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Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m

strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0

3 years ago

Inequalities & problem solving<br>|3-2x/3|>5

shepuryov [24]

Work: 2x/2 > 5/2 =2.5. answer-2.5

5 0

3 years ago

Jack and Jill start driving from the same location. Jack drives 30 miles south while Jill drives west. If Jack and Jill are 78 m

arsen [322]

Answer:

72 mles

Step-by-step explanation:

For better understanding, I'll make use of an attachment.

From the attachment,

O represents the start point

$OB = 30$ --- Jack

$AB = 78$ -- Distance between them

Required

Find OA; Jack's distance

To do this, we make use of Pythagoras theorem

$AB^2 = OA^2 + OB^2$

This gives:

$78^2 = OA^2 + 30^2$

$6084 = OA^2 + 900$

Subtract 900 from both sides

$OA^2= 6084 - 900$

$OA^2= 5184$

Take square roots

$OA = \sqrt{5184}$

$OA = 72$

<em>Hence, Jill drove 72 miles</em>

5 0

3 years ago

Please help me qwq In a reflection, what are corresponding angles?!

zhuklara [117]

Answer: When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called corresponding angles. Example: a and e are corresponding angles. When the two lines are parallel Corresponding Angles are equal.

3 0

3 years ago

Instead of using the values {1,2,3,4,5,6} on a dice, suppose a pair of dice have the following {1,2,2,3,3,4} on one die and {1,3

levacccp [35]

Answer: The answer is $\dfrac{1}{24}.$

Step-by-step explanation: Given that instead of using the values {1,2,3,4,5,6} on a dice, suppose a pair of dice have the numbers {1,2,2,3,3,4} on one die and {1,3,4,5,6,8} on the other. We are to find the probability of getting a sum of 12 by rolling these dice.

Since the first die has 4 distinct numbers, and second has 6 distinct numbers, so total number of combinations possible = 4 × 6 = 24.

Now, number of chances where we get the rolling sum as 12 is only one, i.e., (4,8).

Therefore, the probability of getting sum 12 is

$p=\dfrac{1}{24}.$

Thus, the answer is $\dfrac{1}{24}.$

6 0

3 years ago

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