12 points) how many bit strings of length 12 contain: (a) exactly three 1’s? (b) at most three 1’s? (c) at least three 1’s? (d) an equal number of 0’s and 1’s?

Answers

Answer 1

The number of bit strings that satisfy each condition is:

(a) Exactly three 1's: 220

(b) At most three 1's: 299

(d) An equal number of 0's and 1's: 924.

(a) To count the number of bit strings of length 12 with exactly three 1's, we need to choose 3 positions out of 12 for the 1's, and the rest of the positions must be filled with 0's.

Thus, the number of such bit strings is given by the binomial coefficient:

[tex]$${12 \choose 3} = \frac{12!}{3!9!} = 220$$[/tex]

(b) To count the number of bit strings of length 12 with at most three 1's, we can count the number of bit strings with exactly zero, one, two, or three 1's and add them up.

From part (a), we know that there are [tex]${12 \choose 3} = 220$[/tex]bit strings with exactly three 1's.

To count the bit strings with zero, one, or two 1's, we can use the same formula:

[tex]$${12 \choose 0} + {12 \choose 1} + {12 \choose 2} = 1 + 12 + 66 = 79$$[/tex]

So, the total number of bit strings with at most three 1's is [tex]$220 + 79 = 299$[/tex].

(c) To count the number of bit strings of length 12 with at least three 1's, we can count the complement: the number of bit strings with zero, one, or two 1's.

From part (b), we know that there are 79 bit strings with at most two 1's.

Thus, there are [tex]$2^{12} - 79 = 4,129$[/tex] bit strings with at least three 1's.

(d) To count the number of bit strings of length 12 with an equal number of 0's and 1's, we need to choose 6 positions out of 12 for the 1's, and the rest of the positions must be filled with 0's.

Thus, the number of such bit strings is given by the binomial coefficient:

[tex]$${12 \choose 6} = \frac{12!}{6!6!} = 924$$[/tex]

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Related Questions

solve this differential equation: d y d t = 0.09 y ( 1 − y 100 ) dydt=0.09y(1-y100) y ( 0 ) = 5 y(0)=5

Answers

The solution to the differential equation is y ( t ) = 100 1 + 19 e 0.09 t

How to find the solution to the differential equation?

This is a separable differential equation, which we can solve using separation of variables:

d y d t = 0.09 y ( 1 − y 100 )

d y 0.09 y ( 1 − y 100 ) = d t

Integrating both sides, we get:

ln | y | − 0.01 ln | 100 − y | = 0.09 t + C

where C is the constant of integration. We can solve for C using the initial condition y(0) = 5:

ln | 5 | − 0.01 ln | 100 − 5 | = 0.09 ( 0 ) + C

C = ln | 5 | − 0.01 ln | 95 |

Substituting this value of C back into our equation, we get:

ln | y | − 0.01 ln | 100 − y | = 0.09 t + ln | 5 | − 0.01 ln | 95 |

Simplifying, we get:

ln | y ( t ) | 100 − y ( t ) = 0.09 t + ln 5 95

To solve for y(t), we can take the exponential of both sides:

| y ( t ) | 100 − y ( t ) = e 0.09 t e ln 5 95

| y ( t ) | 100 − y ( t ) = e 0.09 t 5 95

y ( t ) 100 − y ( t ) = ± e 0.09 t 5 95

Solving for y(t), we get:

y ( t ) = 100 e 0.09 t 5 95 ± e 0.09 t 5 95

Using the initial condition y(0) = 5, we can determine that the sign in the solution should be positive, so we have:

y ( t ) = 100 e 0.09 t 5 95 + e 0.09 t 5 95

Simplifying, we get:

y ( t ) = 100 1 + 19 e 0.09 t

Therefore, the solution to the differential equation is:

y ( t ) = 100 1 + 19 e 0.09 t

where y(0) = 5.

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Si lanzo 16 monedas al mismo tiempo ¿cual es la probabilidad de obtener 4 sellos?

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The probability of obtaining exactly 4 heads (or 4 tails) when tossing 16 coins simultaneously is approximately 0.0984, or 9.84%.

When tossing 16 coins simultaneously, the probability of getting 4 heads (or tails, as the probability is the same for both outcomes) can be calculated using the concept of binomial probability.

The formula for binomial probability is given by:

P(X=k) = (nCk) * p^k * q^(n-k)

Where:

P(X=k) is the probability of getting exactly k successes,

n is the total number of trials (in this case, the number of coins tossed),

k is the number of successful outcomes (in this case, 4 heads or 4 tails),

p is the probability of a single success (getting a head or a tail, which is 1/2 in this case),

q is the probability of a single failure (1 - p, which is also 1/2 in this case), and

nCk represents the number of combinations of n items taken k at a time.

Applying the formula to our scenario:

P(X=4) = (16C4) * (1/2)^4 * (1/2)^(16-4)

Using the binomial coefficient calculation:

(16C4) = 16! / (4! * (16-4)!)

= (16 * 15 * 14 * 13) / (4 * 3 * 2 * 1)

= 1820

Now, substituting the values into the formula:

P(X=4) = 1820 * (1/2)^4 * (1/2)^12

= 1820 * (1/2)^16

≈ 0.0984

Therefore, the probability of obtaining exactly 4 heads (or 4 tails) when tossing 16 coins simultaneously is approximately 0.0984, or 9.84%.

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SHOUTOUT FOR CHOSLSTON71!?! THIS QUESTION IS?

Answers

Answer: 31

Step-by-step explanation: 775 divided by 25 = 31

Let X measure the amount of consumption of coffee per day in ounces and Y measure the total number of steps walked during the day. Suppose we know that the correlation coefficient px,y = 0.945. With is information only, which of the following statements are true: a) There is positive association between coffee consumption and physical activity. b) Coffee consumption causes you to be physically active. Increase in coffee consumption is associated with increase in physical activity. c) There is likely a strong linear relationship between coffee consumption and physical activity. d) Decrease in coffee consumption causes decreased physical activity.

Answers

Statements a)"There is a positive association between coffee consumption and physical activity" and c) "There is likely a strong linear relationship between coffee consumption and physical activity" are true.

a) There is a positive association between coffee consumption and physical activity: The correlation coefficient px,y = 0.945 indicates a strong positive correlation between the two variables. This means that as coffee consumption increases, there is a tendency for physical activity to also increase.

c) There is likely a strong linear relationship between coffee consumption and physical activity: The high correlation coefficient value (0.945) suggests a strong linear relationship between coffee consumption and physical activity.

However, statements b) and d) are not necessarily true.

b) Coffee consumption causes you to be physically active. An increase in coffee consumption is associated with an increase in physical activity: The correlation coefficient only indicates that there is a relationship between the two variables, but it does not imply causation. It is possible that people who are already physically active tend to consume more coffee or vice versa.

d) Decrease in coffee consumption causes decreased physical activity: The correlation coefficient cannot be used to determine causation. Therefore, it is impossible to conclude that reducing coffee consumption would lead to decreased physical activity.

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Determine the t critical value for a two-sided confidence interval in each of the following situations. (Round your answers to three decimal places.) (a) Confidence level = 95%, df = 5 (b) Confidence level = 95%, df = 10 (c) Confidence level = 99%, df = 10 (d) Confidence level = 99%, n = 10 (e) Confidence level = 98%, df = 21 (f) Confidence level = 99%, n = 36

Answers

The t critical values are:

(a) 2.571, (b) 2.306, (c) 3.169, (d) 3.250, (e) 2.831, (f) 2.750

We have,

(a) Using a t-table or calculator,

The t critical value for a two-sided confidence interval at a 95% confidence level with df = 5 is 2.571.

(b)