Answer:
Hi there! Please find the answer below.
Explanation:
The program below demonstrates how each of the requirements can be coded in a simple Python script. To copy the array, we can use the copy() method of the array. To calculate the diff, we can use a few different techniques, so I have implemented it using a loop over the arrays and just storing and displaying the difference in the 2 arrays.
new_string.py
def add_test_score(score):
test_score.append(score);
contact_hash = {}
test_score = []
print("Enter 5 test scores: ");
for x in range(0, 5):
input_string = input("Enter test score " + str(x + 1) + ": ")
add_test_score(int(input_string));
print(test_score);
test_score_copy = test_score.copy();
for e in test_score_copy:
if e < 60:
test_score_copy[test_score_copy.index(e)] += 10;
diff = [];
for e in test_score:
print(test_score[test_score.index(e)]);
print(test_score_copy[test_score.index(e)]);
if not test_score[test_score.index(e)] == test_score_copy[test_score.index(e)]:
diff.append(e);
print(diff);
Answer:
The image of truth table is attached.
Explanation:
In the truth table there is a separate table for the expression (A+B).C and for the expression (A.C)+(B.C) you can see in the truth table that the columns of (A+B).C is having same values as the (A.C)+(B.C).Hence we can conclude that (A+B).C is equal to (A.C)+(B.C).
Answer:
It goes like:
public class Program
{
public static void main(String[] args)
{
int j=18;
int sum=0;
for (int i =1; i<7; i++)
{
sum=sum+(i*(j-2));
j=j-2;
}
System.out.println(sum);
}
}
Explanation:
<u>Variables used: </u>
j : controls the first number in product and decreases by 2 each time the loop runs.
sum: saves the values of addition as the loop runs.
Answer: Wired, Wi-fi, Wi-fi, A wired network, Wi-fi
Explanation:
Answer:
c to help build the internets infrastrucuter
Explanation:
