Answer:
public class Main
{
public static void printString(String strText, int intNumber) {
for(int i=0; i<intNumber; i++){
System.out.println(strText);
}
}
public static void main(String[] args) {
printString("Brainly", 3);
}
}
Explanation:
- Define a function called printString that takes two parameters - a String and an int. Since the function will print out the given String, it's type will be void.
- Use for loop to iterate according to the given number and print the given string
- In the main, call the printString function with some parameters.
// Class declaration
public class Printer {
// Define the function and call it printMany
// The return type is void since it doesn't necessarily return any value.
// It just prints to the console.
// It takes two arguments strText of type String and
// intNumber of type int.
public static void printMany(String strText, int intNumber){
// Initialize the loop counter i to 1;
int i = 1;
//Create a while loop that goes
// from i = 1 to i = intNumber.
// Where intNumber is the number of times the string strText
// will be printed.
while(i <= intNumber) {
// At each of the cycle of the loop;
// print out the string strText
// and increment the value of the loop counter by 1
System.out.println(strText);
i++;
} // End of while loop
} // End of method, printMany, declaration
// Write the main method to call the printMany function
public static void main(String[] args) {
// Call the printMany function by supplying sample arguments;
// In this case, strText has been given a value of "Omobowale"
// And intNumber has been given a value of 4
// Therefore, "Omobowale" will be printed 4 times.
printMany("Omobowale", 4);
} // End of main method
} // End of class declaration
When the program above is run, the following will be the output;
----------------------------------------------------
Omobowale
Omobowale
Omobowale
Omobowale
-------------------------------------------------------
Comments:
* The above code has been written in Java
* The code contains comments explaining every part of the code. Kindly go through the comments.
The whole code without comments is re-written as follows;
public class Printer {
public static void printMany(String strText, int intNumber){
int i = 1;
while(i <= intNumber){
System.out.println(strText);
i++;
}
}
public static void main(String[] args) {
printMany("Omobowale", 4);
}
}
Answer:
Since Python is a scripting language, here is code in python.
#prompt user to give check
amount=float(input("Please Enter the check:"))
#prompt user to give service
service=input("Please Enter the service (good, fair or poor):")
# calculate tip on the basis of service
if service =="good":
tip=amount*0.2
elif service=="fair":
tip=amount*0.15
elif service=="poor":
tip=amount*0.05
#calculate total
total=amount+tip
#print tip
print("tip is equal to : {} ".format(tip))
#print total
print("total of the check is : {} ".format(total))
Explanation:
Prompt user to give check and service input.After taking the input from user, based on the service tip will be calculated. if service is "good" then tip will be 20% of the check, tip will be 15% if service is "fair" and tip will be 5% if service is "poor".
Output:
Please Enter the check:125
Please Enter the service (good, fair or poor):good
tip is equal to : 25.0
total of the check is : 150.0
example to help you. The algorithm that you will write should be in everyday language
(no pseudocode or programming language). Write your instructions at the bottom of the
page.
Example: 1. Move
the orange box 2
spaces to the right.
2. Move the green
box one space
down. 3. Move the
green box two
spaces to the left.
Write your instructions. Review the rubric to check your final work.
Rules: All 6 colors (red, green, yellow, pink, blue, purple) must be move to their new location on the grid. Block spaces are
barriers. You cannot move through them or on them – you must move around them
Answer:
Explanation:
Pink: Down 5 then left 2.
Yellow: Left 3 and down 2.
Green: Right 7, down 4 and left 1.
Purple: Up 6 and left 9.
Red: Left 7, down 5 and left 1.
You can do the last one, blue :)
Answer:
Explanation:
u=up, d=down, r=right, l=left
yellow: l3d2
pink: d5l2
green: r7d4l1
purple: u6l9
red: l7d5l1
blue: r2u7l5
Answer:
T3.
Explanation:
A T3 is an acronym for Transmission system 3 and it is also known as Digital Signal Level 3 (DS3). T3 is a point-to-point physical circuit connection which is capable of transmitting up to 44.736Mbps.
This simply means that, when using a Transmission system 3 (T3), it is very much possible or easier to transmit data such as video and audio at the rate of 44.736Mbps.
Please note, Mbps represents megabit per seconds and it is a unit for the measurement of data transmission rate. The Transmission system 3 is an internet connection that has up to 672 circuit channels having 64Kbs.
Also, worthy of note is the fact that a T3 line is made up of twenty-eight (28) Transmission system 1 lines (T1) each having a data transfer rate of 1.544Mbps.
Additionally, Transmission 3 lines are symmetrical and duplex and thus, has equal upload and download speeds. Therefore, transmissions can be done on T3 lines simultaneously without the data lines being clogged or jammed.
If you need speeds of 16 Mbps between two corporate sites in the United States, you would need a T3 leased line.
Generally, the T3 lines are mostly used for multichannel applications and uninterrupted high bandwidth consumptions such as Telemedicine, internet telephony, video conferencing, e-commerce etc.
if (str.indexOf("pea") >= 0)
{
System.out.println("pea");
}
else if (str.indexOf("pear") >= 0)
{
System.out.println("pear");
}
else if (str.indexOf("pearl") >= 0)
{
System.out.println("pearl");
}
II.
if (str.indexOf("pearl") >= 0)
{
System.out.println("pearl");
}
else if (str.indexOf("pear") >= 0)
{
System.out.println("pear");
}
else if (str.indexOf("pea") >= 0)
{
System.out.println("pea");
}
Which of the following best describes the output produced by code segment I and code segment II?
Both code segment I and code segment II produce correct output for all values of str.
Neither code segment I nor code segment II produce correct output for all values of str.
Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pear" but not "pearl".
Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pearl".
Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".
Answer:
Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".
Explanation:
The main issue with the first code segment is the way how the if else if condition are arranged. The "pea" is checked at the earliest time in the first code segment and therefore so long as there is the "pea" exist in the string (regardless there is pear or pearl exist in the string as well), the if condition will become true and display "pea" to terminal. This is the reason why the code segment 1 only work for the values of str that contain "pea".
Internet Protocol (IP) are guideline about how data should be sent across the internet.
Internet Protocol (IP) work hand in hand with packet as they are connected to every packet which therefore means that any internet users who connect to the internet will be allocated with a unique Internet Protocol (IP) address.
This Internet Protocol (IP) address will enables the user devices to connect and communicate across the internet with ease once the packets are connected to the each Internet Protocol (IP) address allocated to them.
Learn more here:
Explanation:
A binary number is converted to hexadecimal number by making a group of 4 bits from the binary number given.Start making the group from Least Significant Bit (LSB) and start making group of 4 to Most Significant (MSB) or move in outward direction.If the there are less than 4 bits in last group then add corresponding zeroes to the front and find the corresponding hexadecimal number according to the 4 bits.For example:-
for this binary number 100111001011011.
100 11100101 1011
There is one bits less in the last group so add 1 zero(0) to it.
0100 1110 0101 1011
4 E 5 B
100111001011011 = 4E5B
101011.1010
0010 1011 . 1010
2 B A
=2B.A