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Vaselesa [24]
3 years ago
9

How many bit strings of length 12 contain the substring 111000?

Computers and Technology
1 answer:
ELEN [110]3 years ago
7 0

Answer:

448 strings

Explanation:

Each bit string can be filled by two ways either 0 or 1. If the bit strings of length contain the substring '111000' we therefore need a substring of length 6.

The selection of bits starts from the 1st bit to 7th bit enclosing the substring 111000 and ensuring that the length 6 are all filled. So, there will be  7 ways to contain this substring.

The total ways to contain this substring = 7*2^6 =448

Therefore 448 strings of length 12 contain the substring '111000'.

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Oksana_A [137]

Answer:

1. E: II and III only

2. A: (int)(Math.random() * (upper − lower) ) + lower

3. A: The value of answer is N

4. E: while( !(userGuess == secretNumber) && numGuesses <= 15 )

5. C: 21

Explanation:

1. Which of the following is equivalent to while(userGuess != secretNumber)?

I. while( userGuess < secretNumber && userGuess > secretNumber)  

NO - This will test until the userGuess is smaller AND greater than the secretNumber, at the same time... so that condition will never be true.

II. while( userGuess < secretNumber || userGuess > secretNumber)

YES - This will test the value of userGuess and see if it's smaller OR greater than secetNumber.  So, it will loop until the user guesses right.

III. while( !(userGuess == secretNumber) )

YES, this will negate the match with the secretNumber.  In order words, if it's not a match, it will return true... so the loop will run until it finds a false condition (a match).

As you can see, only II and III are valid.

2.  If the lower limit were inclusive and the upper limit exclusive, which expression would properly generate values for the secret number?

A: (int)(Math.random() * (upper − lower) ) + lower

Since the lower limit is INCLUSIVE, we mustn't add one to the lower limit.  Also, the Math.random() function returns a value that matches our needs; it returns a value between [0,1[ (meaning the 0 is included, but not the 1).

Assuming the (int) caster does return only the integer portion doing a round down of the result, we'll be perfect.

3. What conclusion can be made about the state of the program when the while loop terminates?

while(!answer.equals( "N"))

{.....

A: The value of answer is N

The condition in the loops reads as "While the negation of the answer being 'N', loop".  If the answer equals 'N' then the method should return true... which will be negated by the '!' operator, causing the condition to be false. Thus we know that if the loop ends, the value of answer contains 'N', any other value will keep the loop going.

4. Assuming numGuesses is initialized to 1, how would the while statement be modified to include an extra criterion limiting the number of guesses to 15?

E: while( !(userGuess == secretNumber) && numGuesses <= 15 )

This modified condition will first test to see if the user has guessed the secretNumber (if he has, the first sub-parenthesis will be true... so the left side of the && operator will be false due to the negation operator.  The right side of the && operator will check to see how many tries have been attempted. Since the counter starts at 1, it needs to go up to 15 inclusively... so the <= is the right comparison operator.

5. After execution of the following code segment, what will be displayed?

int x = 1;

while(x < 18)

{

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}

System.out.println(x);

C: 21

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So after first passage, x = 6

After second passage, x = 11

After third passage, x = 16

After fourth passage, x = 21

Cannot enter the loop again because 21 > 18.

So, it will print out the value of 21.

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