Answer:
def find_max(num_1, num_2):
max_val = 0.0
if (num_1 > num_2): # if num1 is greater than num2,
max_val = num_1 # then num1 is the maxVal.
else: # Otherwise,
max_val = num_2 # num2 is the maxVal
return max_val
max_sum = 0.0
num_a = float(input())
num_b = float(input())
num_y = float(input())
num_z = float(input())
max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)
print('max_sum is:', max_sum)
Explanation:
I added the missing part. Also, you forgot the put parentheses. I highlighted all.
To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)
// This command takes input from the user.
Scanner input = new Scanner(System.in);
int response= input.nextInt();
// Making the variables required
int noCount =0 ;
int yesCount =0;
// Checking the response if it is 1 or 2 and reporting accordingly
if(response == 1 || response ==2){
yesCount+=1;
System.out.println("YES WAS RECORDED");
}
// Checking if the input is 3 or 4 then printing the required lines
else if (response == 3 || response ==4){
noCount+=1;
System.out.println("NO WAS RECORDED");
}
// if the input is not valid than printing INVALID
else{
System.out.println("INVALID");
}
Answer:
Margin of Error=M.E= ± 0.0113
Explanation:
Margin of Error= M.E= ?
Probability that watched network news programs = p = 0.4
α= 95%
Margin of Error =M.E= zₐ/₂√p(1-p)/n
Margin of Error=M.E= ±1.96 √0.4(1-0.4)/7200
Margin of Error=M. E = ±1.96√0.24/7200
Margin of Error=M. E- ±1.96* 0.005773
Margin of Error=M.E= ±0.0113
The Margin of Error is the estimate of how much error is possible as a result of random sampling.
