"Dean wants a quick way to look up staff members by their Staff ID. In cell Q3, nest the existing VLOOKUP function in an IFERROR function. If the VLOOKUP function returns an error result, the text ""Invalid Staff ID"" should be displayed by the formula. (Hint: You can test that this formula is working by changing the value in cell Q2 to 0, but remember to set the value of cell Q2 back to 1036 when the testing is complete.)"

Answer:

I believe it is just a task. Since there exists(on windows) the Task Manager application, where you can stop any running task, I think that they are called tasks

Explanation:

In most operating systems, a running application is typically referred to as a process. A process is an instance of a program that is being executed by the operating system. It represents the execution of a set of instructions and includes the program code, data, and resources required for its execution.

Each process has its own virtual address space, which contains the program's code, variables, and dynamically allocated memory. The operating system manages and schedules these processes, allocating system resources such as CPU time, memory, and input/output devices to ensure their proper execution.

The operating system provides various mechanisms to manage processes, such as process creation, termination, scheduling, and inter-process communication.

Learn more about operating systems here:

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#SPJ6

Answer:

Flaws and limitations identified in this program includes;

1.There was a not necessary usage of variable retrieval. Would have made use of canConvert.

2. Looking at the program, one will notice numerous typos. One of which is the fact that in JAVA we make use of Boolean instead of bool.

3.We rather use Integer.parseInt in JAVA and not Int16, cant make use of Int16.

4. The exception cant be printed

5. JAVA makes use of checkConversion instead of convertNumber as used in the program.

6. It cant work for decimal numbers, 0 and big integers.

Explanation:

See Answer for the detailed explaination of the flaws and limitations identified in the program.

Answer:thats is so easy obviously 69

Explanation:

Answer:

Answer:

see explaination

Explanation:

The data is collected in a comma-separated values (CSV) format that always includes the following fields:

? date: string

? time: string

? client_ip: string

? server_ip: string

? url_stem: string

? url_query: string

? client_bytes: integer

? server_bytes: integer

What should you do?

a. Load the data as lines of text into an RDD, then split the text based on a comma-delimiter and load the RDD into DataFrame.

# import the module csv

import csv

import pandas as pd

# open the csv file

with open(r"C:\Users\uname\Downloads\abc.csv") as csv_file:

# read the csv file

csv_reader = csv.reader(csv_file, delimiter=',')

# now we can use this csv files into the pandas

df = pd.DataFrame([csv_reader], index=None)

df.head()

b. Define a schema for the data, then read the data from the CSV file into a DataFrame using the schema.

from pyspark.sql.types import *

from pyspark.sql import SparkSession

newschema = StructType([

StructField("date", DateType(),true),

StructField("time", DateType(),true),

StructField("client_ip", StringType(),true),

StructField("server_ip", StringType(),true),

StructField("url_stem", StringType(),true),

StructField("url_query", StringType(),true),

StructField("client_bytes", IntegerType(),true),

StructField("server_bytes", IntegerType(),true])

c. Read the data from the CSV file into a DataFrame, infering the schema.

abc_DF = spark.read.load('C:\Users\uname\Downloads\new_abc.csv', format="csv", header="true", sep=' ', schema=newSchema)

d. Convert the data to tab-delimited format, then read the data from the text file into a DataFrame, infering the schema.

Import pandas as pd

Df2 = pd.read_csv(‘new_abc.csv’,delimiter="\t")

print('Contents of Dataframe : ')

print(Df2)

Answer:

(a). 0.1 × 10^-3 = 0.1 msec.

(b). 0.11 × 10^-3 = 0.11 msec.

(c). 1 × 10^-3. = 1 msec.

Explanation:

So, in order to solve this problem or question there is the need to use the Markov queuing formula which is represented mathematically as below;

Mean time delay, t = /(mean frame length,L) × channel capacity, C - frame arrival rate, F. ---------------------------------(1).

(a). Using equation (1) above, the delay experience by 90 frames/sec.

=> 1/(10^-4× 10^8 - 90). = 0.1 × 10^-3= 0.1 msec.

(b). Using equation (1) above, the delay experience by 900 frames/sec.

=> 1/(10^-4× 10^8 - 900). = 0.11 × 10^-3= 0.11 msec.

(c). Using equation (1) above, the delay experience by 90 frames/sec.

=> 1/(10^-4× 10^8 - 9000). = 1 × 10^-3= 1 msec.

Hence, the operating queuing system is; 900/10^-4 × 10^8 = 0.9.

Answer:

vi [ -| -s ] [-l] [-L] [-R] [ -r [ filename ] ] [-S] [-t tag] [-v] [-V]

  [-x] [-w] [-n ] [-C] [+command | -c command ] filename

Explanation:

See attached image file

Answer:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

Explanation:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

An information security policy are set of rules/policies designed to guide employees for the protection of the security of company information and IT systems. The reasons for these policies are:

Answer:

I believe it is science, but before you do anything with this answer, research more on it, just in case I'm wrong! :)

Explanation:

Answer:

4

Explanation:

For easier management of the host records of the company 4 zones should be used because

subdomain is part of a larger domain and the main domain which is the primary domain is the name which the company have decide to use which will represent the company website address and in a situation where the company

have different domain names in which they had registered, they will need to choose one among the domain which will inturn be their main domain.

Therefore for easier , efficient and effective management of the host records 4 zones will be the best zones to be used.

Example of sub domain is north.example.com

For easier management of the host records, 4 zones should be used

Host record or a DNS host, is simply known as record in one's domain's DNS zone file that is said to create the connection between one's domain and its matching IP address.

It is also called an A or Address record. It is said to links a domain to the physical IP address of a computer hosting that has the domain's services.

4 zones in the subdomain is good because a subdomain is part of a larger domain and the main domain and therefore having four will cater for all.

Its gives a creative approach to the management of DNS, DHCP, and IPAM data.

Conclusively, the use of host records helps to manage multiple DNS records, DHCP and IPAM data.

Learn more from

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Answer:

Check the explanation

Explanation:

The Python program that frequently asks the user what pets the user has,

until the user enters "rock", in which case the loop ends can be analysed in the written codes below.

'''

# count of pets

count = 0

# read the user input

pet = input()

# loop that continues till the user enters rock

# strip is used to remove the whitespace

while pet.strip() != 'rock':

  # increment the count of pets

  count += 1

  # output the pet name and number of pets read till now

  print('You have a %s with a total of %d pet(s)' %(pet.strip(),count))

  pet = input() # input the next pet

#end of program

Following are the python program to use the loop to count the number of pets until the user enters the "rock":

c = 0#defining a variable c that initilzes with 0

p = input()#defining a variable p that input value

while p != 'rock':#defining a loop that check p value not equal to rock

  c += 1#using c variable for counts total number of pets

  print('You have a %s with a total of %d pet(s)' %(p,c))#print total number of pet

  p = input() #input value

Output:

Please find the attached file.

Program Explanation:

Find out more about the loop here:

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Answer:

The answer to this question can be described as follows:

Explanation:

Given data:

Performance of the CPU:  

The Fastest Factor Fraction of Work:

[tex]f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65[/tex]

Current Feature Speedup:

[tex]K_1=[/tex] 1.5

CPU upgrade=6000

Disk activity:

The quickest part is the proportion of the work performed:

Current Feature Speedup:

[tex]k_2=3[/tex]

Disk upgrade=8000

System speedup formula:

[tex]s=\frac{1}{(1-f)+(\frac{f}{k})}[/tex]

Finding the CPU activity and disk activity by above formula:

CPU activity:

[tex]S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1) \\[/tex]

Disk activity:

[tex]S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)[/tex]

CPU:

Formula for CPU upgrade:

[tex]= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)[/tex]

DISK:

Formula for DISK upgrade:

[tex]=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)[/tex]

equation (3) and (4),

Thus, for the least money the CPU alternative is the best performance upgrade.

b)

From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

c)

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095

Answer: Font

Explanation:

Answer:

Salesforce Marketing Cloud.

Explanation:

Exact Target was founded by Peter McCormick, Chris Baggott and Scott Dorsey on the 15th of December, 2000 in Indianapolis, Indiana, United States of America and was later acquired by Salesforce in the year 2013 for $2.5 billion. Salesforce in turn converted Exact Target to Salesforce Marketing Cloud in the year 2014.

The Salesforce Marketing Cloud is a company that provide services such as digital marketing automation and analytics software. Salesforce Marketing Cloud softwares are sold primarily on a multi-year subscription basis and based on number of users, features enabled, and level of customer service. It is a hosted, online subscription model that provides users with services such as, consulting, mobile, implementation, email, social and digital marketing.

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

Answer:

at this day and age, any. My questions would focus around other applications at this point.  250 gb or larger.

Explanation:

Office 365 only uses 4 GB "Microsoft's store page", the smallest hard drives commonly available are 250 GB and larger.  

Answer:

This program has been written using JAVA. The code is written beloe:

// import library

import java.util.Scanner;

public category Main operate

public static void main(String[] args)

{

// scanner category

Scanner scan=new Scanner(System.in);

System.out.println("Enter Number:");

int n=scan.nextInt();

//All squares but n

System.out.println("All squares but n are: ");

int temp=0,square=0,i=0;

while(square

range

square=i*i;

if(square>=n)break;

System.out.print(square+" ");

i++;

}

//All positive numbers that are divisible by 10 but less than n

System.out.println(" ");

System.out.println(" ");

System.out.println("All positive numbers are divisible by 10 and less than n are: ");

i=1;

while(i

{

if(i%10==0)

System.out.print(i+" ");

if(i>=n)break;

i++;

}

System.out.println(" ");

System.out.println(" ");

System.out.println("All powers of 2 that are less than n are: ");

//All powers of 2 less than n

i=0;

while(i

mathematics.pow(2, i);

if(num_power

System.out.print(num_power+" ");

i++;

}

}

}

Explanation:

With SimpleWriter out and int n already declared in each case this program makes use of a while loop condition to satisfy the following;

1. All squares less than n

2. positive numbers that are divisible by 10 and less than n.

3. All powers of two less than n.

Answer:

Question (1) the statements (i) and( iv) are logically equivalent and statements (ii) and (iii) are logically equivalent. Question (2) the statements (i) and (iii) are logically equivalent.

Explanation:

Solution

Question (1)

Now,

Lets us p as puppy in the house, and q as i am happy

So,

p : puppy in the house,  and q : i am happy

Thus,

The Statements

(i) so if there is a puppy in the house, I feel happy :  p -> q

(ii) If I am happy, then there is a puppy in the house : q -> p

(iii) If there is no puppy in the house, then I am not happy.   : ~ p -> ~q

(iv) If I am not happy, then there is no puppy in the house : ~q -> ~p

Hence, the statements (i) and( iv) are logically equivalent and statements (ii) and (iii) are logically equivalent.

Question (2)

Let us denote p as i am in school today, and q as i am in CSC231 class, and r as i am in civics class,

So,

p: i am in school today, q: i am in CSC231 class, r: i am in civics class,

Now,

(i) if I am in school today, then I am in CSC231 class  :p -> q

(ii) If I am not in school today, then I am not civics class  :~p -> ~r

(iii) If I am not in CSC231 class, then I am not in school today  :~q -> ~p

(iv) If I am in CSC231 class, then I am in school today  : q -> p

Therefore, the statements i) and iii) are logically equivalent.

Answer:

See explaination

Explanation:

{1, 2}

Runs only once for 1 < 2

{10, 20, 30}

Runs 1 for 10 < 20

Runs 1 for 10 < 30

Runs 1 for 20 < 30

Total 3 times

{30, 20, 10}

Runs 1 for 30 < 20

Runs 1 for 30 < 10

Runs 1 for 20 < 10

Total 3 times

{-4, 7, 1}

Runs 1 for -4 < 7

Runs 1 for -4 < 1

Runs 1 for -7 < 1

​​​​​​​Total 3 times

Generalizing it will run for n*(n-1)/2 so for 2 element it will run for 2*1/2 = 1

For 3 element it will run for 3*2/2 = 3 times

(ii) (3 points) For an array of size , what is the big-oh runtime of this code in the worst case?

Time complexity in Worst case is O(n^2). Reason being it uses two for loop where first loop runs for n time and the second loop runs for n*n time in worst case hence its O(n^2)

Answer:

import java.util.Scanner;

public class StudentScores

{

public static void main(String[] args) {

 Scanner scnr = new Scanner(System.in);

 final int NUM_POINTS = 4;

 int[] dataPoints = new int[NUM_POINTS];

 int controlValue;

 int i;

 controlValue = scnr.nextInt();

 for (i = 0; i < dataPoints.length; ++i) {

     dataPoints[i] = scnr.nextInt();

 }

 for (i = 0; i < dataPoints.length; ++i) {

     System.out.print(dataPoints[i] + " ");

 }

 System.out.println();

 for (i = 0; i < dataPoints.length; ++i) {

     if(dataPoints[i] < controlValue){

         dataPoints[i] = dataPoints[i] * 2;          

     }

     System.out.print(dataPoints[i] + " ");

 }

}

}

Explanation:

*Added parts highligted.

After getting the control value and values for the array, you printed them.

Create a for loop that iterates through the dataPoints. Inside the loop, check if a value in dataPoints is smaller than the contorolValue. If it is, multiply that value with 2 and assign it to the dataPoints array. Print the elements of the dataPoints

Answer:

import java.util.Scanner;

public class numm3 {

   public static void main (String [] args) {

       Scanner scnr = new Scanner(System.in);

       final int NUM_POINTS = 4;

       int[] dataPoints = new int[NUM_POINTS];

       int controlValue;

       int i;

       System.out.println("Enter the control Variable");

       controlValue = scnr.nextInt();

       System.out.println("enter elements for the array");

       for (i = 0; i < dataPoints.length; ++i) {

           dataPoints[i] = scnr.nextInt();

       }

       for (i = 0; i < dataPoints.length; ++i) {

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println();

       //Doubling elements Less than Control Variable

       for (i = 0; i < dataPoints.length; ++i) {

           if (dataPoints[i]<controlValue){

               dataPoints[i] = dataPoints[i]*2;

           }

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println(); } }

Explanation:

See the additional code to accomplish the task in bold

The trick is using an if statement inside of a for loop that checks the condition (dataPoints[i]<controlValue) If true, it multiplies the element by 2

Answer:

the person drinks water if the answer to “thirsty?” is yes

Answer:

C

Explanation:

Answer:

surveys and questionnaires

Answer:

Survey

Explanation:

Answer:

def gradeCalculated(score):

   if score < 60:

       letter = 'F'

   elif score < 70:

       letter = 'D'

   elif score < 80:

       letter = 'C'

   elif score < 90:

       letter = 'B'

   else:

       letter = 'A'

   return letter

def main():

   grade = int(input("enter your grade from 0-100: "))

   print("Letter grade for ",grade,"is:",gradeCalculated(grade))

   if(grade>80):

       print("You are doing good! keep it up!!")

   else:

       print("You have to get work hard")

main()

Explanation:

See attached image for output

Answer:

The queries for each question are explained.

1. Find the sids and snames of students who took "Database" course (i.e., cname = ‘Database’) in Fall 2019.

 The columns sid and sname are taken from students table.

The course name belong to the courses table.

The semester and year belong to the grades table.

To get the final query, all these tables need to be joined in the FROM clause using JOIN ON keywords.

select s.sid, sname

from students s join grades g on g.sid = s.sid

join courses c on g.cid = c.cid

where cname like 'database%'

and semester like 'Fall'

and year=2019

2. Find the average grade of all the students who took "CS327" (i.e., cid=‘CS327’) in Fall 2019. You can assume grade is a numerical attribute.

The grade column is taken from grades table.

The course id belongs to the grades table.

The semester and year also belong to the grades table.

The final query is formed using the grades table only.

 select avg(grade)

from grades

where cid like 'CS327'

and semester like 'Fall'

and year=2019

 3. Find the number of distinct courses offered by each department in Fall 2019.

 The department id belongs to the courses table.

The course id belongs to the grades table.

The semester and year also belong to the grades table.

To get the final query, the courses and the grades tables are joined in the FROM clause using JOIN ON keywords.

select dept_id, count(distinct g.cid)

from courses c join grades g on c.cid = g.cid

where semester like 'Fall'

and year=2019

group by dept_id

Answer:

Aqueous carbon dioxide, CO2 (aq), reacts with water forming carbonic acid, H2CO3 (aq). Carbonic acid may loose protons to form bicarbonate, HCO3- , and carbonate, CO32-. In this case the proton is liberated to the water, decreasing pH. The complex chemical equilibria are described using two acid equilibrium equations.

PLS MARK AS BRAINLIEST

Answer: Provided in the explanation section

Explanation:

Code provided below in a well arranged format

inputNos.txt

70

95

62

88

90

85

75

79

50

80

82

88

81

93

75

78

62

55

89

94

73

82

___________________

StandardDev.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class StandardDev {

public static void main(String[] args) {

//Declaring variables

String filename;

int count = 0, i = 0;

Scanner sc1 = null;

int min, max;

double mean, stdDev;

int nos[] = null;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//Getting the input entered by the user

System.out.print("Enter the name of the filename :");

filename = sc.next();

try {

//Opening the file

sc1 = new Scanner(new File(filename));

//counting no of numbers in the file

while (sc1.hasNext()) {

sc1.nextInt();

count++;

}

sc1.close();

sc1 = new Scanner(new File(filename));

//Creating an Integer based on the Count

nos = new int[count];

//Populating the values into an array

while (sc1.hasNext()) {

nos[i] = sc1.nextInt();

i++;

}

sc1.close();

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.println("Read from file :" + count + " values");

System.out.println("The Minimum Number is :" + min);

System.out.println("The Maximum Number is :" + max);

System.out.printf("The Mean is :%.2f\n", mean);

System.out.printf("The Standard Deviation is :%.2f\n", stdDev);

} catch (FileNotFoundException e) {

e.printStackTrace();

}

}

//This method will calculate the standard deviation

private static double calStandardDev(int[] nos, double mean) {

//Declaring local variables

double standard_deviation = 0.0, variance = 0.0, sum_of_squares = 0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

for (int i = 0; i < nos.length; i++) {

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares += Math.pow((nos[i] - mean), 2);

}

//calculating the variance of an array

variance = ((double) sum_of_squares / (nos.length - 1));

//calculating the standard deviation of an array

standard_deviation = Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(int[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static int findMinimum(int[] nos) {

int min = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static int findMaximum(int[] nos) {

int max = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

_____________________

Output:

Enter the name of the filename :inputNos.txt

Read from file :22 values

The Minimum Number is :50

The Maximum Number is :95

The Mean is :78.45

The Standard Deviation is :12.45

cheers i hope this helped !!!

Answer:

t

Explanation:

Answer: True

Explanation:

Answer:

D

Explanation:

If your options are

a) to demonstrate a sophisticated programming style

b) to demonstrate a flexible programming style

c)to ensure that programs will work on any computer platform

d) to ensure that programs run properly and return expected results

Then yes this is very much correct!

Answer: The UK government offers a global scholarship programme called “Chevening”. It is open to all international students who meet the Chevening eligibility criteria, which states you must have applied for at least three degree courses in the UK, usually one-year Master's courses

Explanation: So yes that should be fine

Answer:

FALSE

Explanation: