Help?

1 answer:

6 0

In stead of going number arithmetic, you could also approach it as string handling. Replace the last lines of your code (right after the scan statement) with:

String s = Integer.toString(num);

System.out.println("Here are the digits: ");

System.out.println(s.charAt(0)+"\n"+s.charAt(1)+"\n"+s.charAt(2));

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It is both true that <span>Technician A says that the excessive length of a heater hose is intended to protect the heater core from undue stress. Technician B says that excessive wear adds to the length of a heater hose, and a replacement heater hose should be roughly three to four inches shorter than its predecessor. So the asnwer is letter B.</span>

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3 years ago

a password to a certain database consists of digits that cannot be repeated. if the password is known to consist of at least 8 d

BaLLatris [955]

Answer:

$\frac{10!}{2}mins$

Explanation:

12 seconds to try one combination will be equivalent to $\frac{1}{12}\times 60 = \frac{1}{5} \ mins$

Password contain at least 8 digit i.e. password can contain 8, 9, 10 digit.

Password cannot contain more than 10 digit because it will give room for repetition which it is clearly stated that digit cannot be repeated.

Possible digit that can be used: 9,8,7,6,5,4,3,2,1,0.

Total number of passwords combination possible for each position in 8 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3. Total number of passwords combination possible in 8 digit is equivalent to $\frac{10!}{2}$ .

Total number of passwords combination possible for each position in 9 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2. Total number of passwords combination possible in 9 digit is equivalent to $\frac{10!}{1}$ .

Total number of passwords combination possible for each position in 10 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2, 10th position = 1. Total number of passwords combination possible in 10 digit is equivalent to 10!.

Adding them up and multiplying by $\frac{1}{5} \ mins$ to get the total number of time needed to guarantee access to database = $[\frac{10!}{2}\times\ \frac{10!}{1} \times 10!] \frac{1}{5}\ mins = \frac{10!}{2}$

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QUESTION 5

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I’m pretty sure D but it should be like $B2

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Answer:

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