\lambda (N)[/tex].

As there are common threads (P− 1) and (Q – 1) which are present throughout the [tex]N-1 = PQ-1 = (P -1)(Q - 1)+ (P-1) + (Q- 1))[/tex], it's also possible, if there are any, for all the common factors [tex](P -1) \ \ \ and \ \ (Q - 1)[/tex]to become very small, if necessary.  

Indication: Its original writers of RSA articles conduct their main age range by choosing E as a modular D-reverse (module [tex]\varphi (N)[/tex]) multiplying. Because a low value (e.g. 65,537) is beneficial for E to improve the testing purpose, existing RSA implementation, such as PKCS#1, rather use E and compute D.

Answer:

10010100001001010100101101010100000000000000000010100101010101111010001001010010100101001

Explanation:

Answer:

string sep = "";

for (i = 0; i < NUM_VALS; i++) {

 cout << sep << hourlyTemp[i];

 sep = ", ";

}

Explanation:

Insert the snippet at the commented location.

With the absence of a join() method as found in other languages, it is always a hassle to suppress that separator comma from appearing at the beginning or the end of your output. This solution is nice since it doesn't require any magic flags to keep track of wether the first element was printed.

Answer:

Communication

Collaboration

ICT literacy

Explanation:

These are some of the skills that are needed in the 21st century to compete and thrive in the information society.

To remain progressive, one needs to have good communication skills. These communication skills can include Active Listening and Passive Listening.

Collaboration is important because you'll have to work with other people as no man is an island, we need someone else so the skill of collaboration is necessary to compete and stay relevant in the information society in the 21st century.

IT literacy is also very important because one needs to have basic computer knowledge such as programming, computer essentials and applications, etc.

Answer:

Here is the function same_person that takes two instances of Person as arguments i.e. p1 and p2 and returns True if they are the same Person, False otherwise.

def same_person(p1, p2):  #definition of function same_person that takes two parameters p1 and p2

   if p1.GTID==p2.GTID:  # if the two instances of Person have same GTID

       return True  #returns true if above condition evaluates to true

   else:  #if the two instances of Person do not have same GTID

       return False #returns false when two persons have different GTID

Explanation:

person1 = Person("David Joyner", 30, 901234567)  #first instance of Person

person2 = Person("D. Joyner", 29, 901234567)  #second instance of Person

person3 = Person("David Joyner", 30, 903987654)  #third instance of Person

print(same_person(person1, person2))  #calls same_person method by passing person1 and person2 instance of Person to check if they are same

print(same_person(person1, person3)) #calls same_person method by passing person1 and person3 instance of Person to check if they are same

The function works as follows:

For function call print(same_person(person1, person2))

The GTID of person1 is 901234567 and that of person2 is 901234567

If condition if p1.GTID==p2.GTID in the function same_person checks if the GTID of person1 is equal to the GTID of person2. This condition evaluates to true because GTID of person1 = 901234567 and GTID of person2 = 901234567

So the output is:

True

For function call print(same_person(person1, person3))

The GTID of person1 is 901234567 and that of person3 is 903987654

If condition if p1.GTID==p2.GTID in the function same_person checks if the GTID of person1 is equal to the GTID of person3. This condition evaluates to false because GTID of person1 = 901234567 and GTID of person2 = 903987654 and they are not equal

So the output is:

False

The complete program along with its output is attached in a screenshot.

Answer:

Answered below

Explanation:

Hardware resources that can be usefully shared and also examples of their sharing include;

A) CPU servers, which carry out computations for clients.

B) Network capacity transmission of data or packet transmission is done using the same circuit, meaning many communication channels share the same circuit.

C) Screen networks windows systems which enable processes in remote computers to update the contents of their local server.

D) Memory cache server.

E) Disk file server, virtual disk server.

Softwares include;

A) Web-page web servers which allow client programs to have read-only access to page contents.

B) Database whose contents can be usefully shared.

C) File servers enable multiple users access to files .

D) Objects can be shared in distributed systems example shared ticket booking, whiteboard etc.

E) Exclusive lock which enables the coordination of accessing a special resource.

Answer:

def pyramid_volume(base_length,base_width,pyramid_height):

      return base_length * base_width * pyramid_height/3

length = float(input("Length: "))

width = float(input("Width: "))

height = float(input("Height: "))

print("{:.2f}".format(pyramid_volume(length,width,height)))

Explanation:

This line declares the function along with the three parameters

def pyramid_volume(base_length,base_width,pyramid_height):

This line returns the volume of the pyramid

      return base_length * base_width * pyramid_height/3

The main starts here

The next three lines gets user inputs for length, width and height

length = float(input("Length: "))

width = float(input("Width: "))

height = float(input("Height: "))

This line returns the volume of the pyramid in 2 decimal places

print("{:.2f}".format(pyramid_volume(length,width,height)))

Answer:

Processor P2 has the highest performance expressed in instructions per second.

Explanation:

Given:

Processors:

P1

P2

Clock rate of P1 = 3 GHz

Clock rate of P2 = 3 GHz

Cycles per instruction = CPI of P1 = 1.5

Cycles per instruction = CPI of P2 = 1.0

To find:

which processor has the highest performance in instructions per second (IPS)

Solution:

Compute CPU time:

CPU time =  (Number of instructions * cycles per instruction) / clock rate

CPU time = (I * CPI) / R

We are given the CPI and clock rate. So

CPU time = (I * CPI) / R

 I / CPU time = R/CPI

Since

Instructions per second = Number of instructions (I) / execution time

                             IPS     = I / CPU time

So

Instructions Per Second = clock rate / cycles per instruction

IPS = R/CPI

Putting the values in above formula:

Instructions Per Second for P1 = IPS (P1)

                                                  = clock rate P1 / CPI (P1)

                                                  = 3 GHz / 1.5

                                     IPS (P1) = 2            

As 1 GHz = 10⁹ Hz

IPS (P1) = 2x10⁹

Instructions Per Second for P2 = IPS (P2)

                                                  = clock rate P2 / CPI (P2)

                                                  = 3 / 1.0

                                     IPS (P2) = 3

As 1 GHz = 10⁹ Hz

IPS (P2) = 3x10⁹

Hence processor P2 has the highest performance expressed in instructions per second i.e. 3x10⁹

Answer:

The correct answer is D

In the case of printing pages when you are not sure of the number of the page the printer will print you should leave some gaps or margins or leave space between the two pages.

Learn more bout the might be considered "old-school,".

brainly.com/question/26057812.

Answer:

Primitive operation

Explanation:

Operations that a given computing agent can perform are called primitive operations.

These operations are basic arithmetic work done by an algorithm such as assigning a variable, calling a method, returning an array, etc. They are sometimes written in pseudo-code.

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