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stiv31 [10]
2 years ago
10

Which statement below correctly differentiates between frames and bits? Frames have more information in them than bits. Frames a

re more basic units of data than bits. Frames are made up of bits but not vice versa. Frames have less information than bits have.
Computers and Technology
2 answers:
Katyanochek1 [597]2 years ago
7 0

Answer:

Frames have more information in them than bits

Frames are made up of bits but not vice-versa.

Explanation:

In the networking world, a frame is a formatted sequence of bits that carry information in them. It is the data protocol unit of the data-link layer of the OSI model. A frame contains about 48 bits. It could be more depending on the type of frame. Groups of bits in a frame carry specific information and perform specific tasks. Some of these bits are;

(i) synchronization bits, which indicate the beginning and the end of the stream of data that is being received at the layer.

(ii) packet payload bits, which are the actual data being transferred. They also contain certain information about the data such as the source and the destination of the data.

(iii) frame sequence check, which is used for detecting errors.

A bit on the other hand is the smallest unit of the data being transmitted in the network. A single bit can hold a single information.

Therefore, a frame has more information that a bit and a frame is made up of one or more bits.

9966 [12]2 years ago
5 0
<span>Frames have more information in them than bits.
</span>
<span>Frames are made up of bits but not vice versa.


A bit (BInary digiT) is the basic unit of digital. It can be 0 (logical false, off) or 1 (not logical false - true, on). Four bits are in a nybble, which can have a value of 0 - 15, eight bits are in a byte which can have a value of 0 - 255. Words vary in size, they consist of multiple bytes and are generally correlated with the system's data bus/processor data width (a 64 bit system has an 8 byte word).
</span>
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A bank in your town updates its customers’ accounts at the end of each month. The bank offers two types of accounts: savings and
Lerok [7]

Answer:

This is your corrected code and the output of each test example. I have also added comments with the provided code to make the code understandable. I have also changed itype variable to from int to String in order to print the account type (Savings or Checking) in output.

import java.util.*;

public class bank{  //class name

public static void main (String [] args)  //start of main function body

{int num,error=1; //declare variables

String itype=" "; //stores Checking or Savings account type

char type =0;  //type variable which is one of savings S or checking C

double min,cur =0,balance =0,rate=0;

//declare variables for minimum balance, current balance, interest rate

Scanner in=new Scanner(System.in);

System.out.println("Enter account number: ");  //prompts user to enter acc no

num=in.nextInt();  //reads input account number

while(error==1) {  

//asks user to enter account type C or S

System.out.println("Enter account type(s-savings or c-checking):");

type=in.next().charAt(0);  //reads the input character of account type

if(type=='c'||type=='C')  //if user inputs c or C

{itype= "Checking";  //set itype to Checking when user input c or C

error=0;  //set value of error to 0 means user entered valid type input

rate=3/100.; }  // Savings accounts receives  3% interest

else if(type=='s'||type=='S')  //if user enters S or s that shows Savings account

{itype= "Savings";  //set itype to Savings when user input s or S

error=0;  //set error to 0 means there is no error

rate=4/100.; }  //Savings accounts receives 4% interest

if(error==1)  //in case of error in giving input

System.out.println("Invalid type-re enter"); }  //asks user to input again

System.out.println("Enter minimum balance: ");  //asks user to enter min bal

min=in.nextDouble();  //reads value of input minimum balance

System.out.println("Enter current balance: ");

// reads value of input current balance

cur=in.nextDouble();

balance = cur;  

if(itype=="Checking") //if the account type is checking

{ if(cur>min+5000) //Checking accounts interest is 5%

{rate=5/100.;

cur=cur+rate*cur; //computes new balance

System.out.printf("New balance: $%.2f\n", cur);} //returns new balance value

/*If a customer’s balance falls below the minimum balance, there is a service charge of $25.00 for checking accounts */

else if(cur<min)

{cur-=25;

System.out.printf("New balance: $%.2f\n", cur);} returns the value of new

}

if(itype=="Savings"){ //if account type is Savings

/*If a customer’s balance falls below the minimum balance, there is a service charge of $10.00 for savings accounts */

if(cur<min)

{cur-=10;

System.out.printf("New balance: $%.2f\n", cur);}

else

//Savings accounts receive 4% interest

{cur=cur+rate*cur;

System.out.printf("New balance: $%.2f\n", cur);}}

/* as the program should output account number, account type, current balance, and new balance so i have commented out the extra print statements below */

//System.out.printf("After interest and fees your balance is = $%.2f\n",cur);

System.out.println("Account Number: " + num);

System.out.println("Account type: " + itype);

System.out.printf("Current balance: $%.2f\n ", balance);  //the result is //displayed up to 2 decimal place .2f   } }

Explanation:

Following is the output of each test example:

46728 S 1000 2700

Account Number: 46728                                            

Account type: Savings                                                                                          

Current balance: $2700.00                                                                                

New balance: $2808.00

87324 C 1500 7689

Account Number: 87324                                                                                      

Account type: Checking                                                                                        

Current balance: $7689.00                                                                                  

New balance: $8073.45

79873 S 1000 800

Account Number: 79873                                                                                    

Account type: Savings                                                                                        

Current balance: $800.00  

New balance: $790

89832 C 2000 3000

Account Number: 89832                                                                                      

Account type: Checking                                                                                        

Current balance: $3000.00                                                                                  

New balance: $3090.00

98322 C 1000 750  

Account Number: 98322

Account Type: Checking  

Current balance: $1000.00

New Balance: $725.00

7 0
2 years ago
The piston engine uses the ________ to convert the reciprocating motion of the piston into rotary motion.
PSYCHO15rus [73]

The piston engine uses the crankshaft to convert the reciprocating motion of the piston into rotary motion.

<span>The crankshaft is used to convert reciprocating motion of the piston into rotary motion, while the conversion process is called torque, which is a twisting force. Aside from crankshaft, there are a total of four parts of the engine that work together in order to convert the reciprocating motion into rotary motion namely cylinder, or also called the chamber of the piston, the piston itself, and the connecting rod.</span>

5 0
2 years ago
Use the single-server drive-up bank teller operation referred to in Problems 1 and 2 to determine the following operating charac
elena-s [515]

Answer:

This question is incomplete, here's the complete question:

1. Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4 customers per minute. 3. Use the single-server drive-up bank teller operation referred to in Problems 1 to determine the following operating characteristics for the system: a.) The probability that no customers are in the system. b.) The average number of customers waiting. c.) The average number of customers in the system. d.) The average time a customer spends waiting. e.) The average time a customer spends in the system. f.) The probability that arriving customers will have to wait for service.

Explanation:

Arrival rate \lambda = 24 customers per hour or 0.4 customers per minute

Service rate \mu​ = 36 customers per hour or 0.6 customers per minute (from problem 1)

a.) The probability that no customers are in the system , P0 = 1 - \lambda / \mu

= 1 - (24/36) = 1/3 = 0.3333

b.) The average number of customers waiting

Lq = \lambda^2 / [\mu(\mu - \lambda)] = 242 / [36 * (36 - 24)] = 1.33

c.) The average number of customers in the system.

L = Lq + \lambda / \mu = 1.33 + (24/36) = 2

d.) The average time a customer spends waiting.

Wq = \lambda / [\mu(\mu - \lambda)] = 24 / [36 * (36 - 24)] = 0.0555 hr = 3.33 min

e.) The average time a customer spends in the system

W = Wq + 1/\mu = 0.0555 + (1/36) = 0.0833 hr = 5 min

f.) The probability that arriving customers will have to wait for service.

= 1 - P0 = 1 - 0.3333 = 0.6667

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