Answer:
(A) IF (IsFound
(afternoonList, child))
{
APPEND (lunchList, child)
}
Hope this helps!
Answer:
Python file with appropriate comments given below
Explanation:
#Take the input file name
filename=input('Enter the input file name: ')
#Open the input file
inputFile = open(filename,"r+")
#Define the dictionary.
list={}
#Read and split the file using for loop
for word in inputFile.read().split():
#Check the word to be or not in file.
if word not in list:
list[word] = 1
#increment by 1
else:
list[word] += 1
#Close the file.
inputFile.close();
#print a line
print();
#The word are sorted as per their ASCII value.
fori in sorted(list):
#print the unique words and their
#frequencies in alphabetical order.
print("{0} {1} ".format(i, list[i]));
<span>BCD only goes from digit 0 (0000) to digit 9 (1001), because for 10 you need two digits, so all you've got to do is make a function that produces high for numbers from 10 (1010) to 15 (1111) as follows:
A3 A2 A1 A0 F
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
...........................
1 0 0 0 0
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
Then simplify the function:
F = A3*A2 + A3*A1
Finally just draw or connect the circuit using NAND</span>
Here you go,
Import java.util.scanner
public class SumOfMax {
public static double findMax(double num1, double num2) {
double maxVal = 0.0;
// Note: if-else statements need not be understood to
// complete this activity
if (num1 > num2) { // if num1 is greater than num2,
maxVal = num1; // then num1 is the maxVal.
}
else { // Otherwise,
maxVal = num2; // num2 is the maxVal.
}
return maxVal;
}
public static void main(String[] args) {
double numA = 5.0;
double numB = 10.0;
double numY = 3.0;
double numZ = 7.0;
double maxSum = 0.0;
/* Your solution goes here */
maxSum = findMax(numA, numB); // first call of findMax
maxSum = maxSum + findMax(numY, numZ); // second call
System.out.print("maxSum is: " + maxSum);
return;
}
}
/*
Output:
maxSum is: 17.0
*/
