Write the definition of a function words_typed, that receives two parameters. The first is a person's typing speed in words per minute (an integer greater than or equal to zero). The second is a time interval in seconds (an integer greater than zero). The function returns the number of words (an integer) that a person with that typing speed would type in that time interval.

Answer:

Check the explanation

Explanation:

Kindly check the attached image for the first step

Note that the -print" statement executes n(n — I)(n — 2) times and the index values for i, j, and k can never be the same.  

Therefore, the algorithm prints out all the possible ways to draw three balls in sequence, without replacement.

Now we need to determine the number of lines this the algorithm print. In this case, we are selecting three different balls randomly from a set of n balls. So, this involves permutation.  

Therefore, the algorithm prints the total  

P(n, 3)  

lines.  

Answer:

There are various ways: Handing out papers/fliers to people, or presenting slides.

Answer:

Explanation:

The solution code is written in Python 3.

Firstly, we need to import math module (Line 1).

Next, create a function standard_deviation that takes one input parameter, which is a list (Line 3). In the function, calculate the mean for the value in the input list (Line 4-8). Next, use the mean to calculate the variance (Line 10-15). Next, use sqrt method from math module to get the square root of variance and this will result in standard deviation (Line 16). At last, return the standard deviation (Line 18).

We can test the function using a sample list (Line 20) and we shall get 4.509249752822894

If we pass an empty list, a ZeroDivisionError exception will be raised.

Answer:

See explaination

Explanation:

dietary_restrictions = input("Any dietary restrictions?: ")

if dietary_restrictions=="lactose intolerant":

print("No cheese")

elif dietary_restrictions == "vegetarian":

print("Veggie burger")

else:

print("No alteration")

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = [tex]\frac{CPI*Instruction Count}{ClockRate}[/tex]

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is [tex]=\frac{2.5*100,000}{750*10^6}[/tex]

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =[tex]\frac{2.5*Instruction}{750*10^6}[/tex]

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

The number of instructions that the program would need to have when compiled for Computer B is; 65000 instructions

Formula for Execution time is;

Execution time = (CPI × Instruction Count)/Clock Time

We are given;

CPI for computer A = 1.3

Instruction Count = 100000

Clock time = 600 MHz = 600 × 10⁶ Hz

Thus;

Execution time = (1.3 * 100000)/(600 × 10⁶)

Execution time(CPU Time) = 216.67 * 10⁻⁶ second

For CPU B;

CPU Time = (2.5 * 100000)/(750 × 10⁶)

CPU Time = 333.33 * 10⁻⁶ seconds

Thus, instructions for computer B for the two computers to have same execution time is;

216.67 * 10⁻⁶ = (2.5 * I)/(750 × 10⁶)

I = (216.67 * 10⁻⁶ * 750 × 10⁶)/2.5

I = 65000 instructions

Read more about programming instructions at; https://brainly.com/question/15683939

Answer:

yes you can do that it's utter logic

Answer:

Yes.

Explanation:

Itunes gift cards do buy you games/movies/In app purchases/ect.

Answer:

see explaination for code

Explanation:

CODE

#include <iostream>

#include <vector>

using namespace std;

vector<int> merge(vector<int> a, vector<int> b) {

vector<int> result;

int k = 0;

int i = 0, j = 0;

while (i < a.size() && j < b.size()) {

if (k % 2 == 0) {

result.push_back(a[i ++]);

} else {

result.push_back(b[j ++]);

}

k ++;

}

while (i < a.size()) {

result.push_back(a[i ++]);

}

while(j < b.size()) {

result.push_back(b[j ++]);

}

return result;

}

int main() {

vector<int> a{1, 4, 9, 16};

vector<int> b{9, 7, 4, 9, 11};

vector<int> result = merge(a, b);

for (int i=0; i<result.size(); i++) {

cout << result[i] << " ";

}

}

Answer:

Check the explanation

Explanation:

a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.

(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.

Explanation:

a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.

(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.

Answer:

Check the explanation

Explanation:

1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Note : Log (m *n) = Log m + Log n : this is property of logarithm

3) f(n) = [tex]O( n^2 )[/tex], since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n

for n times, this makes the complexity – n * n = n2

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Answer:

They can communicate over the phone or have meetings describing what is and isn't working for Joe. It's also very important that Darla makes eye contact and is actively listening to effectively handle their customer.

Explanation:

Answer:

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

Explanation:

For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

Answer:

public class Main {

public static void merge(int[] arr, int l, int m, int r) {

int n1 = m - l + 1;

int n2 = r - m;

int[] L = new int[n1];

int[] R = new int[n2];

for (int i = 0; i < n1; ++i)

L[i] = arr[l + i];

for (int j = 0; j < n2; ++j)

R[j] = arr[m + 1 + j];

int i = 0, j = 0;

int k = l;

while (i < n1 && j < n2) {

if (L[i] <= R[j]) {

arr[k] = L[i];

i++;

} else {

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1) {

arr[k] = L[i];

i++;

k++;

}

while (j < n2) {

arr[k] = R[j];

j++;

k++;

}

printArray(arr, l, r);

}

public static void sort(int[] arr, int l, int r) {

if (l < r) {

int m = (l + r) / 2;

sort(arr, l, m);

sort(arr, m + 1, r);

merge(arr, l, m, r);

}

}

static void printArray(int[] arr, int l, int r) {

System.out.print("{");

for (int i = l; i <= r; ++i)

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

System.out.println("}");

}

public static void main(String[] args) {

int[] arr = {23, 14, 3, 56, 17, 8, 42, 18, 5};

sort(arr, 0, arr.length - 1);

}

}

Explanation:

See answer

Answer:

[tex]5909? \times \frac{?}{?} 10100010 {?}^{?} 00010.222 {?}^{2} [/tex]

Answer:

Title page should be the first page of a report.

hope it helps!

Answer:

Ready to help ☺️

Explanation:

A tatum is a feature of music that has been defined as the smallest time interval between notes in a rhythmic phrase.

Answer:

A tatum bc is a feature of music that has been variously defined as the smallest time interval between successive notes in a rhythmic phrase "the shortest durational value

Answer:A TPS focusing on production and manufacturing to keep production costs low while maintaining quality, and for communicating with other possible vendors. The TPS would later be used to feed MIS and other higher level systems.

Explanation:

Answer:

part (a).

The program in cpp is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>0)

   {

       if(balance<payment)    

         { payment = balance;}

       else

       {  

           std::cout << balance <<"\t\t\t"<< payment << std::endl;

           balance = balance - payment;

       }

   }

   return 0;

}

part (b).

The modified program from part (a), is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   double charge=0.01;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   balance = balance +( balance*charge );

   std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>payment)

   {

           std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

           balance = balance +( balance*charge );

           balance = balance - payment;

       }

   if(balance<payment)    

         { payment = balance;}          

   std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

   return 0;

}

Explanation:

1. The variables to hold the loan balance and the monthly payment are declared as double.

2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.  

3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.

4. The computed values are displayed for each month till the loan balance becomes zero.

5. The output for both part (a) and part (b) are attached as images.

Answer:

See explaination

Explanation:

clc;

clear all;

close all;

x=input(' choose abc or fgh or mno:','s');

to=input('enter intial fluid temperature in celcius:');

t1=input('enter time in minutes:');

ta=input('enter ambient temperature in celcius:');

abc=1;

fgh=2;

mno=3;

if x==1

k=0.01;

elseif x==2

k=0.03;

else

k=0.02;

end

t=ta+((to-ta)*exp((-k)*t1));

X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);

disp(X);

Answer:

b. Risk of a potential breach in the integrity of the data on the flash drive

Explanation:

The corrupted or unreadable file error is an error message generated if you are unable to access the external hard drive connected to the system through the USB port. This error indicates that the files on the external hard drive are no longer accessible and cannot be opened.

There are several reasons that this error message can appear:

Answer:

see explaination

Explanation:

a)Worst Case-time complexity=O(n)

def intersection_list(lst1, lst2):

lst3 = [value for value in lst1 if value in lst2]

return lst3

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele) # adding the element

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele) # adding the element

print(intersection_list(lst1, lst2))

b)Average case-time complexity=O(min(len(lst1), len(lst2))

def intersection_list(lst1, lst2):

return list(set(lst1) & set(lst2))

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele)

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele)

print(intersection_list(lst1, lst2))

Answer:

Think of DMA as a co-processor that is used to quickly transfer data between main memory and peripherals without the intervention of the CPU.

Answer:

Check the explanation

Explanation:

PYTHON CODE: (lesson.py)

import statistics

# function to find standard deviation, mean

def get_stats(csv_string):

   # calling the function to clean the data

   sample=clean(csv_string)

   # finding deviation, mean

   std_dev=statistics.stdev(sample)

   mean=statistics.mean(sample)

   # counting the element in sample

   count=len(sample)

   # return the results in a tuple

   return (count, std_dev, mean)

# function to clean the csv string

def clean(csv_string):

   # temporary list

   t=[]

   # splitting the string by comma

   data=csv_string.split(',')

   # looping item in data list

   for item in data:

       # removing space,single quote, double quote

       item=item.strip()

       item= item.replace("'",'')

       item= item.replace('"','')

       # if the item is valid      

       if item:

           # converting into float

           try:

               t.append(float(item))

           except:

               continue

   # returning the list of float

   return t

if __name__=='__main__':

   csv = "a, 1, '-2', 2.35, None,, 4, True"

   print(clean(csv))

   print(get_stats('1.0,2.0,3.0'))

Answer:

Following are the code to method calling

backwardsAlphabet(startingLetter); //calling method backwardsAlphabet

Output:

please find the attachment.

Explanation:

Working of program:

Answer:.

// Program is written in C++.

// Comments are used for explanatory purposes

// Program starts here..

#include<iostream>

using namespace std;

int main()

{

// Declare Variables

int amount, dollar, quarter, dime, nickel, penny;

// Prompt user for input

cout<<"Amount: ";

cin>>amount;

// Check if input is less than 1

if(amount<=0)

{

cout<<"No Change";

}

else

{

// Convert amount to various coins

dollar = amount/100;

amount = amount%100;

quarter = amount/25;

amount = amount%25;

dime = amount/10;

amount = amount%10;

nickel = amount/5;

penny = amount%5;

// Print results

if(dollar>=1)

{

if(dollar == 1)

{

cout<<dollar<<" dollar\n";

}

else

{

cout<<dollar<<" dollars\n";

}

}

if(quarter>=1)

{

if(quarter== 1)

{

cout<<quarter<<" quarter\n";

}

else

{

cout<<quarter<<" quarters\n";

}

}

if(dime>=1)

{

if(dime == 1)

{

cout<<dime<<" dime\n";

}

else

{

cout<<dime<<" dimes\n";

}

}

if(nickel>=1)

{

if(nickel == 1)

{

cout<<nickel<<" nickel\n";

}

else

{

cout<<nickel<<" nickels\n";

}

}

if(penny>=1)

{

if(penny == 1)

{

cout<<penny<<" penny\n";

}

else

{

cout<<penny<<" pennies\n";

}

}

}

return 0;

}

Answer:

Explanation:

Base on the scenario been described in the question, We are 95% confident that the price of a single hard drive with 33 megahertz speed and 386 CPU falls between $3,943 and $4,987

Answer:

can you give more detail

Explanation:

Answer:

99.6

Explanation:

The cloud provider  provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.

The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.

Answer:

WOW! that does not look easy!

Explanation:

I wish i could help but i have no idea how to do that lol

Answer:

Because then it would break

Explanation:

You achieve this by obtaining correct measurements. When measuring a window, plumb refers to the vertical planes, and level refers to the horizontal planes. So he did not install the window totally plumb