4.7 code practice question 2 edhesive i cant figure out the code for this problem csn anyone help me?
1 answer:
Answer:
See Explanation
Explanation:
<em>See attachment for complete question</em>
The programming language is not stated; I'll answer using Python and Java
Python:
low = 56
high = 70
for i in range(low,high+1):
print(i)
Java:
public class PrintOut{
public static void main(String [] args)
{
int low = 56; int high = 70;
for(int i = low; i<=high;i++)
System.out.print(i+" ");
}
}
For both codes, the explanation is:
The code starts by initializing the range of the print out to 56 and 70
Next, the code segment used an iteration that loops through 56 and 70 and print each digit in this range (both numbers, inclusive).
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We use if-else structure to check the each possible scenario and return the median accordingly in the middleValue() method. The main is also provided so that you can test the method.
Comments are used to explain the each line.
You may see the output in the attachment.
public class Main
{
public static void main(String[] args) {
//call the method for different scenarios
System.out.println(middleValue(1, 2, 3));
System.out.println(middleValue(1, 3, 2));
System.out.println(middleValue(2, 1, 3));
System.out.println(middleValue(2, 3, 1));
System.out.println(middleValue(3, 1, 2));
System.out.println(middleValue(3, 2, 1));
}
//method that takes three int and returns an int
public static int middleValue(int n1, int n2, int n3) {
//set the median as n1
int median = n1;
//check the situation where the n1 is the highest
//if n2 is greater than n2 -> n1 > n2 > n3
//if not -> n1 > n3 > n2
if(n1 > n2 && n1 > n3){
if(n2 > n3)
median = n2;
else
median = n3;
}
//check the situation where the n2 is the highest
//if n3 is greater than n1 -> n2 > n3 > n1
//if not -> n2 > n1 > n3
//note that we set the median as n1 by default, that is why there is no else part
else if(n2 > n1 && n2 > n3){
if(n3 > n1)
median = n3;
}
//otherwise, n3 is the highest
//if n2 is greater than n1 -> n3 > n2 > n1
//if not -> n3 > n1 > n2
//note that we set the median as n1 by default, that is why there is no else part
else{
if(n2 > n1)
median = n2;
}
return median;
}
}
You may see another if-else question at:
Answer:
The program is as follows (See attachment)
Comments are used to explain some lines
Fname = input("Enter Employee First Name: ")
Lname = input("Enter Employee Last Name: ")
Salary = float(input("Enter Employee Salary: "))
Ratings = int(float(input("Enter Employee Ratings: ")))
#Initialize bonus rate to 0; This will be used if ratings is 4 or 4
bonusRate = 0.0
if Ratings == 1:
bonusRate =0.25 #If ratings is 1, bonus rate is 0.25
elif Ratings == 2:
bonusRate = 0.15 #If ratings is 2, bonus rate is 0.15
elif Ratings == 3:
bonusRate = 0.1 #If ratings is 3, bonus rate is 0.1
bonus = Salary * bonusRate
print(Lname+" "+Fname+"'s employee bonus is "+str(bonus))
#End of Program
Explanation:
The program saves the employee's first name and last name in variables Fname and Lname, respectively
The employee's salary is saved as float datatype in variable Salary
The employee's ratings is saved as int datatype in variable Ratings
For ratings greater than or equal to 4, the bonusRate is set to 0.0; this is done on line 6 of the program
Line 7 to 12 check if employee Ratings is 1, 2 or 3 and the respective bonusRate is used
Line 13 calculates the employee's bonus
Line 14 prints the employee's bonus
The program is ended on line 15
