Which of the follow will happen if you miss a monthly credit card payment?

Write on the importance of Computer application to statistics

tatyana61 [14]

Answer:

Another way to categorise the relation statistics-computer is to list the different ways the computer can be used in statistics. The following are examples of such uses: numerical and graphical data analy- sis; symbolic computations; simulations; storing statistical knowledge; presentation of results.

Explanation:

5 0

3 years ago

Look at the slide.

svetlana [45]

Answer:

c

Explanation:

5 0

3 years ago

Read 2 more answers

Description

lapo4ka [179]

Answer:

from datetime import time, datetime, timedelta, date

for _ in iter(list, 0):

t1 = input("Please enter time one: ")

t2 = input("Please enter time two: ")

if t1.isdigit() == True and (int(t1[:2])<=23 and int(t1[2:]) <= 59)\

and t2.isdigit() == True and (int(t2[:2])<= 23 and int(t2[2:])<=59):

time_1 = time(hour=int(t1[:2]), minute=int(t1[2:]))

time_2 = time(hour= int(t2[:2]), minute=int(t2[2:]))

if time_1 > time_2:

diff = datetime. combine (date. today( ), time_1) - datetime. combine(date. today ( ), time_2)

else:

diff = datetime. combine (date. today( ), time_2) -datetime. combine(date. today ( ), time_1)

diff_list = str(diff).split(":")

del diff_list[2]

diff_t = "". join (diff_list)

print(diff_t)

break

if t1.isdigit() == False or t2.isdigit() == False:

print("Military Time Forest Required: Bad input, try again.")

continue

elif int(t1[:2])> 23 or int(t2[:2])> 23:

print("Military Tine Format Required: Hours must be between 00 and 23, try again")

continue

elif int(t1[2:])> 59 or int(t2[2:])> 59:

print("Military Tine Format Required: Minutes must be between me and 59, try again")

continue

Explanation:

The python Datetime package is used to manipulate date and time, creating immutable Datetime objects from the various python data types like integer and string. The block of code above is a conditional for-loop statement that gets two user time value inputs and prints out the difference as a string in Military forest time format.

5 0

3 years ago

Instructions The population of town A is less than the population of town B. However, the population of town A is growing faster

weeeeeb [17]

Answer: The c++ program is given below.

#include <iostream>

using namespace std;

int main() {

float townA, townB, growthA, growthB, populationA, populationB;

int year=0;

cout<<"Enter present population of town A : ";

cin >> townA;

cout<<endl<<"Enter present growth rate of town A : ";

cin >> growthA;

growthA = growthA/100;

cout<<endl<<"Enter present population of town B : ";

cin >> townB;

cout<<endl<<"Enter present growth rate of town B : ";

cin >> growthB;

growthB = growthB/100;

do

{

populationA = townA + (townA * growthA);

populationB = townB + (townB * growthB);

townA = populationA;

townB = populationB;

year++;

}while(populationA < populationB);

cout<<endl<<"After " <<year<< " years, population of town A is "<<populationA << " and population of town B is "<< population<<endl;

return 0;

}

Explanation:

All the variables for population and growth rate are declared with float datatype.

The user inputs the present population of both the towns.

The growth rate entered by the user is the percentage growth.

For example, town A has 10% growth rate as shown in the output image.

The program converts 10% into float as 10/100.

growthA = growthA/100;

growthB = growthB/100;

The above conversion is done to ease the calculations.

The year variable is declared as integer and initialized to 0.

The growth in population is computed in a do-while loop. After each growth is calculated, the year variable is incremented by 1.

The loop continues until population of town A becomes greater than or equal to population of town B as mentioned in the question.

Once the loop discontinues, the final populations of town A and town B and the years needed for this growth is displayed.

The new line is introduced using endl keyword.

The function main has return type int hence, 0 is returned at the end of the program.

6 0

3 years ago

Write a program that allows the user to enter a start time and a future time. Include a function named compute Difference that t

Aleonysh [2.5K]

Step 1

Following is the c++ program:

Variables used:

startHours, startMins are used to store the starting time of time machine.

futureHours, futureMins are used to store the future time of time machine.

xAM variable is used to store meridiem of start and future time.

isSAM is boolean variable that is true if starting time is AM.

isFAM is boolean variable that is true if future time is AM.

Functions:

main() is used to call computeDifference() function that will calculate the difference between the start and future time.

computeDifference() takes as input six variables from the user and then apply the following logic to compute the difference between the start and future time:

If Meridien of both start and future time are the same, then simply subtract the start time from the future time.

Difference = future minutes – start minutes

If Meridien of both start and future time are different, then apply the following formula to calculate difference:

Difference = (12 * 60 – start minutes) + future minutes

Step 2

Program :

#include <iostream> //header file for input and output

using namespace std;

//calculate the time difference(in minutes) between start time and future time.

int computeDifference(int startHours, int startMins, bool isSAM, int futureHours, int futureMins, bool isFAM) {

int difference = 0; //to store difference

startMins = startHours * 60 + startMins; //converting in minutes.

futureMins = futureHours * 60 + futureMins; //converting in minutes.

//applying formula for difference.

if (isSAM == isFAM) {

difference = futureMins - startMins;

} else if (isSAM == true && isFAM == false) {

difference = (12 * 60 - startMins) + futureMins;

}

int main() {

int startHours, startMins; //it will store start time

int futureHours, futureMins; //it will store future time.

string xAM; //it will store meridiem.

bool isSAM; //it will be true if start time is AM.

bool isFAM; //it will be true if future time is in AM.

int difference = 0; //it will store difference.

cout << "Enter start time of time machine as 'HH MM AM/PM': "; //take input from

7 0

3 years ago