They can do different surgery on people. One is called cataract total and hip replacement surgery which is what can be expected
(Count the occurrences of words in a text file) Rewrite Listing 21.9 to read the text from a text file. The text file is passed as a command-line argument. Words are delimited by whitespace characters, punctuation marks (, ; . : ?), quotation marks (' "), and parentheses. Count the words in a case-sensitive fashion (e.g., consider Good and good to be the same word). The words must start with a letter. Display the output of words in alphabetical order, with each word preceded by the number of times it occurs.
I ran my version of the program solution on its own Java source file to produce the following sample output:
C:\Users\......\Desktop>java CountOccurrenceOfWords CountOccurrenceOfWords.java
Display words and their count in ascending order of the words
1 a
1 alphabetical
1 an
2 and
3 args
2 as
1 ascending
1 catch
1 class
4 count
2 countoccurrenceofwords
1 create
1 display
1 else
3 entry
3 entryset
2 ex
1 exception
1 exit
1 file
2 filename
3 for
1 fullfilename
3 get
1 getkey
1 getvalue
1 hasnext
1 hold
5 i
3 if
2 import
2 in
3 input
2 int
1 io
3 java
6 key
3 length
2 line
1 main
3 map
1 matches
3 new
1 nextline
1 null
1 of
2 order
3 out
3 println
1 printstacktrace
2 public
2 put
2 scanner
1 set
1 split
1 static
5 string
4 system
2 the
1 their
1 to
1 tolowercase
2 tree
6 treemap
2 trim
1 try
1 util
1 value
1 void
1 while
9 words
C:\Users\......\Desktop>
Answer:
Explanation:
The following is written in Java. It uses File input, to get the file and read every line in the file. It adds all the lines into a variable called fullString. Then it uses regex to split the string into separate words. Finally, it maps the words into an array that counts all the words and how many times they appear. A test case has been created and the output can be seen in the attached image below. Due to technical difficulties I have added the code as a txt file below.
Which attack form either exploits a software flaw or floods a system with traffic in order to prevent legitimate activities or transactions from occurring?
Answer:
Denial of service attack.
Explanation:
In Cybersecurity, vulnerability can be defined as any weakness, flaw or defect found in a software application or network and are exploitable by an attacker or hacker to gain an unauthorized access or privileges to sensitive data in a computer system.
This ultimately implies that, vulnerability in a network avail attackers or any threat agent the opportunity to leverage on the flaws, errors, weaknesses or defects found in order to compromise the security of the network.
Data theft can be defined as a cyber attack which typically involves an unauthorized access to a user's data with the sole intention to use for fraudulent purposes or illegal operations. There are several methods used by cyber criminals or hackers to obtain user data and these includes DDOS attack, SQL injection, man in the middle, phishing, sniffing, DOS attack, etc.
A denial of service (DOS) attack is a form of cyber attack that involves either the exploitation of a software flaw or overwhelmingly floods a computer system in a network with traffic in order to prevent authorized users from performing legitimate activities or transactions.
Some of the ways to prevent vulnerability or cyber attacks in a network are;
1. Ensure you use a very strong password with complexity through the use of alphanumerics.
2. You should use a two-way authentication service.
3. You should use encrypting software applications or services.
In the forward chaining technique, used by the inference engine component of an expert system, the _____ condition is evaluated first.
Answer:
"if" is the right response.
Explanation:
Forward chaining is being utilized just to split the logical sequential order down as well as operate throughout every component once the preceding one is accomplished, from start to finish.Its whole algorithm begins with proven information that activates all the rules, which are fulfilled as well as makes established realities even more complete.Write a program in Cto define a structure Patient that has the following members
Answer:
Explanation:
The answer should be 1.16 or 61.1
Create and configure databases in oracle database management system operation administration
Answer:
You typically create a database during Oracle Database software installation. However, you can also create a database after installation.
Reasons to create a database after installation are as follows:
You used Oracle Universal Installer (OUI) to install software only, and did not create a database. You want to create another database (and database instance) on the same host computer as an existing Oracle database. In this case, this chapter assumes that the new database uses the same Oracle home as the existing database. You can also create the database in a new Oracle home by running OUI again. You want to make a copy of (clone) a database.
While configuring a wireless access point device, a technician is presented with several security mode options. Which of the following options will provide the most secure access?
Answer:
Bro this is my area the answer is WPA2 and AES
Explanation:
its because AES is a more secure encryption protocol introduced with WPA2. AES isn't some creaky standard developed specifically for Wi-Fi networks, either. It's a serious worldwide encryption standard that's even been adopted by the US government.
There are many possible variations in wireless network design depending on the surroundings.
What is wireless network?Even in a complicated single site, different wireless networks running on the same hardware as part of the broader wireless LAN environment can have different network configuration fundamentals needed for a successful deployment.
At the same time, a number of essential network configuration activities are part of any successful wireless project's typical chores. Let's look at the broad and specific procedures that wireless experts should take while setting up enterprise-grade Wi-Fi networks.
Therefore, There are many possible variations in wireless network design depending on the surroundings.
To learn more about wifi, refer to the link:
https://brainly.com/question/6145118
#SPJ5
????????????????????????????????
(c) 4
Explanation:The function strpos() is a function in php that returns the position of the first occurrence of a particular substring in a string. Positions are counted from 0. The function receives two arguments. The first argument is the string from which the occurrence is to be checked. The second argument is the substring to be checked.
In this case, the string is "Get Well Soon!" while the substring is "Well"
Starting from 0, the substring "Well" can be found to begin at position 4 of the string "Get Well Soon!".
Therefore, the function strpos("Get Well Soon!", "Well") will return 4.
Then, with the echo statement, which is used for printing, value 4 will be printed as output of the code.
Indicate whether the following actions are the actions of a person who will be a victim, or will not be a victim, of phishing attacks.
Replying to an e-mail requesting your user ID and password Phishing victim Not a phishing victim
Answer:
Phishing Victim
Explanation:
Replying to this email could make you a victim of a phishing scam. Phishing attacks are common challenges on security that internet users are faced with. It could lead to others getting hold of sensitive information like your password or bank details. email is one way the hackers used to get this information. replying to such an email could lead one to an unsecure website where information could be extracted
1. What is memory mapped I/O?
Answer:
Memory-mapped I/O and port-mapped I/O are two complementary methods of performing input/output between the central processing unit and peripheral devices in a computer. An alternative approach is using dedicated I/O processors, commonly known as channels on mainframe computers, which execute their own instructions.
Modeling and simulation can enhance the Systems Engineering process by:__________.
a. Helping IPT members to understand the interrelationship of components without physically changing the system.
b. Providing designs that will consistently exceed functional capabilities.
c. Quickly providing physical solutions to meet functional capabilities.
d. Eliminating technical risk before production begins.
Base conversion. Perform the following conversion (you must have to show the steps to get any credit.
3DF16 = ?
Base to be converted to was not included but we would assume conversion to base 10(decimal)
Answer and Explanation:
The 3DF in base 16 is a hex number so we need to convert to its equivalent binary/base 2 form:
We therefore each digit of given hex number 3DF in base 16 to equivalent binary, 4 digits each.
3 = 0011
D = 1101
F = 1111
We then arrange the binary numbers in order
3DF base 16 = 1111011111 in base 2
We then convert to base 10 =
= 1x2^9+1x2^8+1x2^7+1x2^6+0x2^5+1x2^4+1x2^3+1x2^2+1×2^1+1×2^0
= 991 in base 10
Given an array of n distinct integers sorted in ascending order, write a function that returns a Fixed Point in the array, if there is any Fixed Point present in array, else returns -1. Fixed Point in an array is an index i such that arr[i] is equal to i. Note that integers in array can be negative. Examples: #include using namespace std; void removeDivByFive(int *arr,int& size) { int index=0; for(int i=0;i
Answer:
The function is as follows:
int returnsFixed(int arr [],int n){
int retVal = -1;
for(int i = 0; i<n;i++){
if(i == arr[i]){
retVal = i;
break;
}
}
return retVal;
}
Explanation:
This defines the functionl it receives the array and the length of the array
int returnsFixed(int arr [],int n){
This initializes the return value to -1
int retVal = -1;
This iterates through the array
for(int i = 0; i<n;i++){
This checks if i equals arr[i]
if(i == arr[i]){
If yes, the return value (i.e. the fixed point) is set to i
retVal = i;
And the code is exited
break;
} This ends the if condition
} This ends the iteration
This returns the calculated fixed point
return retVal;
}
Cardinality ratios often dictate the detailed design of a database. The cardinality ratio depends on the real-world meaning of the entity types involved and is defined by the specific application. For the binary relationships below, suggest cardinality ratios based on the common-sense meaning of the entity types. Clearly state any assumptions you make.
Entity 1 Cardinality Ratio Entity 2
1. Student SocialSecurityCard
2. Student Teacher
3. ClassRoom Wall
4. Country CurrentPresident
5. Course TextBook
6. Item (that can
be found in an
order) Order
7. Student Class
8. Class Instructor
9. Instructor Office
10. E-bay Auction item E-bay bid
Solution :
ENTITY 1 CARDINALITY RATIO ENTITY 2
1. Student 1 to many Social security card
A student may have more than one
social security card (legally with the
same unique social security number),
and every social security number belongs
to a unique.
2. Student Many to Many Teacher
Generally students are taught by many
teachers and a teacher teaches many students.
3.ClassRoom Many to Many Wall
Do not forget that the wall is usually
shared by adjacent rooms.
4. Country 1 to 1 Current President
Assuming a normal country under normal
circumstances having one president at a
time.
5. Course Many to Many TextBook
A course may have many textbooks and
text book may be prescribed for different
courses.
6. Item Many to Many Order
Assuming the same item can appear
in different orders.
7. Student Many to Many Class
One student may take several classes.
Every class usually has several students.
8. Class Many to 1 Instructor
Assuming that every class has a unique
instructor. In case instructors were allowed
to team teach, this will be many-many.
9. Instructor 1 to 1 Office
Assuming every instructor has only one
office and it is not shared. In case of offices
shared by 2 instructors, the relationship
will be 2-1. Conversely, if any instructor has a joint
appointment and offices in both departments,
then the relationship will be 1-2. In a very general
case, it may be many-many.
10. E-bay Auction item 1-Many E-bay bid
1 item has many bids and a bid is unique
to an item.
Write a removeDuplicates() method for the LinkedList class we saw in lecture. The method will remove all duplicate elements from the LinkedList by removing the second and subsequent elements. If there is only one copy of an element, it is not removed. The list size should shrink accordingly based on the number of duplicates removed.
Answer:
removeDuplicates() function:-
//removeDuplicates() function removes duplicate elements form linked list.
void removeDuplicates() {
//declare 3 ListNode pointers ptr1,ptr2 and duplicate.
//initially, all points to null.
ListNode ptr1 = null, ptr2 = null, duplicate = null;
//make ptr1 equals to head.
ptr1 = head;
//run while loop till ptr1 points to second last node.
//pick elements one by one..
while (ptr1 != null && ptr1.next != null)
{
// make ptr2 equals to ptr1.
//or make ptr2 points to same node as ptr1.
ptr2 = ptr1;
//run second while loop to compare all elements with above selected element(ptr1->val).
while (ptr2.next != null)
{
//if element pointed by ptr1 is same as element pointed by ptr2.next.
//Then, we have found duplicate element.
//Now , we have to remove this duplicate element.
if (ptr1.val == ptr2.next.val)
{
//make duplicate pointer points to node where ptr2.next points(duplicate node).
duplicate = ptr2.next;
//change links to remove duplicate node from linked list.
//make ptr2.next points to duplicate.next.
ptr2.next = duplicate.next;
}
//if element pointed by ptr1 is different from element pointed by ptr2.next.
//then it is not duplicate element.
//So, move ptr2 = ptr2.next.
else
{
ptr2 = ptr2.next;
}
}
//move ptr1 = ptr1.next, after check duplicate elements for first node.
//Now, we check duplicacy for second node and so on.
//so, move ptr1 by one node.
ptr1 = ptr1.next;
}
}
Explanation:
Complete Code:-
//Create Linked List Class.
class LinkedList {
//Create head pointer.
static ListNode head;
//define structure of ListNode.
//it has int val(data) and pointer to ListNode i.e, next.
static class ListNode {
int val;
ListNode next;
//constructor to create and initialize a node.
ListNode(int d) {
val = d;
next = null;
}
}
//removeDuplicates() function removes duplicate elements form linked list.
void removeDuplicates() {
//declare 3 ListNode pointers ptr1,ptr2 and duplicate.
//initially, all points to null.
ListNode ptr1 = null, ptr2 = null, duplicate = null;
//make ptr1 equals to head.
ptr1 = head;
//run while loop till ptr1 points to second last node.
//pick elements one by one..
while (ptr1 != null && ptr1.next != null)
{
// make ptr2 equals to ptr1.
//or make ptr2 points to same node as ptr1.
ptr2 = ptr1;
//run second while loop to compare all elements with above selected element(ptr1->val).
while (ptr2.next != null)
{
//if element pointed by ptr1 is same as element pointed by ptr2.next.
//Then, we have found duplicate element.
//Now , we have to remove this duplicate element.
if (ptr1.val == ptr2.next.val)
{
//make duplicate pointer points to node where ptr2.next points(duplicate node).
duplicate = ptr2.next;
//change links to remove duplicate node from linked list.
//make ptr2.next points to duplicate.next.
ptr2.next = duplicate.next;
}
//if element pointed by ptr1 is different from element pointed by ptr2.next.
//then it is not duplicate element.
//So, move ptr2 = ptr2.next.
else
{
ptr2 = ptr2.next;
}
}
//move ptr1 = ptr1.next, after check duplicate elements for first node.
//Now, we check duplicacy for second node and so on.
//so, move ptr1 by one node.
ptr1 = ptr1.next;
}
}
//display() function prints linked list.
void display(ListNode node)
{
//run while loop till last node.
while (node != null)
{
//print node value of current node.
System.out.print(node.val + " ");
//move node pointer by one node.
node = node.next;
}
}
public static void main(String[] args) {
//Create object of Linked List class.
LinkedList list = new LinkedList();
//first we create nodes and connect them to form a linked list.
//Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.
//Create a Node having node data = 1 and assign head pointer to it.
//As head is listNode of static type. so, we call head pointer using class Name instead of object name.
LinkedList.head = new ListNode(1);
//Create a Node having node data = 2 and assign head.next to it.
LinkedList.head.next = new ListNode(2);
LinkedList.head.next.next = new ListNode(3);
LinkedList.head.next.next.next = new ListNode(2);
LinkedList.head.next.next.next.next = new ListNode(4);
LinkedList.head.next.next.next.next.next = new ListNode(2);
LinkedList.head.next.next.next.next.next.next = new ListNode(5);
LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);
//display linked list before Removing duplicates.
System.out.println("Linked List before removing duplicates : ");
list.display(head);
//call removeDuplicates() function to remove duplicates from linked list.
list.removeDuplicates();
System.out.println("")
//display linked list after Removing duplicates.
System.out.println("Linked List after removing duplicates : ");
list.display(head);
}
}
Output:-
It is common for people to name directories as dir1, dir2, and so on. When there are ten or more directories, the operating system displays them in dictionary order, as dir1, dir10, dir11, dir12, dir2, dir3, and so on. That is irritating, and it is easy to fix. Provide a comparator that compares strings that end in digit sequences in a way that makes sense to a human. First compare the part before the digit as strings, and then compare the numeric values of the digits.
Your program should work with the provided test program
DirectorySortDemo.java.
Call the class you write DirectoryComparator.java.
Submit the two files in your submission.
DirectoryComparator.java
DirectorySortDemo.java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.collections;
public class DirectorySortDemo
{
public static void main (String [] args)
{
String dirNames ("dir12", "dir5", "dir9", "dirl", "dir4",
"lab10", "1ab2", "lab7", "lab17", "lab8",
"quiz8", "quiz10", "quiz11", "quiz12",
"dirll", "dir8", "dir7", "dir15", "dir3");
ArrayList directories = new
ArrayList<> (Arrays.asList (dirNames));
System.out.println ("Unsorted List:");
System.out.println (directories);
Collections.sort (directories, new DirectoryComparator ());
System.out.println ():
System.out-println ("Sorted List: ");
System.out.-println (directories);
Answer:
Here the code is given as follows,
Explanation:
In python,
Here's some fake data.
df = {'country': ['US', 'US', 'US', 'US', 'UK', 'UK', 'UK'],
'year': [2008, 2009, 2010, 2011, 2008, 2009, 2010],
'Happiness': [4.64, 4.42, 3.25, 3.08, 3.66, 4.08, 4.09],
'Positive': [0.85, 0.7, 0.54, 0.07, 0.1, 0.92, 0.94],
'Negative': [0.49, 0.09, 0.12, 0.32, 0.43, 0.21, 0.31],
'LogGDP': [8.66, 8.23, 7.29, 8.3, 8.27, 6.38, 6.09],
'Support': [0.24, 0.92, 0.54, 0.55, 0.6, 0.38, 0.63],
'Life': [51.95, 55.54, 52.48, 53.71, 50.18, 49.12, 55.84],
'Freedom': [0.65, 0.44, 0.06, 0.5, 0.52, 0.79, 0.63, ],
'Generosity': [0.07, 0.01, 0.06, 0.28, 0.36, 0.33, 0.26],
'Corruption': [0.97, 0.23, 0.66, 0.12, 0.06, 0.87, 0.53]}
I have a list of happiness and six explanatory vars.
exp_vars = ['Happiness', 'LogGDP', 'Support', 'Life', 'Freedom', 'Generosity', 'Corruption']
1. Define a variable called explanatory_vars that contains the list of the 6 key explanatory variables
2. Define a variable called plot_vars that contains Happiness and each of the explanatory variables. (Hint: recall that you can concatenate Python lists using the addition (+) operator.)
3. Using sns.pairplot, make a pairwise scatterplot for the WHR data frame over the variables of interest, namely the plot_vars. To add additional information, set the hue option to reflect the year of each data point, so that trends over time might become apparent. It will also be useful to include the options dropna=True and palette='Blues'.
Answer:
Here the answer is given as follows,
Explanation:
import seaborn as sns
import pandas as pd
df = {'country': ['US', 'US', 'US', 'US', 'UK', 'UK', 'UK'],
'year': [2008, 2009, 2010, 2011, 2008, 2009, 2010],
'Happiness': [4.64, 4.42, 3.25, 3.08, 3.66, 4.08, 4.09],
'Positive': [0.85, 0.7, 0.54, 0.07, 0.1, 0.92, 0.94],
'Negative': [0.49, 0.09, 0.12, 0.32, 0.43, 0.21, 0.31],
'LogGDP': [8.66, 8.23, 7.29, 8.3, 8.27, 6.38, 6.09],
'Support': [0.24, 0.92, 0.54, 0.55, 0.6, 0.38, 0.63],
'Life': [51.95, 55.54, 52.48, 53.71, 50.18, 49.12, 55.84],
'Freedom': [0.65, 0.44, 0.06, 0.5, 0.52, 0.79, 0.63, ],
'Generosity': [0.07, 0.01, 0.06, 0.28, 0.36, 0.33, 0.26],
'Corruption': [0.97, 0.23, 0.66, 0.12, 0.06, 0.87, 0.53]}
dataFrame = pd.DataFrame.from_dict(df)
explanatory_vars = ['LogGDP', 'Support', 'Life', 'Freedom', 'Generosity', 'Corruption']
plot_vars = ['Happiness'] + explanatory_vars
sns.pairplot(dataFrame,
x_vars = explanatory_vars,
dropna=True,
palette="Blues")
Large computer programs, such as operating systems, achieve zero defects prior to release. Group of answer choices True False PreviousNext
Answer:
The answer is "False"
Explanation:
It is put to use Six Sigma had 3.4 defects per million opportunities (DPMO) from the start, allowing for a 1.5-sigma process shift. However, the definition of zero faults is a little hazy. Perhaps the area beyond 3.4 DPMO is referred to by the term "zero faults.", that's why Before being released, large computer programs, such as operating systems, must have no faults the wrong choice.
Analyze the following code:
// Enter an integer Scanner input = new Scanner(System.in);
int number = input.nextInt(); if (number <= 0) System.out.println(number);
1) The if statement is wrong, because it does not have the else clause;
2) System.out.println(number); must be placed inside braces;
3) If number is zero, number is displayed;
4) If number is positive, number is displayed.
5) number entered from the input cannot be negative.
Answer:
3) If number is zero, number is displayed;
Explanation:
The code snippet created is supposed to take any input number from the user (positive or negative) and print it to the console if it is less than 0. In this code the IF statement does not need an else clause and will work regardless. The System.out.println() statement does not need to be inside braces since it is a simple one line statement. Therefore, the only statement in the question that is actually true would be ...
3) If number is zero, number is displayed;
Answer:
If number is zero, number is displayed
Explanation:
Given
The above code segment
Required
What is true about the code segment
Analyzing the options, we have:
(1) Wrong statement because there is no else clause
The above statement is not a requirement for an if statement to be valid; i.e. an if statement can stand alone in a program
Hence, (1) is false
(2) The print statement must be in { }
There is only one statement (i.e. the print statement) after the if clause. Since the statement is just one, then the print statement does not have to be in {} to be valid.
Hence, (2) is false
(3) number is displayed, if input is 0
In the analysis of the program, we have: number <=0
i.e. the if statement is true for only inputs less than 0 or 0.
Hence, number will be printed and (3) is false
(4) number is displayed for positive inputs
The valid inputs have been explained in (3) above;
Hence, (4) is false.
(5) is also false
Complete the implementation of the following methods:__init__hasNext()next()getFirstToken()getNextToken()nextChar()skipWhiteSpace()getInteger()
From method names, I am compelled to believe you are creating some sort of a Lexer object. Generally you implement Lexer with stratified design. First consumption of characters, then tokens (made out of characters), then optionally constructs made out of tokens.
Hope this helps.
Assign courseStudent's name with Smith, age with 20, and ID with 9999. Use the print member function and a separate cout statement to output courseStudents's data. End with a newline. Sample output from the given program:
#include
#include
using namespace std;
class PersonData {
public:
void SetName(string userName) {
lastName = userName;
};
void SetAge(int numYears) {
ageYears = numYears;
};
// Other parts omitted
void PrintAll() {
cout << "Name: " << lastName;
cout << ", Age: " << ageYears;
};
private:
int ageYears;
string lastName;
};
class StudentData: public PersonData {
public:
void SetID(int studentId) {
idNum = studentId;
};
int GetID() {
return idNum;
};
private:
int idNum;
};
int main() {
StudentData courseStudent;
/* Your solution goes here */
return 0;
}
Answer:
Replace /* Your solution goes here */
with the following:
courseStudent.SetName("Smith");
courseStudent.SetAge(20);
courseStudent.SetID(9999);
courseStudent.PrintAll();
cout <<courseStudent.GetID();
Explanation:
From the given code segment, we have the following methods defined under the StudentData class;
SetName -> It receives name from the main
SetAge -> It receives age from the main
SetID --> It receives ID from the main and passes it to GetID method
printID --> Prints all the required output.
So, we have:
Pass name to setName
courseStudent.SetName("Smith");
Pass age to setAge
courseStudent.SetAge(20);
Pass ID to setID
courseStudent.SetID(9999);
Print all necessary outputs
courseStudent.PrintAll();
Manually print the student ID
cout <<courseStudent.GetID();
Program to calculate series 10+9+8+...+n" in java with output
Answer:
import java.util.Scanner;
class Main {
public static int calcSeries(int n) {
int sum = 0;
for(int i=10; i>=n; i--) {
sum += i;
}
return sum;
}
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
int n = 0;
do {
System.out.print("Enter n: ");
n = reader.nextInt();
if (n >= 10) {
System.out.println("Please enter a value lower than 10.");
}
} while (n >= 10);
reader.close();
System.out.printf("sum: %d\n", calcSeries(n));
}
}
The next few questions will be based on interpretations of a topographic map from East Brownsville, TX. To answer these questions you will have to access the topographic map of East Brownsville attached below:
What stage of fluvial landscape development is shown across the region displayed here?
a. Youthful
b. Mature
c. Old age
Answer:
a. Youthful
Explanation:
The river gets its water from youth stage fluvial water. The river meanders from side to side. East Brownsville will be less stable in international boundary in terms of physical position on the landscape because its base level drops.
Write a simple nested loop example that sums the even and odd numbers between 5 and 15 and prints out the computation.
Answer:
for e in range(6, 15, 2):
for o in range(5, 15, 2):
calc = e+o
print(e, "+", o, "=", calc)
Explanation:
Calculate how much disk space (in sectors, tracks, and surfaces) will be required to
store 300,000 120-byte logical records if the disk is fixed sector with 512 bytes/
sector, with 96 sectors/track, 110 tracks per surface, and 8 usable surfaces. Ignore
any file header record(s) and track indexes, and assume that records cannot span
two sectors.
It will be great if you could explain the answer to me. :)
Answer:
see your answer in image
mark me brainlist
The amount of disk space in sectors that would be required to store 300,000 records is equal to 75,000 sectors.
Given the following data:
Number of records = 300,000.Size of each logical record = 120 byte.Number of bytes (sector) = 512.Number of sectors per track = 96.Number of tracks per surface = 110.How to calculate the required disk space?First of all, we would determine the number of logical records that can be held by each sector as follows:
Number of logical records per sector = 512/120
Number of logical records per sector = 4.3 ≈ 4.0.
Now, we can calculate the required disk space to store 300,000 records:
Disk space = 300,000/4
Disk space = 75,000 sectors.
For the tracks, we have:
Disk space in tracks = 75,000/96
Disk space in tracks = 781.25 ≈ 782.
Disk space in tracks = 782 tracks.
For the surfaces, we have:
Disk space in surfaces = 782/110
Disk space in surfaces = 7.10 ≈ 8.
Disk space in surfaces = 8 surfaces.
Read more on disk space here: https://brainly.com/question/26382243
Which information can you apply to every page of your document with the page layout options?
header
conclusion
body
footer
introduction
Answer:
Header and footer
Explanation:
The page layout tab usually found in document programs such as Microsoft Word hold options used to configure several page options which is to be applied to a document. The page layout option gives users options to set page margins, orientation, page numbering, indentation and the other page layout options which is primarily how we would like our document to look like when printed. The page layout option also allows the insertion of a header and footer which is a note which can be written at the topmost and bottomost part of the document. Once, a header and footer option is selected, it is applied to every page of the document being worked on at the moment.
How does OOP keep both code and data safe from outside interference and
incuse?
Answer:
it is the process that binds together the data and code into a single unit and keeps both from being safe from outside interference and misuse. In this process, the data is hidden from other classes and can be accessed only through the current class's methods
Batter boards (if properly offset) will allow for the end user to continually re-string the layout to double check accuracy without having to continually set up an instrument True False
Answer:
The given statement is "True".
Explanation:
Batter panels seem to be platform frames that are used to temporarily suspend foundations plan threads. These same batter board members look like barriers once built.Its positioning is important for creating a foundation precisely as the plans state since some components of their development would have to be accurate. Placed correctly batter panels guarantee that the boundaries seem to be at the appropriate temperatures as well as positions.Given main(), complete the Car class (in file Car.java) with methods to set and get the purchase price of a car (setPurchasePrice(), getPurchasePrice()), and to output the car's information (printInfo()).
Ex: If the input is:
2011
18000
2018
where 2011 is the car's model year, 18000 is the purchase price, and 2018 is the current year, the output is:
Car's information:
Model year: 2011
Purchase price: 18000
Current value: 5770
Note: printInfo() should use three spaces for indentation.
Answer and Explanation:
public class Main {
int Carmodel;
int Purchaseprice;
int Currentyear;
Public void setPurchasePrice(int Purchaseprice){
this.Purchaseprice=Purchaseprice;
}
Public void getPurchasePrice(){
return Purchaseprice;
}
static void printInfo(int Carmodel int Currentyear ){
this.Carmodel=Carmodel;
this.Currentyear=Currentyear;
System.out.println(getPurchasePrice() Carmodel Currentyear);
}
}
The above java program defines a class that has three methods, a get method that returns purchase price of the object, a set method that sets purchase price of the object, and a print method that print out the information about the car object(model, year, price). The print method also takes arguments and assigns values of the arguments/parameters to the object, then prints all.
A blogger writes that the new "U-Phone" is superior to any other on the market. Before buying the U-Phone based on this review, a savvy consumer should (5 points)
consider if the blogger has been paid by U-Phone
assume the blogger probably knows a lot about U-Phones
examine multiple U-Phone advertisements
speak to a U-Phone representative for an unbiased opinion
Answer:
Speak to a U-Phone representative.
Explanation:
A representative will tell you the necessary details about the U-phone without any biases whatsoever, as they´re the closest to the company when it comes to detail about their products.
Using Python suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
You are given a target value to search. If found in the array return its index, otherwise return “The input is not on this list”
You may assume no duplicate exists in the array.
Hint: Use a function. Use the built in method .index( ) and/or for loops.
Answer:
Explanation:
class Solution {
public int search(int[] nums, int target) {
int n = nums.length;
int low = 0 , high = n - 1;
While(low < high){//Set virtual node
int mid = (low + high) / 2;
if(nums[mid] > nums[high]){
low = mid + 1;
}else{
high = mid;
}
}
int rot = low;
low = 0;
high = n - 1;
while(low <= high){
int mid = (low + high) / 2;
Int real = (mid + rot) % n;//The virtual node is mapped to the real node.
if(nums[real] == target){
return real;
}else if(nums[real] < target){
low = mid + 1;
}else{
high = mid - 1;
}
}
return -1;
}
}
