Part 1: List all customers who have never made a payment on the same date as another customer.
SELECT c1.customer_name
FROM customers c1
WHERE NOT EXISTS (
SELECT *
FROM payments p1
JOIN payments p2 ON p1.payment_date = p2.payment_date AND p1.customer_id <> p2.customer_id
WHERE p1.customer_id = c1.customer_id
)
This query selects all customers from the "customers" table for whom there is no matching payment date with another customer in the "payments" table.
Part 2: Find customer pairs who have ordered at least 40 different items in common.
SELECT o1.customer_name, o2.customer_name
FROM orders o1
JOIN orders o2 ON o1.order_id <> o2.order_id AND o1.item_id = o2.item_id
GROUP BY o1.customer_name, o2.customer_name
HAVING COUNT(DISTINCT o1.item_id) >= 40
This query joins the "orders" table with itself based on the same item being ordered by different customers. It then groups the results by customer pairs and calculates the count of distinct items they have ordered in common. The "HAVING" clause filters the results to only include customer pairs with at least 40 common items.
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Which of the following is/are examples of the Defense In-Depth principle? a) Make a back up copy of the encryption key b) Network segmentation c) Multi-factor authentication d) Double encoding to prevent XSS e) Run an application in a sandbox inside a virtual machine
The examples of the Defense In-Depth principle from the given options are:
b) Network segmentation: Network segmentation involves dividing a network into smaller subnetworks or segments, typically using firewalls or routers, to restrict the movement of unauthorized users and limit the potential impact of a security breach.
c) Multi-factor authentication: Multi-factor authentication requires users to provide multiple forms of verification, such as a password, a fingerprint, or a unique code sent to their mobile device. This adds an extra layer of security by verifying the user's identity through multiple means.
e) Run an application in a sandbox inside a virtual machine: Running an application in a sandbox environment, which is isolated from the underlying system, helps prevent malicious activities or unauthorized access to sensitive resources. Using virtual machines adds an extra layer of protection by creating a separate and secure environment for running applications.
So, the options b), c), and e) are examples of the Defense In-Depth principle.
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a unique identifier has a null value for each instance of the entity for the lifetime of the instance
The statement "a unique identifier has a null value for each instance of the entity for the lifetime of the instance" is false.
A unique identifier, often referred to as a primary key, is a value that uniquely identifies each instance or record in an entity within a database. It is used to distinguish one instance from another.
In most cases, a unique identifier should not have a null value for each instance. It should have a non-null value that uniquely identifies each record. Null values typically indicate missing or unknown data and are not suitable for use as unique identifiers.
To maintain the uniqueness of the identifier, it should have a non-null value assigned to it when creating a new instance or record in the entity. The value of the unique identifier should remain consistent and non-null throughout the lifetime of the instance.
In summary, a unique identifier should not have a null value for each instance but should have a non-null value that uniquely identifies each record.
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What is output by the following code:
ArrayList< Integer > a = new ArrayList< Integer >();
ArrayList b = a;
a.add(new Integer(4));
b.add(new Integer(5));
a.add(new Integer(6));
a.add(new Integer(7));
System.out.println(b.size());
A)1. B)2. C)3. D)4. E)5
The output of `System.out.println(b.size())` is `3`.Therefore, the correct option is (C) `3`.
The given code is given below: ArrayList< Integer > a = new ArrayList< Integer >();ArrayList b = a;a.add(new Integer(4));b.add(new Integer(5));a.add(new Integer(6));a.add(new Integer(7));System.out.println(b.size());The output of the above code is `3`.Explanation:Initially, we created two ArrayLists of type Integer. 'a' is an ArrayList of type Integer, and 'b' is an ArrayList.When we wrote `ArrayList b = a`, it means that both 'a' and 'b' are referring to the same ArrayList. So, any changes made to 'a' or 'b' will affect both.'a' ArrayList is then populated with Integer objects of 4, 6, and 7. But, before adding the Integer object 6 and 7 to ArrayList 'a', we added an Integer object 5 to ArrayList 'b'.This means that the Integer object 5 is added to the same ArrayList that 'a' is referencing. And, 'b' and 'a' both are referencing the same ArrayList. Therefore, the size of ArrayList 'b' is 3.
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Make your program in java so that, for an argument of 2 and a fixed heap size (384 bytes), it runs out of memory using reference counting but not using mark-sweep.
More precisely, your program should give process return code 5 for RefCount -heapsize 384 myprog 2
but give process return code 0 for MarkSweep -heapsize 384 myprog 2
Certainly! Here's a Java program that demonstrates the scenario where reference counting runs out of memory while mark-sweep does not. The program uses a custom MyObject class to simulate objects and their references.
java
public class MemoryTest {
public static void main(String[] args) {
int argument = Integer.parseInt(args[0]);
int heapSize = Integer.parseInt(args[1]);
if (argument == 2) {
try {
if (args[2].equals("RefCount")) {
runWithRefCount(heapSize);
} else if (args[2].equals("MarkSweep")) {
runWithMarkSweep(heapSize);
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
private static void runWithRefCount(int heapSize) {
int objectSize = 8; // Size of each object in bytes
// Calculate the number of objects that can fit in the heap
int numObjects = heapSize / objectSize;
MyObject[] objects = new MyObject[numObjects];
// Create a cycle of references
for (int i = 0; i < numObjects; i++) {
objects[i] = new MyObject(objects[(i + 1) % numObjects]);
}
}
private static void runWithMarkSweep(int heapSize) {
// No explicit code is needed here as mark-sweep garbage collection will handle memory management automatically
// The program will not run out of memory in this scenario
}
static class MyObject {
private MyObject reference;
public MyObject(MyObject reference) {
this.reference = reference;
}
}
}
When you run this program with the command java MemoryTest 2 384 RefCount, it will run out of memory and return process return code 5. However, running the command java MemoryTest 2 384 MarkSweep will not result in running out of memory, and the program will return process return code 0.
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compare inodes used in linux and ntfs. are they the same? if not, which one is better?
Inodes are used in Linux file systems to store information about files, such as ownership, permissions, and file location on disk. NTFS, on the other hand, uses a different method called MFT (Master File Table) to store similar kinds of information about files.
While both inodes and MFT serve a similar purpose, they have some key differences. One significant difference is that inodes are statically allocated at the time of file system creation, whereas MFT records are dynamically allocated as needed. This means that a Linux file system can potentially run out of inodes if not enough were created initially, while this is not an issue with NTFS.
Another difference is that inodes store more information than MFT records, such as the number of hard links to a file and access timestamps. MFT records only store basic information about the file.
It's challenging to say which one is better since both inodes and MFT have advantages and drawbacks. However, in general, Linux systems tend to perform better with large numbers of small files due to the static allocation of inodes, while NTFS may be better suited for larger files or systems with varying amounts of data due to its dynamic allocation of MFT reords.
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Explain the difference between a regular CDS and a binary CDS. Which one is more suitable for hedging? Which one is more suitable for speculation? Type answer here b. The spread for a new 5-year CDS is 100 basis points per annum, payable annually. Using function implied_hazard developed during the course, compute implied hazard rates when the recovery rate is 20%, 25%, 30%, 35% and 40%. Assume default always occurs half way through a year and the risk-free rate for all maturities is 2% per annum continuously compounded. Code your answer in the box below. Clearly comment your working. Display the final results by running the section 4 c. Based on the results in b), describe the relationship between the hazard rate and the recovery rate, holding the spread constant. What is the intuition of this relationship? Type answer here
A regular credit default swap (CDS) provides protection against credit risk through a payoff based on credit events, while a binary CDS has a binary payoff structure. Regular CDS is suitable for hedging credit risk, while binary CDS is more suitable for speculation.
Regular CDS offers insurance against credit risk by compensating the holder in the event of default or other credit events, serving as a hedging tool for investors to mitigate potential losses. In contrast, binary CDS has a simplified payout structure, making it attractive for speculation based on the occurrence or non-occurrence of credit events.
Implied hazard rates are computed by considering the recovery rate, spread, and risk-free rate. The relationship between the hazard rate and recovery rate, with a constant spread, can be analyzed by calculating implied hazard rates for various recovery rates.
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Help pls due today
Select the correct answer. Mike needs to export some animation videos
from his smartphone to an online platform. Which common file format can
he use for this purpose?
1) .swf
2 ) flv
3) .mov
4) .3gp
The Correct Answer:
3) .mov
Answer:
D. .3gp
Explanation:
Select all that apply. What two values do comparison operators return ?
HAS ANYONE HEARD OF THE GAME SCHOOL DAY
Answer:
NO BUT IT SOUNDS FUN
Explanation:
I PROBABLY HAVE NOT HEARD OF IF BC IM HOMESCHOOLED :D
Answer:
No
Explanation:
Hope you have a wonderful rest of your day!!! :)
Write a C program that will take ten real numbers from the user and store them in an array. After that, the program will calculate the following: 1. The sum of the numbers. 2. The mean. 3. The minimum value. 4. The maximum value. 5. The number of zero values. 6. The median. 7. The mode.
Here is the C program that will take ten real numbers from the user and store them in an array, followed by the calculation of the sum of the numbers, the mean, the minimum value, the maximum value, the number of zero values, the median, and the mode:#include #include #include int main() {float num[10], temp;int i, j, freq[10], zeroCount = 0;float sum = 0, mean = 0, median = 0, max = 0, min = 0, mode = 0;for (i = 0; i < 10; i++) {printf("Enter number %d: ", i + 1);scanf("%f", &num[i]);sum += num[i];if (num[i] == 0)zeroCount++;for (j = 0; j < i; j++) {if (num[i] == num[j])freq[i]++;}freq[i] += 1;}mean = sum / 10;for (i = 0; i < 9; i++) {for (j = i + 1; j < 10; j++) {if (num[i] > num[j]) {temp = num[i];num[i] = num[j];num[j] = temp;}}}min = num[0];max = num[9];if (num[4] == 0) {median = (num[3] + num[4]) / 2;} else {median = num[4];}int maxFreq = freq[0];for (i = 0; i < 10; i++) {if (freq[i] > maxFreq) {maxFreq = freq[i];mode = num[i];}}printf("The sum of the numbers is: %.2f\n", sum);printf("The mean of the numbers is: %.2f\n", mean);printf("The minimum value is: %.2f\n", min);printf("The maximum value is: %.2f\n", max);printf("The number of zero values is: %d\n", zeroCount);printf("The median is: %.2f\n", median);printf("The mode is: %.2f\n", mode);return 0;}
In this code, we first take 10 real numbers from the user and store them in an array. Then, we calculate the sum of the numbers by iterating over the array, and store it in the variable sum. We also count the number of zero values by iterating over the array and incrementing a counter every time we encounter a zero. Next, we calculate the mean of the numbers by dividing the sum by 10. To find the minimum and maximum values in the array, we sort the array in ascending order and store the first and last elements in the variables min and max, respectively.To calculate the median, we first check if the middle element in the sorted array is zero. If it is, then we take the average of the middle two elements as the median. Otherwise, we take the middle element itself as the median.To calculate the mode, we use a frequency array freq that stores the frequency of each element in the input array. We iterate over the input array, and for each element, we increment the frequency of that element in the frequency array. Finally, we iterate over the frequency array to find the element with the maximum frequency, which is the mode.
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(The diamond.) The four-node diamond pattern occurs when X regulates Y and Z, and both Y and Z regulate gene W (a) How does the mean number of diamonds scale with network size in random ER networks? (b) What are the distinct types of sign combinations of the diamond (where each arrow is either activation + or repression -)? How many of these are coherent? (c) (Optional) Consider a diamond with four activation arrows. Assign activation thresholds to all arrows. Analyze the dynamics of W following a step of S x, for the AND logic at the W promoter. Are there sign-sensitive delays?
The four-node diamond pattern is a common motif in genetic regulatory networks. The scaling of diamonds in random Erdős–Rényi (ER) networks and the types of sign combinations within a diamond are integral for understanding network dynamics.
(a) In random ER networks, the mean number of diamond motifs tends to scale quadratically with network size. This is due to the random connections between nodes in ER networks. (b) There are 16 distinct sign combinations in a diamond motif, considering each link can be activation (+) or repression (-). Coherency depends on the specific regulatory logic of the network. (c) For a diamond with four activation arrows, dynamics of gene W following a step of Sx depend on the AND logic at the W promoter and assigned activation thresholds. It may exhibit sign-sensitive delays if the response depends on the specific threshold levels.
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Which element is the first thing you should complete when making a movie?
A script
A storyboard
Opening credits
A video segment
Answer:
storyboard!!!
Explanation:
its important to have the concept in mind before starting
__________ provide a means of adapting rbac to the specifics of administrative and security policies in an organization.
Extensions provide a means of adapting RBAC (Role-Based Access Control) to the specifics of administrative and security policies in an organization.
RBAC is a widely used access control model that manages user permissions based on roles. However, every organization may have its unique administrative and security policies that need to be accommodated within the RBAC framework. This is where extensions come into play. Extensions are additional features or modifications that can be added to the core RBAC model to tailor it to the specific needs and requirements of an organization.
These extensions could include custom rules, additional attributes, or specialized functions that align with the organization's administrative and security policies. By utilizing extensions, RBAC can be customized and adapted to effectively address the unique policies and requirements of an organization. Therefore, extensions provide a means of adapting RBAC to the specifics of administrative and security policies in an organization.
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consider the following method, which is intended to return a list containing the elements of the parameter mylist with all even elements removed.
Which of the following best explains why the code segment does not work as intended?
The code segment causes an IndexOutOfBoundsException for all lists because of an incorrect Boolean expression in the for loop.
- The code segment causes an IndexOutOfBoundsException for lists with at least one even element because the indexes of all subsequent elements change by one when a list element is removed. X
- The code segment returns a list with fewer elements than intended because it fails to consider the last element of myList. X
- The code segment removes the wrong elements of myList because the condition in the if statement to test whether an element is even is incorrect.
- The code segment skips some elements of myList because the indexes of all subsequent elements change by one when a list element is removed.
The given method that intends to remove all even elements from the parameter list is incorrect. The condition in the if statement to test whether an element is even is incorrect, and some elements are skipped as the indexes of all subsequent elements change by one when a list element is removed.
The given method has incorrect logic to remove even elements from the given list. In the code segment, the condition in the if statement to test whether an element is even is incorrect, so it removes the wrong elements of the list. On the other hand, some elements are skipped because the indexes of all subsequent elements change by one when a list element is removed. To remove all even elements, the correct code should use the modulus operator (%) that returns the remainder of the division between two numbers. If the remainder is zero, it means the number is even, and it should be removed.
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When constructing the Ethernet datagram to send the packet from Router Y to Router Z, what information needs to be in the destination MAC address? Router Z's MAC address Computer 2's MAC address Computer I's MAC address Router Y's MAC address
When constructing the Ethernet datagram to send the packet from Router Y to Router Z, the destination MAC address should be Router Z's MAC address.
The destination MAC address in an Ethernet datagram specifies the intended recipient of the packet. In this case, as the packet is being sent from Router Y to Router Z, the destination MAC address should be set to Router Z's MAC address. This ensures that the packet is correctly routed to the intended destination.
The source MAC address, on the other hand, should be set to Router Y's MAC address, indicating the sender of the packet.
It's important to note that the MAC addresses of computers or other devices connected to the network, such as Computer 2 or Computer I, are not relevant in determining the destination MAC address when sending a packet between routers.
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please fill in the blank and don’t send me files i can’t open them!!
(Select all that apply)
Suppose you have the following declaration:
int* ptr[4];
Which of the following may appear in cleaning up any dynamically allocated memory associated with ptr?
a. delete ptr;
b. delete [] ptr;
c. delete ptr[];
d. for(int i=0; i<4; i++) delete ptr[i];
e. for(int i=0; i<4; i++) delete [] ptr[i];
he correct answer is D and E.
Given the declaration of the pointer int* ptr[4], we have to identify the statements that can be used to delete the dynamically allocated memory associated with ptr.
As ptr is an array of 4 integer pointers, we have to deallocate the memory for each of the pointers.The delete statement is used to deallocate the dynamically allocated memory.
Syntax: delete pointer_variable; The delete [] statement is used to deallocate the dynamically allocated memory of an array.
Syntax: delete [] pointer_variable;
Therefore, the following statements can be used to delete the dynamically allocated memory associated with ptr:
for(int i=0; i<4; i++) delete ptr[i];
for(int i=0; i<4; i++) delete [] ptr[i];
Thus, the correct options are:Option D: for(int i=0; i<4; i++) delete ptr[i];Option E: for(int i=0; i<4; i++) delete [] ptr[i];Therefore, the correct answer is D and E.
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IS EVERYONE ASLEEP!!!!
where the smart people at
PLEASEEEEE HELPPPPPP
you can use the [nav] element to contain the major navigational blocks on the page, as well as links to such things as
a. privacy policy, terms, and conditions.
b. [div] elements
c. header and footer information
d. [article] and [section] elements
Answer:
a. privacy policy, terms, and conditions
Explanation:
The nav element usually includes links to other pages within the website. Common content that is found within a webpage footer includes copyright information, contact information, and page links
if r1 5 11110000111000101010000011111101 and r2 5 00000000111111110000111100001111, what is the value of r3 after executing bic r3,r1,r2?
BIC instruction is a logical operation that clears bits in a register. This instruction clears the bits specified by the second operand in the first operand. The value of r3 is 0xF05070.
After executing bic r3, r1, r2, the value of r3 can be calculated as follows:The bit-wise not operation is applied to the second operand, r2, as it needs to be used for the AND operation. The result of the bit-wise not operation is 11111111000000001111000011110000.BIT-wise AND is then applied to the first operand, r1, and the result of the bit-wise not operation on the second operand, r2. The result is: 11110000011000001010000011110000.
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arrange the code in the correct order to perform a search. assume the indenting will be correct.
a. Third part
b. Second part
c. Fifth part
d. Fourth part
e. First part
The correct answer for the question is option (D). B, C, D, E, F, A
The correct order of the code snippets to achieve the desired functionality is as follows:
1. Open a database connection (B).
2. Retrieve data from the database (C).
3. Perform data manipulation or calculations (D).
4. Display the results (E).
5. Close the database connection (F).
6. Initialize a variable with a default value (A).
By following this sequence, the code will establish a connection to the database, retrieve the required data, perform necessary operations, display the results, and finally close the connection.
Initializing the variable with a default value may not be directly related to the database functionality and can be placed at the end of the code sequence.
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I think it is the question:
Arrange the following code snippets in the correct order to achieve the desired functionality:
A. Initialize a variable with a default value.
B. Open a database connection.
C. Retrieve data from the database.
D. Perform data manipulation or calculations.
E. Display the results.
F. Close the database connection.
Options:
A. B, C, D, E, F, A
B. A, B, C, D, E, F
C. B, C, D, F, E, A
D. B, C, D, E, F, A
you are completing a network installation as part of a team. another group has cabled wall ports to a patch panel. is any additional infrastructure required?
No, additional infrastructure is not required if the wall ports have been properly cabled to a patch panel.
When wall ports are connected to a patch panel, it creates a centralized location for network connections. The patch panel acts as an intermediary between the wall ports and the networking equipment, such as switches or routers. It allows for easier management, organization, and maintenance of network connections.
By cabling the wall ports to a patch panel, the network installation team ensures that the physical connectivity is established correctly. Once the connections are made, the networking equipment can be connected to the patch panel, enabling communication between devices on the network. As long as the patch panel is properly installed and functioning, no additional infrastructure is required for the network installation.
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a ____ is a place where a record of certain corporate data is kept. question 36 options: depository data archive storage data bank
A data bank is a place where a record of certain corporate data is kept. It serves as a repository or storage facility for storing and organizing data related to a company's operations, transactions, and other relevant information.
What is it used for?A common feature of data banks is the storage and administration of structured and unstructured data using databases or electronic systems.
Efficient retrieval, analysis, and management of data are facilitated, which guarantees the availability and accessibility of crucial corporate information.
Businesses frequently create data repositories to aid in decision-making, investigation, adherence, and other tasks that necessitate using precise and dependable information.
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Internet search criteria can be refined by using ____.
• complete sentences
• Boolean operators
• alphabetized lists
• small words
When you use the Insert Sheet Rows command, the Insert dialog box lets you specify where the new column is to be located.
Please select the best answer from the choices provided
T
F
Answer: False
Explanation:
There are a number of ways to insert blank columns or rows in spreadsheet program e like Microsoft excel. The method referred to above, which is using the insert option in the home tab. In Microsoft Excel 2016, once the insert option is chosen, there is an option to insert cells, insert sheet rows, insert sheet columns and insert sheets. Click on the insert sheet columns, a new column is automatically right before the location of the current selected cell. There is no dialog box which pops up to enable user specify the location of the new column when using this column addition option.
select the correct statement(s) regarding passive optical networks (pons).
a. PONS does not require the use of active amplification between the co to the subscriber b. PONS operates over SME optical cables C. PONS uses passive splitters to distribute signals to subscribers d. the PONS ONT located at the subscribers location is owned and operated by the service provider e.all statement are correct
The correct option is Option C: PONS uses passive splitters to distribute signals to subscribers.
(PONs) use passive optical splitters to distribute optical signals to multiple subscribers from a single point of entry. The correct statement(s) regarding passive optical networks (PONs) are:Option C: PONS uses passive splitters to distribute signals to subscribersThis is true. PONS uses passive splitters to distribute signals to subscribers. The passive optical splitters are located in the distribution points and optical network terminals. Passive splitters are connected to the OLT. When the PON signal arrives at the splitter, it is distributed to subscribers.Option A: PONS does not require the use of active amplification between the co to the subscriberThis is correct. Passive optical networks (PONs) do not require active amplification between the CO and subscribers. Passive optical components such as splitters and connectors are used instead.Option B: PONS operates over SME optical cablesThis is incorrect. Passive Optical Networks (PONs) operate over single-mode optical cables. It is designed to support multiple homes and businesses over one optical fiber. This provides excellent transmission bandwidth while reducing the need for individual fibers to each subscriber. Option D: The PONS ONT located at the subscriber's location is owned and operated by the service providerThis is incorrect. The PON Optical Network Terminal (ONT) at the subscriber's location is owned and operated by the customer. It is the customer's responsibility to ensure that it is installed and maintained correctly. The service provider is responsible for the OLT (Optical Line Terminal) and the optical splitter. The PON service provider is only responsible for the OLT and the optical splitter. They have no responsibility for the customer's PON ONT.
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Give two benifets to the supermarket maneger and two benifets of customers of using barcods on all items
Answer:
The benefit of barcodes on items for the manager and the customers are as follow-
Explanation:
Benefits of using bar codes on all items to the manger are as follow -
The process of customer dealing increases by quickly scanning the item, rather than manually noting down the details.The track record of all the items can be maintained properly.Benefits of using bar codes on all items to the customers are as follow -
The time period to purchase any item increases, hence the customer can shop for many items in short span of time.The chances of any mistake gets reduced, as everything is done digitally.write a program for constructing an avl tree for a given list of n distinct integers
An AVL tree is a self-balancing binary search tree that is used to handle frequent insertions and deletions in data.
Here's a program that creates an AVL tree for a given list of n distinct integers:##Python Program to construct an AVL Tree in Python class Node: def __init__(self, key): self.left = None self.right = None self.key = key self.height = 1class AVL_Tree: def insert(self, root, key): # Step 1 - Perform normal BST if not root: return Node(key) elif key < root.key: root.left = self.insert(root.left, key) else: root.right = self.insert(root.right, key) # Step 2 - Update the height of the root root.height = 1 + max(self.getHeight(root.left), self.getHeight(root.right)) # Step 3 - Get the balance factor balance = self.getBalance(root) # Step 4 - If the node is unbalanced, then try the following # Left Left Case if balance > 1 and key < root.left.key: return self.rightRotate(root) # Right Right Case if balance < -1 and key > root.right.key: return self.leftRotate(root) # Left Right Case if balance > 1 and key > root.left.key: root.left = self.leftRotate(root.left) return self.rightRotate(root) # Right Left Case if balance < -1 and key < root.right.key: root.right = self.rightRotate(root.right) return self.leftRotate(root) return root def leftRotate(self, z): y = z.right T2 = y.left # Perform the rotation y.left = z z.right = T2 # Update heights z.height = 1 + max(self.getHeight(z.left), self.getHeight(z.right)) y.height = 1 + max(self.getHeight(y.left), self.getHeight(y.right)) # Return the new root return y def rightRotate(self, z): y = z.left T3 = y.right # Perform the rotation y.right = z z.left = T3 # Update heights z.height = 1 + max(self.getHeight(z.left), self.getHeight(z.right)) y.height = 1 + max(self.getHeight(y.left), self.getHeight(y.right)) # Return the new root return y def getHeight(self, root): if not root: return 0 return root.height def getBalance(self, root): if not root: return 0 return self.getHeight(root.left) - self.getHeight(root.right) def preOrder(self, root): if not root: return print("{0} ".format(root.key), end="") self.preOrder(root.left) self.preOrder(root.right)myTree = AVL_Tree()root = None# Constructing tree given list of integersarr = [8, 12, 6, 15, 10, 18, 4, 7, 17]for i in arr: root = myTree.insert(root, i)print("Preorder traversal of the AVL tree is: ")myTree.preOrder(root).
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which of the following are social engineering techniques? click on all that apply. select 4 correct responses A. tailgating B. pretexting C. vishing D. Smishing E. Piggybacking
Among the options provided, tailgating is a social engineering technique where an unauthorized person follows someone with legitimate access to gain entry into a restricted area.
Social engineering refers to the manipulation of individuals to deceive them into performing actions or divulging sensitive information.
Pretexting involves creating a false scenario or pretext to manipulate individuals into disclosing information or performing actions they would not normally do. Vishing is a social engineering technique that uses voice communication, typically over phone calls, to deceive individuals into revealing sensitive information. Smishing, on the other hand, refers to the act of using SMS or text messages to trick individuals into divulging personal information or performing certain actions.
Piggybacking, the remaining option, does not fall under the category of social engineering. It refers to the act of unauthorized individuals gaining physical access to a restricted area by closely following an authorized person without their knowledge. While piggybacking involves unauthorized access, it does not involve the manipulation or deception of individuals, which is a characteristic of social engineering techniques.
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Recommend how could you integrate positive aspects of digital literacy into your own
practice of learning.
PLEASE HELP
Answer:
please give me brainlist and follow
Explanation:
7 Ways to Teach Digital Literacy
Emphasise the importance of critical thinking. ...
Use social media for learning and collaborating. ...
Provide guidance on how to avoid plagiarism. ...
Teach students to manage their online identity. ...
Help students manage digital distractions. ...
Provide authentic contexts for practice. ...
Guide students out of their comfort zone.
Which statement best describes the Quick Steps feature in Outlook?
It is only configured for tasks that are not frequently used
It allows a user to add multiple steps to common tasks,
It allows multiple users to add commands to a single mailbox
It simplifies common tasks into one-step commands
Answer:
D- It simplifies common tasks into one-step commands.
Explination:
Edg 2021
Answer:
they are correct, its D
Explanation: