)finding an unused location in the hash table is called
1 answer:
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Answer:
The Matlab code is as given in the explanation, paste the code in the new script file, save it and run.
Explanation:
The code is as follows
Code:
<em>%% Initialization of the code</em>
<em>% The code takes an array of six AIS scores of the form [AIS1 AIS2 AIS3</em>
<em>% AIS4 AIS5 AIS6] of order 1x6.</em>
<em>AIS=input('Please enter the AIS array of the form [AIS1 AIS2 AIS3 AIS4 AIS5 AIS6]');%taking input of the array</em>
<em>AISs=sort(AIS,'descend');%sorting the array in the descending order</em>
<em>iss=(AISs(1,1)^2)+(AISs(1,2)^2)+(AISs(1,3)^2);%Calculating the ISS</em>
<em>disp(['The injury severity score (ISS) for the entered array is ', num2str(iss)])%Displaying the output</em>
Output:
<u><em>Please enter the AIS array of the form [AIS1 AIS2 AIS3 AIS4 AIS5 AIS6][3 0 4 5 3 0];</em></u>
<u><em>The injury severity score (ISS) for the entered array is 50</em></u>
<u><em /></u>
Answer:
Explanation:
The code does not fail on the first step since 1900 divided by 4 is actually 475 and has no remainder, meaning that it should return True. The code won't work because the if statements need to be nested in a different format. The correct algorithm would be the following, which can also be seen in the picture attached below that if we input 1900 it would output is not a leap year as it fails on the division by 400 which gives a remainder of 0.75
input_year = int(input())
if input_year % 4 == 0:
if input_year % 100 == 0:
if input_year % 400 == 0:
print(input_year, "is a leap year")
else:
print(input_year, "- not a leap year")
else:
print(input_year, "is a leap year")
else:
print(input_year, "- not a leap year")
