Processing information involves ____.a. Performing arithmetic or logical operations on binary data accepting information from th
1 answer:
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Answer:
Check the explanation
Explanation:
//Bugs are highlighted in bold text
class Invoice
Declarations
private num invoiceNumber
private string customer
private num balanceDue
private num tax
public void setInvoiceNUMBER(num number)
Declarations
num LOW_NUM = 1000
num HIGH_NUM = 9999
if number > HIGH_NUM then
invoiceNumber = 0
else
if number < LOW_NUM then
invoiceNumber = 0
else
invoiceNumber = num
endif
return
public void setCustomer(string cust)
customer = cust
return
public void setBalanceDue(num balance)
//Bug balanceDue is Invoice class varible
//but it is assigned to balance .it gives error
balance = balanceDue
setTax()
return
private void setTax()
Declarations
//Bug TAX_RATE is declared as string
//but assigned a double value
string TAX_RATE = 0.07
tax = tax * TAX_RATE
return
public void displayInvoice()
output "Invoice #", invoiceNumber
output "Customer: ", customer
output "Due: ", balanceDue
output "Tax: ", taxDue
//Bug
//Invoice class has no variable called balance .it should be balanceDue
output "Total ", balance + taxDue
return
endClass
start
Declarations
Invoice inv1
Invoice inv2
Invoice inv3
//Warning
//it gives warning object taken but not initilaized
Invoice inv4
inv1.setInvoiceNumber(1244)
inv1.setCustomer("Brown")
inv1.setBalanceDue(1000.00)
inv1.displayInvoice()
inv2.setInvoiceNumber(77777)
inv2.setCustomer("Jenkins")
inv2.setBalanceDue(2000.00)
inv2.displayInvoice()
inv3.setInvoiceNumber(888)
inv3.setCustomer("Russell")
inv3.setBalanceDue(3000.00)
//Bug
//setTax method of Invioce doesnot take any arguments
inv3.setTax(210.00)
inv3.displayInvoice()
stop
//Here is the complete program in c++
//Run the program using Microsoft visual studio 2010 vc++
#include<iostream>
#include<iomanip>
#include<string>
using namespace std;
class Invoice
{
//class varibales
private:
int invoiceNumber;
string customer;
double balanceDue;
double tax;
//class methods
public:
void setCustomer(string cus);
void displayInvoice();
void setBalanceDue(double balance);
void setInvoiceNUMBER(int number);
void setTax();
};
void Invoice::displayInvoice()
{
cout<< setw(10)<<"Invoice #"<<setw(5)<<invoiceNumber<<endl;
cout<<setw(10)<<"Customer: "<<setw(5)<<customer<<endl;
cout<<setw(10)<<"Due: "<<setw(5)<<balanceDue<<endl;
cout<<setw(10)<<"Tax: "<<setw(5)<<tax<<endl;
//Bug
//Invoice class has no variable called balance .it should be balanceDue
cout<< "Total "<< balanceDue + tax<<endl;
}
void Invoice::setCustomer(string cust)
{
customer = cust;
}
void Invoice::setInvoiceNUMBER(int number)
{
const int LOW_NUM = 1000;
const int HIGH_NUM = 9999;
if( number > HIGH_NUM )
invoiceNumber = 0;
else
if (number < LOW_NUM )
invoiceNumber = 0;
else
invoiceNumber = number;
}
void Invoice::setBalanceDue(double balance)
{
balanceDue = balance;
}
void Invoice::setTax()
{
double TAX_RATE = 0.07;
tax = balanceDue * TAX_RATE;
}
int main()
{
Invoice inv1;
Invoice inv2;
Invoice inv3;
inv1.setInvoiceNUMBER(1244);
inv1.setCustomer("Brown");
inv1.setBalanceDue(1000.00);
inv1.setTax();
inv1.displayInvoice();
inv2.setInvoiceNUMBER(77777);
inv2.setCustomer("Jenkins");
inv2.setBalanceDue(2000.00);
inv2.setTax();
inv2.displayInvoice();
inv3.setInvoiceNUMBER(888);
inv3.setCustomer("Russell");
inv3.setBalanceDue(3000.00);
inv3.setTax();
inv3.displayInvoice();
system("pause");
return 0;
}
Kindly check the output image below.
Answer:
the average arrival rate \lambda in units of packets/second is 15.24 kbps
the average number of packets w waiting to be serviced in the buffer is 762 bits
Explanation:
Given that:
A single channel with a capacity of 64 kbps.
Average packet waiting time in the buffer = 0.05 second
Average number of packets in residence = 1 packet
Average packet length r = 1000 bits
What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?
The Conservation of Time and Messages ;
E(R) = E(W) + ρ
r = w + ρ
Using Little law ;
r = λ × T_r
w = λ × T_w
r / λ = w / λ + ρ / λ
T_r =T_w + 1 / μ
T_r = T_w +T_s
where ;
ρ = utilisation fraction of time facility
r = mean number of item in the system waiting to be served
w = mean number of packet waiting to be served
λ = mean number of arrival per second
T_r =mean time an item spent in the system
T_w = mean waiting time
μ = traffic intensity
T_s = mean service time for each arrival
the average arrival rate \lambda in units of packets/second; we have the following.
First let's determine the serving time T_s
the serving time T_s
= 0.015625
now; the mean time an item spent in the system T_r = T_w +T_s
where;
T_w = 0.05 (i.e the average packet waiting time)
T_s = 0.015625
T_r = 0.05 + 0.015625
T_r = 0.065625
However; the mean number of arrival per second λ is;
r = λ × T_r
λ = r / T_r
λ = 1000 / 0.065625
λ = 15238.09524 bps
λ ≅ 15.24 kbps
Thus; the average arrival rate \lambda in units of packets/second is 15.24 kbps
b) Determine the average number of packets w waiting to be serviced in the buffer.
mean number of packets w waiting to be served is calculated using the formula
w = λ × T_w
where;
T_w = 0.05
w = 15238.09524 × 0.05
w = 761.904762
w ≅ 762 bits
Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits
Answer:
SELECT college, city_state, accre_agency, distance LIMIT 10
Explanation:
Given
Table name: College
See attachment for table
Required
Retrieve top 10 college, state, agency and school distance from the table
To retrieve from a table, we make use of the SELECT query
The select statement is then followed by the columns to be selected (separated by comma (,))
So, we have:
SELECT college, city_state, accre_agency, distance
From the question, we are to select only first 10 records.
This is achieved using the LIMIT clause
i.e. LIMIT 10 for first 10
So, the complete query is:
SELECT college, city_state, accre_agency, distance LIMIT 10
