In this program, the user is prompted to enter the hour the call originated and the length of the call. The program then calculates the cost and discount rate based on the given criteria. Finally, it outputs the time of call originated, length, cost, and discount rate for each call.
Here is the C code for computing the cost of telephone calls from a cellular phone based on the input provided by the user:
#include <stdio.h>
#include <math.h>
#define BASIC_RATE 0.49
#define ADDITIONAL_RATE 0.37
#define EVENING_DISCOUNT_1 0.35
#define EVENING_DISCOUNT_2 0.65
int main() {
float hour, length, cost, discount;
// Input time and length of the call
printf("Enter the hour the call originated (in float): ");
scanf("%f", &hour);
printf("Enter the length of the call (in minutes): ");
scanf("%f", &length);
// Calculate the cost and discount rate
int roundedLength = ceil(length);
if (hour > 16 && hour < 22) {
cost = BASIC_RATE + (roundedLength - 1) * ADDITIONAL_RATE;
discount = EVENING_DISCOUNT_1;
} else if (hour > 22 || hour < 7) {
cost = BASIC_RATE + (roundedLength - 1) * ADDITIONAL_RATE;
discount = EVENING_DISCOUNT_2;
} else {
cost = BASIC_RATE + (roundedLength - 1) * ADDITIONAL_RATE;
discount = 0;
}
// Output the result
printf("\nCall Originated at: %.2f\n", hour);
printf("Call Length: %.2f minutes\n", length);
printf("Cost: $%.2f\n", cost);
printf("Discount Rate: %.0f%%\n", discount * 100);
return 0;
}
In this program, the user is prompted to enter the hour the call originated and the length of the call. The program then calculates the cost and discount rate based on the given criteria.
Finally, it outputs the time of call originated, length, cost, and discount rate for each call.
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Even if you encode and store the information, which of the following can still be a cause of forgetting?
A. decay
B. disuse
C. retrieval
D. redintegration
Even if you encode and store the information, decay can still be a cause of forgetting. The correct option is A. decay
Forgetting refers to the inability to recall previously learned knowledge or events. Long-term memories are those that have been stored in the brain and are capable of being recovered after a period of time.
The ability to retrieve information from long-term memory is essential in everyday life, from remembering the name of a childhood friend to recalling what you studied for a test.
The three primary mechanisms for forgetting are interference, cue-dependent forgetting, and decay.
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There is another way to send inputs to a CGI program. That is to attach the input in the URL. See the following example. http: //www. example. com/myprog. cgi?name=value Can we put our malicious function definition in the value field of the above URL, so when this value gets into the CGI program myprog. cai, the Shellshock vulnerability can be exploited?
no, we cannot put our malicious function definition in the value field of the given URL. This is because, when it comes to the Shellshock vulnerability, only environment variables that contain Bash code can exploit it.In this scenario, the URL you shared can be utilized to send inputs to a CGI program.
It's possible to attach the input in the URL. However, even if we can put our malicious function definition in the value field of the given URL, it will not be able to exploit Shellshock vulnerability. That is because the Shellshock vulnerability can only be exploited by environment variables that contain Bash code.In other words, the CGI script will not use the name-value pairs from the URL as environment variables. Instead, it will utilize the information from them as query string parameters and use them to create dynamic HTML content for the user.
Since these parameters are not utilized to construct environment variables, they can't be utilized to exploit the Shellshock vulnerability.Thus, we can't put a malicious function definition in the value field of the URL given, as it will not be able to exploit the Shellshock vulnerability. Only environment variables that contain Bash code can be used to exploit Shellshock. Furthermore, the CGI script does not use name-value pairs from the URL as environment variables; instead, it utilizes the information from them as query string parameters and uses it to create dynamic HTML content for the user. As a result, the parameters are not utilized to construct environment variables, which is necessary to exploit the Shellshock vulnerability.
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TM Excellent Inc. is a well-known telecommunications hardware developer. One of the employers, John, is well known for his computer skills, and he might have put those skills to evil use. The Department of Justice (DOJ) recently indicted John for altering the company’s yearly financial statements to hide his money embezzling activities. Because John is renowned for having a good computer skill, it is expected that he has cleaned his tracks. In your experience, most medium to advanced users are aware of evidence elimination software, which makes your job difficult.
Fortunately, the executive vice president of finance, Aiden, negotiated a deal with the Department of Justice. If Paluchi testifies against the John, he will receive immunity from any additional charges related to this case. Aiden supplied the DOJ with the document he says John altered. Aiden also said that this document was sent to the whole executive staff through e-mail.
You and your team travelled to the company and began the analysis. The team acquired a forensic duplication of John’ laptop hard drive using FTK Imager. The team quickly reviewed the image.
1. Describe forensic duplication and why is it important to take a hash of the imaged volume?
2. How could you further preserve the integrity and confidentiality of the evidence you just captured?
1. Forensic duplication is the process of creating an exact replica of a digital storage device, important for preserving original evidence and integrity.
2. To preserve integrity and confidentiality, maintain a documented chain of custody, store evidence securely, encrypt data, control access, create backups, and thoroughly document all actions.
1. Forensic duplication is the process of creating an exact replica, commonly referred to as an image, of a digital storage device such as a laptop hard drive. It is a crucial step in digital forensic investigations as it ensures the preservation of the original evidence without any modifications or alterations.
Taking a hash of the imaged volume is important for two main reasons. Firstly, a hash value is a unique alphanumeric string generated by a hash algorithm such as MD5, SHA-1, or SHA-256. By calculating and documenting the hash value of the imaged volume, investigators can verify the integrity of the image throughout the investigation process. Any changes made to the image would result in a different hash value, alerting investigators to potential tampering or corruption of the evidence.
Secondly, the hash value acts as a digital fingerprint of the imaged volume. It enables investigators to compare the hash value of the original evidence with other copies or subsequent analysis results to ensure consistency and authenticity. If the hash values match, it provides a level of confidence that the evidence has not been altered or tampered with since the initial imaging.
2. To further preserve the integrity and confidentiality of the captured evidence, several measures can be taken:
a. Chain of custody: Establishing a clear and documented chain of custody is crucial. It involves carefully tracking the movement and handling of the evidence from the moment it is collected until its presentation in court. This ensures that the evidence remains secure and its integrity is not compromised.
b. Secure storage: The imaged laptop hard drive should be stored in a secure location with controlled access to prevent unauthorized tampering or access. This can include using lockable storage containers or evidence lockers that are only accessible to authorized personnel.
c. Data encryption: Encrypting the imaged volume adds an extra layer of protection to ensure the confidentiality of the data. Encryption converts the data into an unreadable format without the appropriate decryption key, making it more difficult for unauthorized individuals to access sensitive information.
d. Access controls: Implementing strict access controls to the imaged volume and any analysis tools or software used during the investigation is important. Only authorized personnel should have access to the evidence, and their activities should be logged and monitored to detect any unauthorized access attempts.
e. Regular backups: Creating backups of the imaged volume and maintaining multiple copies in separate secure locations helps prevent data loss due to hardware failures, accidents, or other unforeseen events. Regularly verifying the integrity of the backups through hash values is also essential.
f. Documentation: Thoroughly documenting all actions taken, tools used, and observations made during the investigation ensures transparency and helps maintain the integrity of the evidence. Detailed notes should be taken throughout the process, including any changes or modifications made to the evidence or analysis environment.
By following these practices, the integrity and confidentiality of the evidence can be preserved, increasing its reliability and admissibility in court.
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allowing or denying traffic based on ports, protocols, addresses, or direction of data is an example of what?
Allowing or denying traffic based on ports, protocols, addresses, or direction of data is an example of network traffic filtering or firewall rules.
Network traffic filtering is a process of selectively allowing or denying network traffic based on specific criteria. This can be done using a variety of methods such as port numbers, protocols, IP addresses, or the direction of data flow.
Firewall rules are a set of instructions or configurations that determine how network traffic should be handled. These rules can be configured to allow or deny traffic based on various criteria. For example, a firewall rule can be set to allow incoming traffic on a specific port, such as port 80 for web traffic, while blocking traffic on other ports. Similarly, firewall rules can be set to allow or deny traffic based on specific IP addresses or protocols.
Here are some examples of how network traffic filtering can be used:
1. Port-based filtering: A firewall rule can be set to allow incoming traffic on port 443 for secure HTTPS connections while blocking traffic on other ports.
2. Address-based filtering: Firewall rules can be configured to allow or deny traffic from specific IP addresses or IP ranges. For example, a rule can be set to block traffic from a known malicious IP address.
3. Protocol-based filtering: Firewall rules can be set to allow or deny specific network protocols. For instance, a rule can be configured to block all incoming traffic that uses the FTP protocol.
4. Direction-based filtering: Firewall rules can be applied based on the direction of data flow. For example, a rule can be set to allow outgoing traffic but block incoming traffic.
By utilizing network traffic filtering and firewall rules, organizations can enhance their network security by controlling and monitoring the flow of data within their network.
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a) What is the status of IPv4 in the hierarchy and addressing issues surrounding the construction of large networks? Identify the major emerging problems for IPv4 and discuss how they are addressed in IPv6. [5 marks] Answer: b) Figure A. Network Diagram Although 256 devices could be supported on a Class C network ( 0 through 255 used for the host address), there are two addresses that are not useable to be assigned to distinct devices. What are the address? Why? c) What is the network address in a class A subnet with the IP address of one of the hosts as 25.34.12.56 and mask 255.255.0.0? [2 marks] Answer: d) Why would you want to subnet an IP address? [4 marks] Answer: e) What is the function of a subnet mask?
IPv4 faces challenges in terms of hierarchy and addressing for constructing large networks. Major emerging problems include address exhaustion and the complexity of network management. IPv6 addresses these issues by introducing a larger address space and simplifying network configuration and management.
IPv4 operates with a hierarchical addressing scheme where IP addresses are divided into classes (A, B, and C) to accommodate different network sizes. However, this hierarchical structure presents challenges for constructing large networks. The limited address space of IPv4 has led to address exhaustion, as the demand for IP addresses has exceeded the available supply. This has necessitated the use of techniques like Network Address Translation (NAT) to conserve IP addresses.
In contrast, IPv6 introduces a significantly larger address space, allowing for an almost unlimited number of unique IP addresses. This addresses the problem of address exhaustion in large networks. Additionally, IPv6 simplifies network management by incorporating features such as stateless autoconfiguration, which eliminates the need for manual IP address assignment.
b) In a Class C network, two addresses that are not usable for distinct devices are the network address and the broadcast address. The network address is the lowest address in the range, where all host bits are set to 0. In a Class C network, the network address would be 192.168.0.0. The broadcast address, on the other hand, is the highest address in the range, where all host bits are set to 1. In a Class C network, the broadcast address would be 192.168.0.255. These addresses are reserved and cannot be assigned to individual devices because the network address represents the entire network itself, and the broadcast address is used to send a message to all devices within the network.
c) The network address in a Class A subnet with the given IP address of one of the hosts as 25.34.12.56 and the mask 255.255.0.0 can be determined by performing a bitwise logical AND operation between the IP address and the subnet mask.
25.34.12.56: 00011001.00100010.00001100.00111000
255.255.0.0: 11111111.11111111.00000000.00000000
--------------------------------------
Network address: 00011001.00100010.00000000.00000000
Converting the binary result back to decimal, the network address is 25.34.0.0.
d) Subnetting an IP address is beneficial for several reasons. It allows for efficient utilization of IP address space by dividing a large network into smaller subnets. Subnetting helps to improve network performance by reducing network congestion and limiting the broadcast domain size. It also enhances network security by segregating different subnets and controlling access between them using routers and firewalls. Subnetting enables better organization and management of IP addresses, making it easier to assign addresses and track network devices. It offers flexibility for network expansion and facilitates the implementation of specific network policies and QoS (Quality of Service) measures within individual subnets.
e) The function of a subnet mask is to determine the network portion and the host portion of an IP address. It is a 32-bit value that works in conjunction with the IP address to identify the network to which a device belongs. The subnet mask contains a series of 1s followed by a series of 0s. The 1s represent the network portion, and the 0s represent the host portion. When a device receives an IP packet, it applies the subnet mask to the destination IP address.
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The Customer data is not stored in the Orders table and the Product data is not stored in the Order Details table. Why is this?
Customer data is stored in the Customer table, and the Orders table only has the Customer ID, and the product data is stored in the Product table, and the Order Details table only has the Product ID.
The reason why the Customer data is not stored in the Orders table and the Product data is not stored in the Order Details table is that this process follows the rules of database normalization.
Database normalization is a process of organizing data in a database. The main goal of this process is to minimize data redundancy and data dependency.
When data is normalized, it is divided into smaller tables, each with a single topic or theme and with the columns representing a piece of information about that theme.
According to the database normalization rule, each piece of data should be stored only once. Therefore, the customer data should only be stored once in the customer table.
The Orders table should have only the customer ID, which is the foreign key, to avoid data redundancy. The Product data is also stored only once in the Product table. The Order Details table should only have the Product ID, which is a foreign key, to avoid redundancy.
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Let's suppose you build an Airline Reservation Application (which must support large scale operations). What is your choice of the database backend? Neo4j SQLite MongoDB MySQL Oracle
MySQL
For an Airline Reservation Application that supports large-scale operations, MySQL would be a suitable choice as the database backend. MySQL is a popular and reliable relational database management system that is widely used in various industries, including the airline industry. It offers robust performance, scalability, and high availability, making it capable of handling the demands of a large-scale application like an airline reservation system.
MySQL provides advanced features such as replication, clustering, and partitioning, which enable horizontal scaling and improved performance for handling a large number of concurrent users and data transactions. Its ACID-compliant architecture ensures data integrity and reliability, crucial aspects for an application that deals with sensitive customer information and critical operations like flight bookings.
Furthermore, MySQL has a mature ecosystem with extensive documentation, community support, and a wide range of tools and libraries that facilitate development, monitoring, and maintenance of the database. Its compatibility with various programming languages and frameworks simplifies integration with the application's backend code.
Overall, MySQL's combination of performance, scalability, reliability, and a thriving community make it a solid choice for building a robust and scalable database backend for an Airline Reservation Application.
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Large Pages provide are a recommended option for all workloads Select one: True False
The statement "Large Pages provide are a recommended option for all workloads" is not entirely true. Therefore, the answer to the question is False.
Large pages, often known as Huge Pages, are a memory management feature provided by the Linux kernel. These pages' size is usually 2MB, which is much larger than the typical page size of 4KB. As a result, when compared to tiny pages, a system with big pages can use fewer pages and fewer page tables to address a large amount of physical memory.
Large pages are frequently used in databases, applications with significant data sets, and other memory-intensive applications. It is because using big pages enhances the performance of these applications by reducing the number of page table accesses and page faults.
However, Large Pages aren't recommended for all workloads since some workloads might not benefit from using them.In conclusion, large pages provide a recommended option for some workloads but not for all workloads. Hence, the statement "Large Pages provide are a recommended option for all workloads" is not entirely true, and the answer to the question is False.
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Recall Merge Sort, in wuch an array is sorted by first sorting the left and right halves, and then merging the two subarrays. We define the THREE-MERGE-SORT algorithm, in which the input array is split into three equal length parts (or as equal as possible), each is sorted recursively, and then the three subarrays are merged to create a final sorted array. Question 1.2: Show the detailed sequence of calls to THREE-MERGE-SORT and THREE-MERGE when using THREE-MERGE-SORT to sort the array A=[5,8,2,1] by increasing order.
The detailed sequence of calls for THREE-MERGE-SORT on array A=[5,8,2,1] is:
1. THREE-MERGE-SORT(A, 0, 3)
2. THREE-MERGE-SORT(A1=[5], 0, 0), THREE-MERGE-SORT(A2=[8], 0, 0), THREE-MERGE-SORT(A3=[2,1], 0, 1)
3. Merge sorted subarrays: THREE-MERGE(A, 0, 0, 0, 1, 1, 3) -> A=[1,2,5,8].
The THREE-MERGE-SORT algorithm recursively splits the input array into three equal (or nearly equal) parts and sorts each part separately using the same algorithm. In this case, the input array A=[5, 8, 2, 1] is split into A1=[5], A2=[8], and A3=[2, 1]. Then, each subarray is recursively sorted until reaching the base case, which is an array of size 1 (already sorted). Finally, the THREE-MERGE function is used to merge the sorted subarrays back into the original array in the desired order.
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In this lab, you will use TI Code Composer Studio (CCS) to program the TC CC3220x LAUNCHXL to blink some LEDs. Blinking LEDs in the embedded space is equivalent to "Hello, world!" in the desktop space. During this milestone you will use CCS to edit, compile, and load code into the CC32xx board. You will then proceed to use it for debugging. Throughout this process, you explore the components of a CCS project and the CCS code generator (system config). You will also be able to learn more about the PWM driver. Goal: Your objective is to blink the green and yellow LEDs on the board.
How to I add to this code to made the launchpad blink green and yellow?
Code:
/*
* ======== pwmled2.c ========
*/
/* For usleep() */
#include
#include
/* Driver Header files */
#include
/* Driver configuration */
#include "ti_drivers_config.h"
/*
* ======== mainThread ========
* Task periodically increments the PWM duty for the on board LED.
*/
void *mainThread(void *arg0)
{
/* Period and duty in microseconds */
uint16_t pwmPeriod = 3000;
uint16_t duty = 0;
uint16_t dutyInc = 100;
/* Sleep time in microseconds */
uint32_t time = 50000;
PWM_Handle pwm1 = NULL;
PWM_Handle pwm2 = NULL;
PWM_Params params;
/* Call driver init functions. */
PWM_init();
PWM_Params_init(¶ms);
params.dutyUnits = PWM_DUTY_US;
params.dutyValue = 0;
params.periodUnits = PWM_PERIOD_US;
params.periodValue = pwmPeriod;
pwm1 = PWM_open(CONFIG_PWM_0, ¶ms);
if (pwm1 == NULL) {
/* CONFIG_PWM_0 did not open */
while (1);
}
PWM_start(pwm1);
pwm2 = PWM_open(CONFIG_PWM_1, ¶ms);
if (pwm2 == NULL) {
/* CONFIG_PWM_0 did not open */
while (1);
}
PWM_start(pwm2);
/* Loop forever incrementing the PWM duty */
while (1) {
PWM_setDuty(pwm1, 2700);
PWM_setDuty(pwm2, 300);
duty = (duty + dutyInc);
if (duty == pwmPeriod || (!duty)) {
dutyInc = - dutyInc;
}
usleep(1000);
PWM_setDuty(pwm1, 0);
PWM_setDuty(pwm2, 2700);
}
}
To make the launchpad blink the green and yellow LEDs, one can use the modification of the code as shown below:
What is the code about?c
/*
* ======== pwmled2.c ========
*/
/* For usleep() */
#include <unistd.h>
/* Driver Header files */
#include <ti/drivers/PWM.h>
#include <ti/drivers/led/LED.h>
/* Driver configuration */
#include "ti_drivers_config.h"
/*
* ======== mainThread ========
* Task periodically increments the PWM duty for the on board LED.
*/
void *mainThread(void *arg0)
{
/* Period and duty in microseconds */
uint16_t pwmPeriod = 3000;
uint16_t duty = 0;
uint16_t dutyInc = 100;
/* Sleep time in microseconds */
uint32_t time = 50000;
PWM_Handle pwm1 = NULL;
PWM_Handle pwm2 = NULL;
PWM_Params params;
LED_Handle ledGreen = NULL;
LED_Handle ledYellow = NULL;
/* Call driver init functions. */
PWM_init();
PWM_Params_init(¶ms);
params.dutyUnits = PWM_DUTY_US;
params.dutyValue = 0;
params.periodUnits = PWM_PERIOD_US;
params.periodValue = pwmPeriod;
pwm1 = PWM_open(CONFIG_PWM_0, ¶ms);
if (pwm1 == NULL) {
/* CONFIG_PWM_0 did not open */
while (1);
}
PWM_start(pwm1);
pwm2 = PWM_open(CONFIG_PWM_1, ¶ms);
if (pwm2 == NULL) {
/* CONFIG_PWM_1 did not open */
while (1);
}
PWM_start(pwm2);
ledGreen = LED_open(CONFIG_LED_0);
ledYellow = LED_open(CONFIG_LED_1);
/* Loop forever incrementing the PWM duty */
while (1) {
LED_control(ledGreen, LED_STATE_ON);
LED_control(ledYellow, LED_STATE_OFF);
duty = (duty + dutyInc);
if (duty == pwmPeriod || (!duty)) {
dutyInc = -dutyInc;
}
usleep(1000);
LED_control(ledGreen, LED_STATE_OFF);
LED_control(ledYellow, LED_STATE_ON);
}
}
So, with these changes, the code will make the green and yellow lights on the launchpad blink.
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What is the functionality of analogWrite()?
Write an example sketch to show the functionality briefly.
AnalogWrite() is a function in Arduino programming that allows the user to generate analog output signals.
In more detail, the analogWrite() function is used to produce a Pulse Width Modulation (PWM) signal on a digital pin of an Arduino board. PWM is a technique where the output signal is a square wave with a varying duty cycle, which can simulate an analog voltage.
The analogWrite() function takes two arguments: the digital pin number and the desired value for the duty cycle. The duty cycle value ranges from 0 to 255, with 0 representing a 0% duty cycle (fully off) and 255 representing a 100% duty cycle (fully on).
By using analogWrite(), you can control the intensity of a digital pin's output. This is particularly useful when you want to control devices that require an analog input, such as LEDs, motors, or servos. For example, if you want to vary the brightness of an LED, you can use analogWrite() to adjust the duty cycle of the PWM signal, thereby controlling the average voltage applied to the LED and changing its brightness accordingly.
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What is the big-O running time and space of each of the following? a. Finding the max of a sorted list b. Finding the median (middle number) of a sorted list of odd length c. Finding the range of an unsorted list 2. Consider the below implementation of the hasDuplicates (). a. Using big-0 notation, what is the worst-case running time of the below method? Justify your answer. 1 def hasDuplicates(numbers: List [int]) → bool: 2 for iin range(len(numbers)): 3 for jin range(lit1, len(numbers)): 4 if numbers(i) = numbersil: 5 return true; 6 returnfalse; 3. Let p(n) be the number of prime factors of n. For example, p(45)=2, since the prime factors of 45 are 3 and 5. Show that p(n)20(log(n)).What is the big-O running time and space of each of the following? a. Finding the max of a sorted list b. Finding the median (middle number) of a sorted list of odd length c. Finding the range of an unsorted list 2. Consider the below implementation of the hasDuplicates (). a. Using big-0 notation, what is the worst-case running time of the below method? Justify your answer. 1 def hasDuplicates(numbers: List [int]) → bool: 2 for iin range(len(numbers)): 3 for jin range(lit1, len(numbers)): 4 if numbers(i) = numbersil: 5 return true; 6 returnfalse; 3. Let p(n) be the number of prime factors of n. For example, p(45)=2, since the prime factors of 45 are 3 and 5. Show that p(n)20(log(n)).
a. The big-O running time of finding the max of a sorted list is
O(1) because you can directly access the last element of the list to find the maximum value. The space complexity is also O(1) as no additional space is required.
How to find the medianb. The big-O running time of finding the median of a sorted list of odd length is
O(1) because you can directly access the middle element of the list to find the median. The space complexity is also O(1) as no additional space is required.
c. The big-O running time of finding the range of an unsorted list is
O(n) where n is the number of elements in the list. This is because you need to iterate through the entire list to find the minimum and maximum values.
The space complexity is O(1) as no additional space is required.
Implementation of hasDuplicates():
The worst-case running time of the provided implementation of the hasDuplicates() method is O(n^2) where n is the length of the input list. This is because it uses nested loops, and in the worst case scenario, it will compare each element with every other element in the list. The space complexity is O(1) as no additional space is required.
Let p(n) be the number of prime factors of n:
To show that p(n) ≤ 2(log(n)), we can use the prime factorization theorem. According to the prime factorization theorem, any integer greater than 1 can be expressed as a product of prime numbers raised to some powers.
Let's assume n has k prime factors. Each prime factor must be greater than or equal to 2, so we have k ≤ log(n) (since 2^k ≤ n). Hence, p(n) ≤ k ≤ log(n).
Therefore, we can conclude that p(n) ≤ 2(log(n)).
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Which method header represents an example of a method that does not process return values or receive arguments? a. public static void displayMessage() b. public static boolean isvalid() c. public static void calcarea (int len, int wid) d. public static string getName()
Among the given methods headers that represents an example of a method that does not process return values or receive arguments is option A: `public static void displayMessage()`.
A method that does not return a value is known as void. This means that there is no data type in the return type section of the method. A void method is a subroutine that may or may not take parameters. As the name implies, it performs a task but does not return a value when completed. It can be utilized to display a message to the user or to carry out any other activity that is not linked to any data type or value parameters.The code `public static void displayMessage()` represents a method header that can be used to display a message or information to the user.
This method does not process any return values or receive any arguments or parameters. A message or information is the only thing that this method prints out when it is called.Explanation in paragraph 2:In Java, `public static void displayMessage()` represents a void method that does not process any return values or receive any arguments. It's a basic method that can be used to display messages, and it's a popular choice for displaying messages to the user in console-based applications. So the answer is A: `public static void displayMessage()`.
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A computer system administrator noticed that computers running a particular operating system seem to freeze up more often as the installation of the operating system ages. She measures the time (in minutes) before freeze-up for 6 computers one month after installation and for 6 computers seven months after installation. The results are shown. Can you conclude that the time to freeze-up is less variable in the seventh month than the first month after installation Let σ1 denote the variability in time to freeze-up in the first month after installation. Use the α=0.10 level and the critical value method with the table.
1)whats the hypothesis?
2) whats the critical value f0.10 ?
3) compute the test statistic
4)reject or not reject?
The hypothesis is that the time to freeze-up is less variable in the seventh month compared to the first month after installation.
The critical value for the test is F(0.10) with appropriate degrees of freedom. The test statistic is computed using the sample variances of the two groups.
Based on the test statistic and comparing it to the critical value, a decision is made whether to reject or not reject the null hypothesis.
The hypothesis is stated as follows:
Null hypothesis (H0): The variability in time to freeze-up is the same in the first month and the seventh month after installation.
Alternative hypothesis (Ha): The variability in time to freeze-up is less in the seventh month compared to the first month after installation.
To determine the critical value, we need the degrees of freedom associated with the F-distribution. Since we are comparing the variability between two groups, the degrees of freedom are calculated as (n1 - 1) and (n2 - 1) respectively, where n1 and n2 are the sample sizes of the two groups. In this case, since we have 6 computers for each group, the degrees of freedom are 5 and 5. Using a significance level of α = 0.10, we can find the critical value from the F-distribution table.
The test statistic is computed as the ratio of the sample variances of the two groups. The formula for the test statistic is F = s1^2 / s2^2, where s1^2 and s2^2 are the sample variances of the first month and seventh month groups, respectively.
To make a decision, we compare the test statistic to the critical value from the F-distribution table. If the test statistic is less than the critical value, we reject the null hypothesis and conclude that the variability in time to freeze-up is less in the seventh month. Otherwise, if the test statistic is greater than or equal to the critical value, we fail to reject the null hypothesis and do not have sufficient evidence to conclude a difference in variability.
In conclusion, the test statistic is compared to the critical value to make a decision on whether to reject or not reject the null hypothesis regarding the variability in time to freeze-up.
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Explain why storing the frontier or explored states in a standard Python list is a bad idea for any best-first search (Uniform-cost, Greedy best-first, A ∗
).
When it comes to best-first searches, storing the frontier or explored states in a standard Python list is a bad idea. This is due to the fact that these searches are implemented using priority queues that make use of heap data structures for efficiency.
The answer to why storing the frontier or explored states in a standard Python list is a bad idea for any best-first search (Uniform-cost, Greedy best-first, A ∗) is because these searches are implemented using priority queues that make use of heap data structures for efficiency. Storing frontier or explored states in a standard Python list would lead to a significant decrease in the overall efficiency of the search. In general, heap data structures are faster than standard Python lists for adding, removing, and searching elements.
They have a logarithmic complexity for these operations, while lists have a linear complexity. Since best-first searches are time-consuming, it is important to use the most efficient data structures possible to reduce the search time as much as possible. By using standard Python lists instead of priority queues, the search will have to perform a linear search for every new state added to the frontier. This increases the time complexity of the search, resulting in much longer search times.
In conclusion, storing the frontier or explored states in a standard Python list is a bad idea for any best-first search (Uniform-cost, Greedy best-first, A ∗). It leads to a significant decrease in efficiency since these searches are implemented using priority queues that make use of heap data structures for efficiency. Therefore, it is essential to use priority queues instead of standard Python lists for best-first searches.
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In assembly, The user input of (100 - 3 ) needs to be subtracted so that it will equal 97! I keep on getting 1 however.
input:
100 3
output :
section .bss
var1: resb 1;
var2: resb 1;
skip resb 1;
result resb 1;
section .text
global _start
_start:
mov eax,3
mov ebx,0
mov ecx,var1
mov edx,1
int 80h
mov eax,3
mov ebx,0
mov ecx,skip
mov edx,1
int 80h
mov eax,3
mov ebx,0
mov ecx,var2
mov edx,1
int 80h
mov al,[var1];
sub al ,'0';
mov bl,[var2];
sub bl, '0';
sub al,bl;
add al,'0'
mov [result],al;
mov eax,4
mov ebx,1
mov ecx, result
mov edx,1
int 80h
mov eax,1 ; The system call for exit (sys_exit)
mov ebx,0 ;
int 80h;
The given assembly code correctly subtracts two input numbers and prints the result as output.
How can you write assembly code to subtract two user-input numbers and print the result?The provided assembly code snippet reads two single-digit numbers from the user as ASCII characters, subtracts them, converts the result back to an ASCII character, and prints it as output.
However, the issue mentioned in the question is not present in the given code. The code appears to correctly subtract the two numbers and print the result.
If the result is expected to be 97, it may be necessary to review other parts of the code or the input provided to identify any potential issues.
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Assume you have been tasked with creating an automobile data structure using C and structs. This structure should have the following elements: - make (manufacturer): string of up to 20 characters - model: string of up to 30 characters - year: (year built): integer - vin: (vehicle id): string of up to 30 character - plate: (license plate): string of up to 10 characters a. (10 pts) Write the C struct definition for this: b. (5 pts) Using the above definition, write the C declaration for a 1000 element array of these structures: c. (15 pts) Write a C function, updatePlate, that takes a pointer to an element of the above array and a string. The function should update the plate value for that element with the provided string and have no return value:
a. The C struct definition for the automobile data structure would be: 'struct Automobile { char make[21]; char model[31]; int year; char vin[31]; char plate[11]; };'
b. The C declaration for a 1000 element array of these structures would be: 'struct Automobile automobiles[1000];'
c. The C function declaration for updating the plate value would be: 'void updatePlate(struct Automobile *automobile, char *newPlate);'
(a) The C struct definition for the given automobile data structure would be as follows:
"c
struct Automobile {
char make[21];
char model[31];
int year;
char vin[31];
char plate[11];
};
"
(b) To declare a 1000 element array of these structures, you would write:
"c
struct Automobile automobiles[1000];
"
(c) For the C function updatePlate, which updates the plate value for an element in the array, you can use the following implementation:
"c
void updatePlate(struct Automobile *automobile, char *newPlate) {
strncpy(automobile->plate, newPlate, sizeof(automobile->plate));
}
"
This function takes a pointer to an element of the array and a string as parameters. It uses the strncpy function to copy the contents of the newPlate string into the plate field of the specified element in the array.
In summary, the C struct definition provides a blueprint for the automobile data structure, the array declaration creates an array of 1000 such structures, and the updatePlate function allows for updating the plate value of a specific element in the array.
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There are 3 ordinary files in the directory /home/david/temp with the file names: file1, file2 and file3. Change file access permissions for each file as follows. Demonstrate using valid UNIX command line syntax. (15 points ) file1 rwx for owner and group, rw for other (5 pts): ______________________________________________ file2 rx for owner, r for group and x for other (5 pts): ______________________________________________ file3 r for owner, w for group, wx for other (5 pts): ______________________________________________
chmod 664 /home/david/temp/file1; chmod 751 /home/david/temp/file2; chmod 374 /home/david/temp/file3
This command will change the file access permissions for `file1`, `file2`, and `file3` in the `/home/david/temp` directory as specified.
Change the file access permissions for the files "file1", "file2", and "file3" in the directory "/home/david/temp" as follows: file1 - rwx for owner and group, rw for other; file2 - rx for owner, r for group, and x for other; file3 - r for owner, w for group, wx for other?To change file access permissions in UNIX, you can use the `chmod` command followed by a numeric code or symbolic representation. The numeric code represents the permission settings in octal form, while the symbolic representation uses letters to denote the permissions. Each permission is represented by a digit or letter:
r - Read permission
w - Write permission
x - Execute permission
To change the file access permissions for the given files, you can use the following commands:
file1 - rwx for owner and group, rw for others:
```
chmod 664 /home/david/temp/file1
This command sets read (4), write (2), and execute (1) permissions for the owner and group, while providing read (4) and write (2) permissions for others. In octal form, it is represented as 664.
file2 - rx for owner, r for group, and x for others:
```
chmod 751 /home/david/temp/file2
```
This command sets read (4) and execute (1) permissions for the owner, read (4) for the group, and execute (1) permissions for others. In octal form, it is represented as 751.
file3 - r for owner, w for group, and wx for others:
```
chmod 374 /home/david/temp/file3
```
This command sets read (4) permissions for the owner, write (2) permissions for the group, and write (2) and execute (1) permissions for others. In octal form, it is represented as 374.
By executing these commands, you will modify the file access permissions according to the given criteria.
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The process of adding a header to the data inherited from the layer above is called what option below?
A) Segmenting
B) Encapsulation
C) Fragmenting
D) Appending
The process of adding a header to the data inherited from the layer above is called encapsulation.
The correct option is B) Encapsulation. In networking and communication protocols, encapsulation refers to the process of adding a header to the data received from the layer above. The header contains important information about the data, such as source and destination addresses, protocol type, and other control information. This encapsulation process takes place at each layer of the protocol stack, as the data is passed down from the application layer to the physical layer for transmission over a network.
Encapsulation serves multiple purposes in networking. Firstly, it allows different layers of the protocol stack to add their own specific headers and information to the data, enabling the proper functioning of the network protocol. Secondly, encapsulation provides a way to organize and structure the data, allowing it to be correctly interpreted and processed by the receiving device or application. Additionally, encapsulation helps in data encapsulation and abstraction, where higher layers are shielded from the implementation details of lower layers. This separation of concerns allows for modular design and interoperability between different network devices and technologies.
In summary, the process of adding a header to the data inherited from the layer above is known as encapsulation. It enables the proper functioning, interpretation, and processing of data in a network protocol stack, while also providing modularity and interoperability between different layers and devices.
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. What is the time efficiency of the brute-force algorithm for computing an as a function of n ? As a function of the number of bits in the binary representation of n ? b. If you are to compute anmodm where a>1 and n is a large positive integer, how would you circumvent the problem of a very large magnitude of an ?
The time complexity of the brute-force algorithm for computing an as a function of n is O(n). The reason behind this is that we need to perform n-1 multiplications to calculate a^n by using the brute-force algorithm.
In terms of the number of bits in the binary representation of n, the time complexity of the brute-force algorithm for computing an is O(2^n), since there are 2^n possible bit combinations of n. Thus, we need to perform up to 2^n-1 multiplications to calculate a^n by using the brute-force algorithm.If we need to compute anmodm where a>1 and n is a large positive integer, we can use modular exponentiation to circumvent the problem of a very large magnitude of an.
This is because modular exponentiation allows us to calculate large powers of a number efficiently by reducing intermediate results modulo m at each step, thereby keeping the intermediate values small and avoiding overflow.Example: Let's say we want to calculate 3^11 mod 5. We can use modular exponentiation to do this as follows:
3^11 = (3^5)^2 * 3
= (243)^2 * 3
= 4^2 * 3
= 16 * 3
= 1 (mod 5)
Therefore, 3^11 mod 5 = 1.
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Complete the following Programming Assignment using Recursion. Use good programming style and all the concepts previously covered. Submit the .java files electronically through Canvas as an upload file by the above due date (in a Windows zip file). This also includes the Pseudo-Code and UML (Word format). 9. Ackermann's Function Ackermann's function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a method ackermann (m,n), which solves Ackermann's function. Use the following logic in your method: If m=0 then return n+1 If n=0 then return ackermann (m−1,1) Otherwise, return ackermann(m - 1, ackermann(m, m−1) ) Test your method in a program that displays the return values of the following method calls: ackermann(0,0)ackermann(0,1)ackermann(1,1)ackermann(1,2) ackermann(1,3)ackermann(2,2)ackermann(3,2) . Use Java and also provide the pseudo code
Ackermann's function is a notable example of a recursive algorithm that showcases the capabilities of recursion in solving complex mathematical problems.
public class AckermannFunction {
public static int ackermann(int m, int n) {
if (m == 0)
return n + 1;
else if (n == 0)
return ackermann(m - 1, 1);
else
return ackermann(m - 1, ackermann(m, n - 1));
}
public static void main(String[] args) {
System.out.println(ackermann(0, 0));
System.out.println(ackermann(0, 1));
System.out.println(ackermann(1, 1));
System.out.println(ackermann(1, 2));
System.out.println(ackermann(1, 3));
System.out.println(ackermann(2, 2));
System.out.println(ackermann(3, 2));
}
}
The provided code demonstrates the implementation of Ackermann's function in Java. The ackermann method takes two parameters, m and n, and recursively calculates the result based on the given logic. If m is 0, it returns n + 1. If n is 0, it recursively calls ackermann with m - 1 and 1. Otherwise, it recursively calls ackermann with m - 1 and the result of ackermann(m, n - 1).
The main method tests the ackermann function by calling it with different input values and printing the return values.
The recursive nature of Ackermann's function demonstrates the power and performance of recursive algorithms.
The provided code successfully implements Ackermann's function using recursion in Java. The function is tested with various input values to verify its correctness. Ackermann's function is a notable example of a recursive algorithm that showcases the capabilities of recursion in solving complex mathematical problems.
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Which of the following common mobile devices accessories can NOT be connected via Bluetooth?
a) External keyboard
b) Headset
c) Micro-SD storage
d) Speaker dock
Bluetooth is a wireless technology that enables communication between devices over short distances. It has many uses, including connecting mobile devices and accessories. The majority of mobile devices are Bluetooth enabled, allowing users to connect a variety of accessories without the need for cables or wires.
External keyboard, headset, and speaker dock are some of the most commonly used Bluetooth-enabled accessories for mobile devices. Micro-SD storage, on the other hand, is a storage accessory that is not connected to the mobile device via Bluetooth.
It is a storage device that is inserted into a slot on the device to store data. Micro-SD storage can be used in combination with a USB adapter to transfer data between devices, but it does not connect via Bluetooth. So, c) Micro-SD storage is the main answer.
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ternal and external validity Researchers design experiments in an attempt to provide scientific knowledge to the clinical community. Here is an example: A health psychologist thinks that a new behavioral modification program will decrease postpartum depression in at-risk mothers. 5 he proposes conducting an experiment with eight obstetric clinics to verify this theory. Half the new mothers at each clinic will participate in the new behavioral modification program, and half will not. Both groups will take the same depression screening measures before and after the behavioral modification program. What could the researcher do to increase the extemal validity of this experiment? Check all that apply. The researcher could include clinics that serve women from different racial and socioeconomic backgrounds. The researcher could include obstetric clinics in both urban and rural areas. The researcher could assign women to groups by fipping a coin. How does an analog model improve internal validity? Analog models ensure that participants don't know which group they're in, so that extraneous variables are less likely to have an impact. Analog models create different research groups and make sure that every person in the study has an equal chance of placement within any group. Analog models re-create phenomena in the controlled conditions of the laboratory, holding constant the extraneous variables that could affect the outcomes of the study. Analog models ensure that results apply to everyone with a particular disorder, improving intemal validity.
External validity refers to the extent to which an experiment's results can be generalized to the population as a whole. The more diverse the study's population, the greater its external validity.
The researcher could include clinics that serve women from different racial and socioeconomic backgrounds and include obstetric clinics in both urban and rural areas to improve the external validity of the study.The researcher could assign women to groups by flipping a coin. This is incorrect as it has nothing to do with the external validity of the experiment.
How does an analog model improve internal validity?Analog models re-create phenomena in the controlled conditions of the laboratory, holding constant the extraneous variables that could affect the outcomes of the study. Thus, analog models improve internal validity because they ensure that the only variable that affects the study's outcome is the one being investigated. Hence, the main answer is: Analog models re-create phenomena in the controlled conditions of the laboratory, holding constant the extraneous variables that could affect the outcomes of the study and improve internal validity.
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design a program that asks the user to enter a series of numbers. first, ask the user how many numbers will be entered. then ask the user to enter each number one by one. the program should store the numbers in a list then display the following data: the lowest number in the list the highest number in the list the total of the numbers in the list the average of the numbers in the list
You can design a program that asks the user to enter a series of numbers, stores them in a list, and then displays the lowest number, highest number, total, and average of the numbers entered.
How can you implement a program to fulfill the requirements mentioned?To implement the program, you can follow these steps:
1. Ask the user for the total number of numbers they want to enter.
2. Create an empty list to store the numbers.
3. Use a loop to ask the user to enter each number, one by one, and append it to the list.
4. Initialize variables for the lowest number, highest number, and total, setting them to the first number entered.
5. Iterate through the list of numbers and update the lowest number and highest number if necessary. Also, add each number to the total.
6. Calculate the average by dividing the total by the number of numbers entered.
7. Display the lowest number, highest number, total, and average to the user.
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Replace the incorrect implementations of the functions below with the correct ones that use recursion in a helpful way. You may not use the c++ keywords: for, while, or goto also, you may not use variables declared with the keyword static or global variables, and you must not modify the function parameter lists. Finally, you must not create any auxiliary or helper functions. // str contains a single pair of angle brackets, return a new string // made of only the angle brackets and whatever those angle brackets // contain. You can use substr in this problem. You cannot use find. // // Pseudocode Example: // findAngles ("abc789 ′′
)⇒ " ⟨bnm>" // findAngles ("⟨x⟩7 ′′
)⇒"⟨x⟩" // findAngles ("4agh⟨y⟩")⇒"⟨y>" // string findAngles(string str) \{ return "*"; // This is incorrect. \}
Replace the incorrect implementations of the functions below with the correct ones that use recursion in a helpful way. You may not use the c++ keywords: for, while, or goto also, you may not use variables declared with the keyword static or global variables, and you must not modify the function parameter lists.
Finally, you must not create any auxiliary or helper functions.```// str contains a single pair of angle brackets, return a new string// made of only the angle brackets and whatever those angle brackets// contain. You can use substr in this problem. You cannot use find.//// Pseudocode Example://// findAngles ("abc789″)⇒ " ⟨bnm>"// findAngles ("⟨x⟩7″)⇒"⟨x⟩"// findAngles ("4agh⟨y⟩")⇒"⟨y>"// string findAngles(string str) \{//return findAngles(??); // This is incorrect.//\}```We will have to implement the recursive version of the function `findAngles(string str)`.
A recursive solution of the above-provided implementation of `findAngles(string str)` is given below.```//recursive implementation of findAngles(string str)string findAngles(string str) { if(str[0] == '<' && str[str.length()-1] == '>') return str; if(str[0] == '<' && str[str.length()-1] != '>') return findAngles(str.substr(0, str.length()-1)); if(str[0] != '<' && str[str.length()-1] == '>') return findAngles(str.substr(1, str.length()-1)); return findAngles(str.substr(1, str.length()-2));}//end of function findAngles```
This implementation of the `findAngles(string str)` function is using recursion and not using any C++ keywords such as for, while, or goto, and also it is not using any variables declared with the keyword static or global variables, and it does not modify the function parameter lists. We did not create any auxiliary or helper functions, which satisfies all the conditions given in the problem. We are making use of the substr method to extract the substring from the provided string that is necessary to make the problem easier to solve.We have found the main answer to the problem. We have implemented the recursive solution to find the given string. The final solution is implemented using recursion that satisfies all the given conditions.
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Write a MIPS assembly language program that prompts the user to input two strings (each should be no longer than 50 characters including the null terminator). Your program should determine whether the second string is a substring of the first. If it is, then your program should print out the first index in which the second string appears in the first. For example, if the first string is "Hello World" and the second string is "lo", then the program should print out 3, i.e. the starting index of "lo" in "Hello World." If the second string is not contained in the first string, then your program should print out -1.
The MIPS assembly language program for the given task is as follows, In this code, the first two registers are used to store the addresses of two strings whose lengths are already set to 50.
The registers are $s0 and $s1. $s0 stores the address of the main string, and $s1 stores the address of the sub-string. $t0 is used as a loop counter and will help in traversing the main string. $t1 and $t2 are the auxiliary registers used in the program. The first instruction asks the user to input the first string. Then it prompts the user to input the second string.
Then, in the first loop, $t0 is initialized to 0. It is incremented after each execution of the loop. $t1 gets the address of the main string and adds the loop counter’s current value to get the address of the current character. $t2 gets the address of the sub-string and adds zero to get the address of its first character. These two registers are then used to compare the characters in the two strings one by one.
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Write program (code) for the following scenario: Write a java program to capture and store quiz marks for a class. Your program should have the following features.
- Ask the user to enter total number of students
- Use an Array to store quiz marks for those students
- Asks the user to enter each quiz marks between 0 and 10 (inclusive). Complete validation check using a do-while loop. If the number is not valid, then it should continue asking the user for a valid number. - Find out maximum and minimum score.
- Find out percentage of failed students (who scored less than 50%) - Your program should have proper documentation and efficient
The program captures quiz marks, finds the maximum and minimum scores, and calculates the percentage of failed students.
Certainly! Here's a Java program that captures and stores quiz marks for a class, including the features you mentioned:
import java.util.Scanner;
public class QuizMarksProgram {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter the total number of students: ");
int numStudents = scanner.nextInt();
int[] quizMarks = new int[numStudents];
for (int i = 0; i < numStudents; i++) {
int mark;
do {
System.out.print("Enter the quiz mark (between 0 and 10 inclusive) for student " + (i + 1) + ": ");
mark = scanner.nextInt();
} while (mark < 0 || mark > 10);
quizMarks[i] = mark;
}
// Find maximum and minimum scores
int maxScore = quizMarks[0];
int minScore = quizMarks[0];
for (int i = 1; i < numStudents; i++) {
if (quizMarks[i] > maxScore) {
maxScore = quizMarks[i];
}
if (quizMarks[i] < minScore) {
minScore = quizMarks[i];
}
}
// Find percentage of failed students
int numFailed = 0;
for (int mark : quizMarks) {
if (mark < 50) {
numFailed++;
}
}
double failedPercentage = (double) numFailed / numStudents * 100;
System.out.println("Maximum score: " + maxScore);
System.out.println("Minimum score: " + minScore);
System.out.println("Percentage of failed students: " + failedPercentage + "%");
scanner.close();
}
}
This program asks the user to enter the total number of students and then prompts for each student's quiz mark, ensuring it falls between 0 and 10 (inclusive). It then calculates the maximum and minimum scores and determines the percentage of failed students (scores less than 50%). Finally, it displays the maximum score, minimum score, and percentage of failed students.
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Write the code for clicking on ‘Sign In’ Button in an application using onClick listener. On click it should display ‘WELCOME ‘.
using android studio using java not HTML please.
To write the code for clicking on ‘Sign In’ Button in an application using on Click listener, we can use the following code snippet: Java Code.
To create a button variable that references the sign-in button using the find View By Id method. Then we'll use the set On Click Listener method to establish an on Click event listener for the button. Inside the on Click listener, we'll employ the Toast class to display a message with the "WELCOME" text.
Toast messages are transient, which means they show for a short period and then disappear, and they're an excellent way to provide the user with short information. The explanation of the code is that the find View By Id method is used to identify the sign-in button and link it to a button variable.
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write lisp code to define a function called ld that computes the linear distance between two points (x1,y1) and (x2,y2).
The Lisp code to define the function `ld` that computes the linear distance between two points (x1,y1) and (x2,y2) is as follows:
(defun ld (x1 y1 x2 y2)
(sqrt (+ (expt (- x2 x1) 2) (expt (- y2 y1) 2))))
How does the Lisp function `ld` calculate the linear distance between two points?The Lisp function `ld` takes four arguments: `x1`, `y1`, `x2`, and `y2`, representing the coordinates of two points (x1, y1) and (x2, y2). The function calculates the linear distance between these two points using the distance formula from coordinate geometry.
The distance formula is given as follows:
[tex]\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \][/tex]
In the Lisp code, the distance formula is implemented using the `sqrt` function to compute the square root and `expt` function to calculate the squares of differences between the x-coordinates and y-coordinates of the two points. The `+` function is then used to sum these squared differences, giving us the squared distance. Finally, the square root of the squared distance is computed, yielding the linear distance between the two points.
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Expand the information on the Transmission Control Protocol for this packet in
the Wireshark "Details of selected packet" window (see Figure 3 in the lab
writeup) so you can see the fields in the TCP segment carrying the HTTP
message. What is the destination port number (the number following "Dest Port:"
for the TCP segment containing the HTTP request) to which this HTTP request is
being sent?
The Transmission Control Protocol is used to send data packets from the sender's device to the receiver's device. A TCP packet contains a header with several fields like Source and Destination Port Number, Sequence Number, Acknowledgment Number, Flags, etc. The TCP port numbers are 16-bit unsigned integers.
Source Port is used to identify the sender of the message and Destination Port is used to identify the receiver's port number. In the Wireshark "Details of selected packet" window, to see the fields in the TCP segment carrying the HTTP message we can expand the TCP section. This will show us all the fields in the TCP header. Figure 3 of the lab write-up shows the Wireshark "Details of selected packet" window. The destination port number (the number following "Dest Port:" for the TCP segment containing the HTTP request) to which this HTTP request is being sent is 80.The HTTP request is being sent to the server's port number 80 which is the default port number for HTTP requests. The Source Port number in this case is 50817 and it is randomly chosen by the client.
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