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A group consists of 10 kids and 2 adults. On a hike, they must form a line with an adult at the front and an adult at the back.

blsea [12.9K]

Answer:

b. 2.9!

Explanation:

There are is a mistake in the question.

Suppose the group consist of 10 kids and 2 adults, the number of ways in which they can form the line is:

= 2! 10!

= 2× 1× 10!

= 2.10!

But since that is not in the given option.

Let assume that the group consists of 9 kids and 2 adults, the number of ways in which they can form the line is:

No of ways the kids can be permutated = 9 ways

No of ways the adult can be permutated = two ways.

Thus; the number of ways in which they can form the line = 2! 9!

= 2 × 1× 9!

= 2.9!

6 0

3 years ago

"Write a program that calculates the balance of a savings account at the end of a period of time. It should ask the user for the

Akimi4 [234]

Answer:

#include <iostream>

#include <fstream>

using namespace std;

int main() {

// Definitions

double annualInterest; // The annual interest rate

double balance; // The account balance

int months; // Total number of months

double totalDeposit = 0.0; // Total deposited value

double totalWithdraw = 0.0; // Total withdrawn value

double earnedInterest= 0.0; // The earned amount

double deposited; // Monthly deposited value

double withdrawn; // Monthly withdrawn value

// Get the initial required data.

cout<<"Please enter the annual interest rate,\n";

cout<<" Interest rate: ";

cin >>annualInterest;

cout<<"--------------------------------------------\n";

cout<<"Please enter the starting balance,\n";

do{

cout<<"[the number could not be negative]\n";

cout<<"Starting Balance: ";

cin >>balance;

}while(balance<0);

cout<<"--------------------------------------------\n";

cout<<"Please enter the number of months,\n";

do{

cout<<"[the number could not be less than 0]\n";

cout<<"Number of months: ";

cin >>months;

}while(months<0);

// Iterated Loop to get all the months data.

for(int month=1; month <= months; month++){

// Get the deposited value during that month.

cout<<"Enter the deposited value during month "

<<month<<",\n";

do{

cout<<"[the value could not be less than 0]\n";

cout<<" Deposited value: ";

cin >>deposited;

}while(deposited<0);

totalDeposit += deposited;

balance += deposited;

// Get the withdrawn value during that month.

cout<<"Enter the withdrawn value during month "

<<month<<",\n";

do{

cout<<"[the value could not be less than 0]\n";

cout<<" Withdrawn value: ";

cin >>withdrawn;

}while(withdrawn<0);

totalWithdraw += withdrawn;

balance -= withdrawn;

// Negative balance close the program.

if(balance < 0){

cout<<"Sorry, the account has been closed due to\n";

cout<<"the negative balance.\n";

break;

// Calculate value due to the interest rate.

}

else {

earnedInterest += (annualInterest/12) * balance;

// monthly interest rate

balance += (annualInterest/12) * balance;

}

// Display the statistics,

ofstream file;

file.open ("Report written to Report.txt");

file << "Report written to Report.txt\n";

file.close();

cout<<"The ending balance: "<<balance<<std::endl;

cout<<" Total deposited: "<< totalDeposit<<std::endl;

cout<<" Total withdrawals: "<< totalWithdraw<<std::endl;

cout<<" Earned interest: "<< earnedInterest<<std::endl;

cout<<"============================================\n";

return 0;

}

6 0

3 years ago

Write a short reflection piece (it may consist of three bulleted items, with one explanatory sentence) on three things you learn

nydimaria [60]

Answers

OS(The Operating System) sends interrupts to the processor to stop whatever is being processing at that moment and computer architecture send data bus. This bus sends data between the processor,the memory and the input/output unit.

The operating system is a low-level software that supports a computer’s basic functions, such as scheduling tasks and controlling peripherals while the computer architecture has the address bus bar. This bus carries signals related to addresses between the processor and the memory.

The interface between a computer’s hardware and its software is its Architecture while An operating system (OS) is system software that manages computer hardware and software resources and provides common services for computer programs.

Explanation:

In short explanation,the Computer Architecture specifically deals with whatever that's going on in the hardware part of the computer system while the Operating System is the computer program which has been program to execute at some instances depending on the programming instructions embedded in it. An example is the MS Office.

5 0

3 years ago

Word wrap is the same as __________ return and means you let the ______________ control when it will go to a new line.

dusya [7]

Word wrap is the same as; Soft Return

Word wrap means that you let the; Computer Control when it will go to a new line

<h3>What is Text Wrapping?</h3>

Text wrapping is simply defined as a process used in MS Word to Wrap a Text around an Image.

Now, the way it is done is by selecting the image you want to wrap text around and then On the Format tab, click the Wrap Text command in the Arrange group, then select the desired text wrapping option to wrap the text.

Finally Word wrap is also same as using soft return and letting the computer control when it goes to the next line.

Read more about text wrapping at; brainly.com/question/5625271

5 0

2 years ago

A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co

tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k ...(1)

probability of 2 hops = k(1-k) ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

= k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

∞ n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

n-1

= 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet = $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

6 0

3 years ago