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3 years ago

15

(4 points) A bit string is a sequence of 0s and 1s.How many bit strings of length 12 either start with 0000 or end with 1111

1 answer:

marta [7]3 years ago

8 0

Answer:

The answer is "496"

Step-by-step explanation:

It begins at 0000 or finishes at 1111:

$= 2^8 + 2^8 - 2^4\\\\= 256 +256 -16\\\\= 512 - 16 \\\\= 496$

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What are the x and the y in this function

Marianna [84]

Answer:

X = 3

Y = all real numbers

Step-by-step explanation:

x is the input

y is the output

The equation would be x = 3

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2 years ago

Solve the equation for the variable.

Flura [38]

1/4x -2 = -6 + 5/12x
3x - 24 = -72 +5x
3x -5x = -72 +24
-2x = -48
x = 24

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3 years ago

I need help with question 6 please, and thank you so much! ;)

trasher [3.6K]

Answer:

A) Ms. Hynnes'

B) Mr. Charles'

Step-by-step explanation:

I start by finding the LCM (lowest common multiple) of 5, 9, and 7, which is 315. Then, to calculate the top number, I divide 315 by the denominator to determine how many times I need to multiply the numerator. For 5, it's 63, giving us a numerator of 126. For 9, it's 35, giving us a numerator of 140. For 7, it's 45, giving us a numerator of 135. I then compare the numbers to find the answer.

5 0

3 years ago

Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

6 0

2 years ago

How do you simplify -3x²ym + 7x - 5ymx² + 16x

brilliants [131]

you cant add or subtract or multiply or divide if the variables arent the same so the simplified version of <span>-3x²ym + 7x - 5ymx² + 16x would be
</span> -3x²ym <span>- 5ymx² + 23x</span>

3 0

3 years ago

Read 2 more answers