help, ill mark brainliest

The correct statements are that the interval was produced by a technique that captures mu 95% of the time, 95% of all college students work between 4.63 and 12.63 hours a week, 95% of all samples will have x-bar between 4.63 and 12.63, and the probability that mu is between 4.63 and 12.63 is .95.

The correct statements are:

These statements are correct because a confidence interval is a range of values that is likely to contain the true population mean. With a confidence coefficient of .95, we can say that there is a 95% confidence level that the population mean falls within the interval.

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Answer:

Standard deviation of a normal data distribution is a measure of data dispersion.

Step-by-step explanation:

Standard deviation is used to measure dispersion which is present around the mean data.

The value of standard deviation will never be negative.

The greater the spread, the greater the standard deviation.

Steps-

1. At first, the mean value should be discovered.

2.Then find out the square of it's distance to mean value.

3.Then total the values

4.Then divide the number of data point.

5.the square root have to be taken.

Formula-

SD=[tex]\sqrt{\frac{(\sum{x-x)^2} }{n-1}[/tex]

   Advantage-

It is used to measure dispersion when mean is used as measure of central tendency.

Standard deviation of a normal data distribution is a measure of data dispersion.

A normal distribution is a probability distribution that is symmetric around the mean of the distribution. This means that the there are more data around the mean than data far from the mean. When shown on a graph, a normal distribution is bell-shaped.

Standard deviation is a measure of variation. It measures the dispersion of data from its mean. It can be calculated by determining the value of the square root of variance.

To learn more about standard deviation, please check: brainly.com/question/12402189

The phrases you would like to be written as expressions are not listed. I would nevertheless, explain how to write phrases as expressions so that the same approach could be applied to you own question.

Phrases are dynamic, depending on the problem. They do not necessarily take a particular form.

The constant thing about phrases is the operators connecting the words in the phrases. Theses operators are:

Addition (+), Subtraction (-), Division (÷), and Multiplication (×).

In word problems, it is a matter of interpretation, these operators can be written in many ways.

ADDITION

plus

the sum of

increase

grow

add

profit

And so on.

SUBTRACTION

minus

loss

decrease

reduce

subtract

And so on

MULTIPLICATION

times

multiply

triple

And so on

DIVISION

split

share

divide

distribute

And so on.

Examples

(1) 56 is added to a number to give 100

Interpretation: x + 56 = 100

(2)The difference between Mr. A and Mr. B is 5

Interpretation: A - B = 5

(3) This load (L1) is three times heavier than that one (L2)

Interpretation: L1 = 3L2

(4) Share this orange (P) equally between the three children

Interpretation: P/3

To write expressions using the numbers 54 and 67 with operations, you can add (54 + 67), subtract (54 - 67), multiply (54 x 67), or divide (54 ÷ 67). The order of operations is important and parentheses may be used to dictate the sequence of calculations. Practice with expressions can be enhanced by working through problems symbolically and then substituting numbers.

The question asks to write expressions using the numbers 54 and 67 with operations. When creating expressions, operations such as addition (+), subtraction (-), multiplication (x or ×), and division (÷) are typically used. For example, if we want to add 54 and 67, the expression would be 54 + 67.

If we wanted to multiply them, the expression would be 54 x 67 (or 54 × 67). Remember that in writing expressions, the order of operations is important, so if we want to perform different operations, we may need to use parentheses to ensure the correct sequence of calculations.

For more practice with expressions and operations, you can approach problems by covering up one number and solving for it using the remaining information (as mentioned in the given examples). This technique helps you solve various types of problems using the same principle. Moreover, when working with algebraic expressions, it's beneficial to solve the equation symbolically first and then substitute the numbers.

Substituting numbers and parentheses placement can significantly alter the result of the expression, as shown in the examples given in the question. The expressions created through experimentation, such as (74) to the third power, can be solved symbolically or using a calculator to understand how different operations and powers affect the result.

Answer:

46

Step-by-step explanation:

-Let b be the constant in the linear equation.

#The linear equation can be expressed as:

[tex]p(x)=100x+b[/tex]

Substitute the values in the equation to find b:

[tex]p(x)=100x+b\\\\-4=100(0.1)+b\\\\b=-14\\\\\#or\\\\36=100(0.5)+b\\\\b=-14[/tex]

We know have the constant value b=-14, substitute the values of b and x in the p(x) function:

[tex]p(x)=100x+b\\\\p(x)=100(0.6)-14\\\\p(x)=60-14\\\\p(x)=46[/tex]

Hence, the profit when selling price is $0.60 is $46

#From our calculations, it's evident that the cookies production has a very high fixed cost which can only be offset by raisng the selling price or the number of units sold at any given time.

If the students sell the cookies for $0.60 each, they will make a profit of $46.

To solve this problem, let's first define the variables and set up the linear function p(x)  that represents the net profit based on the selling price x per cookie.

Given information:

- Selling each cookie for $0.10 results in a net loss of $4.

- Selling each cookie for $0.50 results in a net profit of $36.

From this information, we can set up two equations based on the net profit:

1. When selling each cookie for $0.10:

[tex]\[ R = 100 \cdot 0.10 = 10 \] \[ P(0.10) = R - C = 10 - C = -4 \] \[ C = 10 + 4 = 14 \][/tex]

(Total cost of ingredients)

2. When selling each cookie for $0.50:

[tex]\[ R = 100 \cdot 0.50 = 50 \] \[ P(0.50) = R - C = 50 - C = 36 \] \[ C = 50 - 36 = 14 \][/tex]

Total cost of ingredients)

So, the total cost of ingredients C is $14 regardless of the selling price, since it's consistent in both scenarios.

Now, let's define the linear function  P(x) :

[tex]\[ P(x) = R - C \][/tex]

Where ( R = 100x ) (total revenue from selling 100 cookies at x dollars each), and ( C = 14 ) (total cost of ingredients).

Therefore,

[tex]\[ P(x) = 100x - 14 \][/tex]

This function  P(x) gives us the net profit when each cookie is sold for x dollars.

Now, to find out how much profit the students will make if they sell the cookies for $0.60 each:

[tex]\[ x = 0.60 \]\[ P(0.60) = 100 \cdot 0.60 - 14 \]\[ P(0.60) = 60 - 14 \]\[ P(0.60) = 46 \][/tex]

So, if the students sell each cookie for $0.60, they will make a profit of $46.

Explanation of Reasonableness:

The function [tex]\( P(x) = 100x - 14 \)[/tex] is a linear function that accurately represents the relationship between the selling price x and the net profit ( P(x) ). The function is derived from the given conditions where selling at $0.10 results in a loss and selling at $0.50 results in a profit, confirming the slope and intercept of the function.

Answer: The probability of drawing a red marble the sixth time is 1/2

Step-by-step explanation:

Here is the complete question:

A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?

Explanation:

Since the sampling at random from the box containing the marbles is with replacement, that is, after picking a marble, it is replaced before picking another one, the probability of picking a red marble is the same for each sampling. Probability, P(A) is given by the ratio of the number of favourable outcome to the total number of favourable outcome.

From the question,

Number of favourable outcome = number of red marbles =10

Total number of favourable outcome = total number of marbles = 10+10= 20

Hence, probability of drawing a red marble P(R) = 10 ÷ 20

P(R) = 1/2

Since the probability of picking a red marble is the same for each sampling, the probability of picking a red marble the sixth time is 1/2

The probability is [tex]\frac{1}{2}[/tex]

First, we have to calculate the probability of drawing a red marble on any given try using the formula of probability:

Probability of an event = [tex]\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}[/tex]

P(red) = [tex]\frac{Number\ of\ red\ marbles}{Total\ number\ of\ marbles} =\frac{10}{20} = \frac{1}{2}[/tex]

Since we are replacing after every time we pick up a marble, each event of picking a marble is an independent event

Thus, every time a marble is to be picked, the probabilities remain same.

Hence the probability of drawing a red marble the sixth time is = [tex]P(red) = \frac{1}{2}[/tex]

The complete question is:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?

Answer:

Therefore, we conclude that 2/15  of the team work part-time.

Step-by-step explanation:

We know that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. We calculate what proportion of the team work part-time.

So, we will divide the number of those people who work part-time, by the number of people employed in  a softball team. We get:

x=2/15

Therefore, we conclude that 2/15  of the team work part-time.

Answer:

Mean 71

Standard deviation 3

Step-by-step explanation:

We use the Central Limit Theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution with a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 71, \sigma = 24, n = 64[/tex]. So

Mean 71

Standard deviation [tex]s = \frac{24}{\sqrt{64}} = 3[/tex]

The answers to mentioned questions are conditional probabilities and proportions, requiring more detailed numerical data to provide exact values. But they can be calculated using simple formulas of probability and proportion.

This is a statistics problem and we would need more details to fully answer these questions. But here are some general insights:

Without explicit numbers provided for each category of writer, it's impossible to give numerical solutions to these problems. Instead, we can only provide the formulas to solve them. The same reasoning applies to all subsequent questions.

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Average number of books published in the last five years  = 1.1 books

Proportion of sci-fi writers who published 2 books =  0.2

Probability that a romantic novel writer had not published in the last five years =  0.04

Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book = 0.3

Probability a writer who published in the last years writes romantic novels = 0.2

Probability that 5 writers didn't publish any books = .00243.

Probability that a non-romantic writer published more than 1 book= 0.32

Independence of book types and number published= independent events

Independence of writing sci-fi and not publishing = independent events.

Probability 2 of 4 have published romantic novels= 0.1536

Proportion of writers did not write sci-fi and published 1 or 2 books=0.42

Let's breakdown this complex problem step-by-step.

1. Average number of books published in the last five years

Given that 40% published 2 books and 30% published 1 book, we can find the average as follows:

Average = (0.4 * 2) + (0.3 * 1) + (0.3 * 0) = 0.8 + 0.3 + 0 = 1.1 books

2. Proportion of sci-fi writers who published 2 books

Since the number of sci-fi writers publishing 1 and 2 books is the same, and sci-fi comprises 40% of all writers:

Proportion = 0.4 * 0.5 = 0.2 or 20%

3. Probability that a romantic novel writer had not published in the last five years

20% of total writers wrote romantic novels, and 20% of those did not publish in the last five years:

Probability = 0.2 * 0.2 = 0.04 or 4%

4. Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book

If 60% did not publish romantic or sci-fi, half of these published 2 books:

Probability = 0.6 * 0.5 = 0.3 or 30%

5. Probability a writer who published in the last years writes romantic novels

40% published 2 books and 30% published 1 book, so 70% published. Romantic novelists comprise 20% of all writers:

Probability = (0.2 * 0.7) / 0.7 = 0.2 or 20%

6. Probability that 5 writers didn't publish any books

The probability that one writer didn't publish is 0.3:

Probability = 0.3^5 = 0.00243 or 0.243%

7. Probability that a non-romantic writer published more than 1 book

80% are non-romantic, and 40% of total published 2 books:

Probability = (0.4 * 0.8) = 0.32 or 32%

8. Independence of book types and number published

We need to see if P(A ∩ B) = P(A)*P(B). Since specific data does not align well, these events are not independent.

9. Independence of writing sci-fi and not publishing

40% wrote sci-fi and 30% did not publish:

P(A ∩ B) = 0.4 * 0.3 = 0.12 or 12%

These are independent events.

10. Probability 2 of 4 have published romantic novels

Using binomial distribution:

Probability P(X = 2) = 4C2 * (0.2)^2 * (0.8)^2 = 0.1536 or 15.36%

11. Proportion of writers did not write sci-fi and published 1 or 2 books

60% did not write sci-fi, and of those published, 70% published 1 or 2 books:

Proportion = 0.6 * 0.7 = 0.42 or 42%

Answer:

72¹⁶ possible passwords

10¹⁰ years

Step-by-step explanation:

For each of the 16 characters, the number of possible outcomes is 10 numbers, 52 letters, or 10 special characters, totaling 72 possible values. The number of total different 16 characters passwords is:

[tex]n = 72^{16}[/tex]

If you can test 1 trillion passwords per second, the number of passwords per year is:

[tex]P = 10^{12} * 3,600*24*365\\P=3.1536*10^{19}[/tex]

The time in years that would take to test all passwords is:

[tex]T=\frac{72^{16}}{3.1536*10^{19}}\\T = 1.65*10^{10}\ years[/tex]

Rounding to the nearest power of 10, it would take 10¹⁰ years

The question concerns combinatorics in Mathematics, calculating the total possible passwords given 72 character options for a 16-character length (72^16). Given a rate of 1 trillion tests per second, the time it would take to test all these combinations depends on this total, which we express in years.

The subject of your question is Combinatorics, which falls under Mathematics. It requires finding the total number of valid passwords that can be comprised of certain types of characters, then finding how long it would take to test all those passwords under a certain rate.

If each character in the password can be one of 10 digits, 52 letters (uppercase and lowercase) or 10 special characters, there are overall 72 possible characters. Given the password length is 16 characters, the total number of possibilities would be 72^16. This represents the total number of valid passwords.

With the ability to test 1 trillion (10^12) passwords per second, to find out how long it would take to test all passwords, you divide the total number of passwords by the testing rate. Expressing this in years (seconds in a year being approximately 3.15 x 10^7), you would have 72^16 divided by (10^12 x 3.15 x 10^7) years. Hence, the time required in the worst-case scenario is ultimately dependent on the total number of valid passwords (72^16).

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The cognitive percentile for someone who has played for 7 years is 78.62

The cognitive percentile for someone who has played for 16 years is 48.56

Step-by-step explanation:

From the question the regression line for predicting percentile from years is given as

                 percentile = [tex]102 - 3.34 \ Years[/tex]

For someone who has played for 7 years his cognitive percentile would be

           [tex]Percentile = 102 - 3.34(7)[/tex]

                             [tex]=78.62[/tex]

For someone who has played for 16 years his cognitive percentile would

be   [tex]Percentile = 102 - 3.34(16)[/tex]

                        [tex]=48.56[/tex]

Final answer:

The predicted cognitive percentile for someone who has played football for 7 years is 78.62, while for someone who has played for 16 years, it is 48.56, using the regression equation Percentile = 102 - 3.34 Years.

Explanation:

To predict the cognitive percentile for a football player who has played for 7 years, we use the given regression equation Percentile = 102 - 3.34 Years. By plugging 7 into the equation for Years, we get:

Percentile = 102 - 3.34 × 7

Percentile = 102 - 23.38

Percentile = 78.62

So, for someone who has played football for 7 years, the predicted cognitive percentile would be 78.62.

Similarly, for someone who has played football for 16 years:

Percentile = 102 - 3.34 × 16

Percentile = 102 - 53.44

Percentile = 48.56

Therefore, the predicted cognitive percentile for someone who has played for 16 years is 48.56.

Answer and Step-by-step explanation:

The answer is attached below

To identify a second solution y2 from a given differential equation and its solution y1, it is necessary to extract P(x) from the differential equation, compute integral -∫P(x) dx, and multiply the result by the function y1.

First, we need to find the function P(x) in the equation y2 = y1(x) e−∫P(x) dx (5). Looking at the given differential equation (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1, we can rearrange terms and find that P(x) is equal to -2(1+x)/(1-2x-x2). Then, we can calculate the integral -∫P(x) dx, and multiply this by our given solution y1 to find the second solution y2. It's important to remember that when carrying out these steps, accuracy is crucial since each step builds on the last.

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Answer:

[tex] z = \frac{X -\mu}{\sigma}[/tex]

And we can solve for the value of X like this:

[tex] X = \mu + z*\sigma[/tex]

And since we know that z=2 we can replace and we got:

[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we can assume the distribution for X is given by:

[tex]X \sim N(49,20.7)[/tex]  

Where [tex]\mu=49[/tex] and [tex]\sigma=20.7[/tex]

And for this case the z score is given by:

[tex] z = \frac{X -\mu}{\sigma}[/tex]

And we can solve for the value of X like this:

[tex] X = \mu + z*\sigma[/tex]

And since we know that z=2 we can replace and we got:

[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]

To find the test score corresponding to a z-score above z=2.00, use the formula x = (z * standard deviation) + mean. Plugging in the values, the test score is approximately 90 when rounded to the nearest whole number.

The teacher wants to identify students who have z-scores above z=2.00. To find the corresponding test score,

we can use the formula for z-score:

z = (x - mean) / standard deviation

Rearranging the formula, we get:

x = (z * standard deviation) + mean

Substituting z=2.00, standard deviation=20.7, and mean=49, we have:

x = (2.00 * 20.7) + 49

Simplifying the equation, we get:

x = 41.4 + 49 = 90.4

Therefore, the test score corresponding to a z-score above z=2.00 is approximately 90 when rounded to the nearest whole number.

48 is what 4x4 + 8x3 + 4x2 equals So which means my assumption would definitely be that 'F' is 24 So it would be like- 24x, Like 24 x 2..? I'm so so sorry if i'm wrong but i'm 95.0% sure i'm right! OwO

Answer:

f(x) = 48

Multiplicity: 24 * 2

Step-by-step explanation:

Evaluate the function

Rather Simple

And i believe multiplicity you mean as in the equation you would use to get 48 right?

so that would be 24 * 2

Hope this helps~

Answer:

Data

Step-by-step explanation:

We are given the following in the question:

We want to measure height of all psychology majors at the university.

Thus, the resulting raw scores of each individual are called the data.

Data point:

Height of each psychology majors at the university

Data:

Collection of all heights of all psychology majors at the university

These value are constants but comprises a data.

They are neither coefficients nor statistic because they do not describe a sample.

Thus, the correct answer is

Data

Answer:

254.34 in^3

Step-by-step explanation:

Volume of a cylinder is V=pi*r^2*h

Since r=6/2=3, and h=3r=3*3=9, then we solve for V:

V=3.14*3^2*9

V=3.14*9*9

V=3.14*81

V=254.34

So the volume of the cylinder is 254.34 in^3

Answer: volume = 254.3 inches³

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Diameter = 6 inches

Radius = diameter/2 = 6/2 = 3 inches

Height = 3 × radius = 3 × 3 = 9 inches

Therefore,

Volume = 3.14 × 3² × 9

Volume = 254.3 inches³ to the nearest tenth

Answer:

c) NORM.INV(R, 1200, 350)

Step-by-step explanation:

Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.

X = demand for gasolene at a local gas station is N(1200, 350)

R is any random number between 0 and 1.

Daily demand for gasolene would be

X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.

The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X

Hence correct choice would be

Option c) NORM.INV(R, 1200, 350)

Answer:

Yes, its continuous

Step-by-step explanation:

We use the formula:

x^2-y^2=(x-y)(x+y),

And we know that 16=4^2, so we have:

[tex]\frac{x^2-16}{(x+4)(x-5)}=\frac{(x-4)(x+4)}{(x+4)(x-5)}=\frac{x-4}{x-5}[/tex]

So for x=-4 we have -8/-9,i.e, it is 8/9, so it is continuous.

I dont know what is x=1, because for x=1 the function has value 3/4.

But function is not continuous in x=5 becaus for that x we will get 1/0, and that is not definite.

:)

Answer:

The standard error is  0.0456 for the mean finish time of 30 randomly selected participants.            

Step-by-step explanation:

We are given the following in the question:

Population mean, [tex]\mu[/tex] = 1.67 hours

Population standard deviation, [tex]\sigma[/tex] = 0.25 hours

Sample mean, [tex]\bar{x}[/tex] = 1.59 hours

Sample standard deviation, s = 0.30 hours

Sample size, n = 30

We have to find the standard error for the mean finish time of 30 randomly selected participants.

Formula:

[tex]\text{Standard error} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{0.25}{\swqrt{30}} = 0.0456[/tex]

Thus, the standard error is  0.0456 for the mean finish time of 30 randomly selected participants.

The standard error for the mean finish time of 30 randomly selected participants is 0.0549 hours.

The standard error for the mean finish time of 30 randomly selected participants can be calculated using the formula:

Standard Error = Standard Deviation / √(Sample Size)

Plugging in the given values, the standard error would be:

Standard Error = 0.30 / √(30) = 0.0549 hours

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Answer:

The measure of the other angle is 59°

Step-by-step explanation:

Supplementary Angles

Two angles [tex]\alpha[/tex] and [tex]\beta[/tex] are supplementary when they add up to 180 degrees, i.e.

[tex]\alpha+\beta=180^o[/tex]

One notable property is that together they make a straight angle although they don't have to be together to be supplementary.

We are given one of two supplementary angles with a value of 121 degrees, we can compute the measure of the other angle, say [tex]\alpha[/tex] as

[tex]\alpha=180^o-\beta=180^o-121^o=59^o[/tex]

The measure of the other angle is 59°

Answer:

(a) 3.432

(b) 3.175

Step-by-step explanation:

The given data and corresponding weights is:

                                                  [tex]\begin{array}{cc}x_i&w_i\\3.2&6\\3.0&3\\2.5&2\\4.0&8\\\end{array}[/tex]

(a) The weighted mean is determined by the following expression:

[tex]M_W=\sum\frac{x_i*w_i}{w_i}\\M_W=\frac{3.2*6+3.0*3+2.5*2+4.0*8}{6+3+2+8}\\M_W=3.432[/tex]

(b) The simple mean of the given data is determined by adding up the four values dividing the result by 4:

[tex]M_S = \frac{3.2+3.0+2.5+4.0}{4}\\ M_S=3.175[/tex]

The value is lower than the weighted mean.

The weighted mean is 3.455, while the sample mean without weighting is 3.175.

To compute the weighted mean, we multiply each data value by its corresponding weight, then add up the results. We then divide this sum by the sum of the weights. In this case, the weighted mean can be calculated as follows:

Weighted Mean = (3.2 * 6 + 3.0 * 3 + 2.5 * 2 + 4.0 * 8) / (6 + 3 + 2 + 8) = 3.455

To compute the sample mean without weighting, we simply add up all the data values and divide the sum by the number of data values. In this case, the sample mean can be calculated as follows:

Sample Mean = (3.2 + 3.0 + 2.5 + 4.0) / 4 = 3.175

The difference between the two computations is that the weighted mean takes into account the importance of each data value by assigning weights to them, whereas the sample mean without weighting treats all data values equally.

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Answer:

97.92% probability that the mean score of your sample is between 22 and 28

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 25, \sigma = 6.5, n = 25, s = \frac{6.5}{\sqrt{25}} = 1.3[/tex]

What is the probability that the mean score of your sample is between 22 and 28

This is the pvalue of Z when X = 28 subtracted by the pvalue of Z when X = 22. So

X = 28

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{28 - 25}{1.3}[/tex]

[tex]Z = 2.31[/tex]

[tex]Z = 2.31[/tex] has a pvalue of 0.9896

X = 22

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{22 - 25}{1.3}[/tex]

[tex]Z = -2.31[/tex]

[tex]Z = -2.31[/tex] has a pvalue of 0.0104

0.9896 - 0.0104 = 0.9792

97.92% probability that the mean score of your sample is between 22 and 28

a) 21 sample points

b) Sum 2: (1, 1)

- Sum 3: (1, 2)

- Sum 4: (1, 3), (2, 2)

- Sum 5: (1, 4), (2, 3)

- Sum 6: (1, 5), (2, 4), (3, 3)

- Sum 7: (1, 6), (2, 5), (3, 4)

- Sum 8: (2, 6), (3, 5), (4, 4)

- Sum 9: (3, 6), (4, 5)

- Sum 10: (4, 6), (5, 5)

- Sum 11: (5, 6)

- Sum 12: (6, 6)

c) Probability of obtaining sum of 5 is 2/21

d) Probability of obtaining 8 or greater is 3/7

e) Probability of even is higher that the probability of odd, so even sum are expect to have more appear.

f) classic method

What is probability?

(a) There are 21 possible sample points when rolling a pair of dice.

(b) Here are the sample points for each sum:

- Sum 2: (1, 1)

- Sum 3: (1, 2)

- Sum 4: (1, 3), (2, 2)

- Sum 5: (1, 4), (2, 3)

- Sum 6: (1, 5), (2, 4), (3, 3)

- Sum 7: (1, 6), (2, 5), (3, 4)

- Sum 8: (2, 6), (3, 5), (4, 4)

- Sum 9: (3, 6), (4, 5)

- Sum 10: (4, 6), (5, 5)

- Sum 11: (5, 6)

- Sum 12: (6, 6)

(c) The probability of obtaining a sum of 5 is

: (1, 4), (2, 3)

P(sum of 5) = no of sum of 5 /total number of our

= 2/21

(d) The probability of obtaining a sum of 8 or greater is (2, 6), (3, 5), (4, 4) (3, 6), (4, 5) (4, 6), (5, 5),(5, 6) (6, 6)

P(sum of 8 or greater) = no of sum of 8 and above

P(sum => 8) = 9/21 = 3/7

(e) Yes!

Out of 21 sample points 12 are even while 9 are odd.

P(even) = 12/21 = 4/7

P(odd) = 1 - 4/7 = 3/7

Probability of even is higher that the probability of odd, so even sum are expect to have more appear.

(f) We used the classical method, which involves counting the number of favorable outcomes and dividing by the total number of possible outcomes.

Answer:

5278.0

21112

2927

Step-by-step explanation:

P = Po[2^(t/5)]

8000 = Po(2^⅗)

Po = 5278.0

P = 5278(2^(10/5))

P = 21112

P = Po[2^(t/5)]

ln(P/Po) = (t/5)ln2

ln(P) - ln(Po) = (t/5)ln2

1/P . dP/dt = ln2/5

dP/dt = P(ln2)/5

At t = 10, P = 21112

dP/dt = 2927

Using differentials, the estimated maximum error in the calculated surface area of a sphere with a measured circumference of 74 cm and a possible error of 0.5 cm is 24 cm². The relative error is approximately 5%.

The subject concerns the application of differentials in estimating the maximum error in the calculated surface area of a sphere. Given the circumference C = 74 cm with a possible error δC = 0.5 cm, we can calculate the radius r = C / (2π). With the surface area formula of a sphere A = 4πr², differentiating this equation gives dA = 8πr dr. By substituting the values, the maximum error in calculated surface area δA = dA = 8πr δr = 8π(C/2π) (δC/2π) = 2C δC / π. Plugging the values of C = 74 cm and δC = 0.5 cm, we get δA ≈ 24 cm² which is the maximum error in the calculated surface area. For the relative error, it is the absolute error divided by the actual measurement, hence, the relative error is δA/A = δA / 4πr² = (2C δC / π) / 4π(C/2π)² ≈ 0.05 or 5%.

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To find a formula for the moose population, calculate the rate of change and use it in the formula P = 190t + 4360. The model predicts the moose population to be 7710 in 2003.

To find a formula for the moose population, we need to determine the rate of change in the population. We can do this by finding the slope of the line that represents the change in population from 1991 to 1999. First, we calculate the change in population: 5880 - 4360 = 1520. Then, we calculate the change in time: 1999 - 1991 = 8. Next, we divide the change in population by the change in time to find the rate of change: 1520/8 = 190. So, the formula for the moose population, P, is P = 190t + 4360, where t represents the years after 1991.

To predict the moose population in 2003, we substitute t = 12 (since 2003 is 12 years after 1991) into the formula: P = 190(12) + 4360 = 7710. Therefore, the model predicts the moose population to be 7710 in 2003.

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The minimum sample size required to estimate μ within 3 with a confidence level of 0.99, given a standard deviation of 8, is approximately 48. This was determined by plugging the values into the sample size formula and rounding up to the nearest whole number.

To find the minimum sample size, we need to use the formula for sample size n, = (Z_α/2 * σ / E)^2. In this problem, you want to estimate μ correct to within 3 with a probability of 0.99. In other words, you want the error E to be 3 and the confidence level to be 0.99.

The Z value corresponding to a confidence level of 0.99 is approximately 2.576 (you can find this value from a standard Z-table). The measurements range from 8 to 40, so we can estimate the standard deviation σ as (40 - 8) / 4 = 8.

Plugging these values into the formula, we get n = (2.576 * 8 / 3)^2 = 47.36. This number must be rounded up to the nearest whole number because the sample size cannot be a fraction. So, the minimum sample size required is 48.

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Answer:

The distribution with n = 225 will give a smaller standard error.

Since sigma x = sigma/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of sigma.

Step-by-step explanation:

Standard error is given by standard deviation (sigma) divided by square root of sample size (√n).

The distribution with n = 225 would give a smaller standard error because the square root of 225 is 15. The inverse of 15 multiplied by sigma is approximately 0.07sigma which is smaller compared to the distribution n = 100. Square of 100 is 10, inverse of 10 multiplied by sigma is 0.1sigma.

0.07sigma is smaller than 0.1sigma

The right format of the number is (45^8)(88^5).

Answer:

There are 8 zeros

Step-by-step explanation:

Using the unique factorization of integers theorem, we can break any integer down into the product of prime integers.

So breaking it down we have;

(45^8) = (3 x 3 x 5)^(8)

(88^5) = (2 x 2 x 2 x 11)^(5)

Now, if we put it back together as separate factors, we'll get;

(3^(16)) x (5^(8) ) x (2^(15)) x (11^(5))

Now let's find the number of zeroes by figuring out how many factors of 10 (which equals 2 x 5) we can make. Thus, we can make 8 factors of 10 so it looks like;

(3^(16)) x (2^(7)) x (11^(5)) x (10^(8))

Thus, we can see that there will be 8 zeros as the end is (10^(8))

The number of trailing zeros in the product of 457 and 885 is determined by the factors of 10 (2 and 5) in these numbers. In this specific case, there are no trailing zeros. This also applies in scientific notation - the number of significant figures after the decimal in scientific notation indicates the quantity of zeros at the end of the number.

To determine the number of zeroes at the end of the number 457 · 885, consider the factors in the product. Each zero at the end of a number results from a factor of 10, which contains a factor of 2 and a factor of 5. Looking at the numbers 457 and 885, we notice that neither has a factor of 5, therefore there are no trailing zeroes in the product of 457 and 885.

This approach also applies to the scientific notation. The number of significant figures after the decimal in the scientific notation of a number corresponds to the quantity of zeros at the end of it. In such cases, leading zeros are not significant and only serve as placeholders to locate the decimal point. For instance, in the case of the number 1.300 × 10³, the scientific notation shows that there are three significant figures after the decimal and therefore, three zeros at the end of the number.

Overall, understanding how to find the number of trailing zeroes in a product, such as 457 · 885, without actual computation involves a knowledge of the factors of the numbers being multiplied and the principles of significant figures in scientific notation.

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Answer:

2.4948ml

Step-by-step explanation:

First, we change the child's weight into kilograms:

[tex]22pounds=9.9790kgs[/tex]

From the info, the dose is recommended as 2.5mg/kg. Let x be the number of mg administered to the child:

[tex]1kg=2.5mg\\9.9790kg=x\\\\x=2.5\times9.9790\\\\x=24.9475mg[/tex]

#The drug contains 10mg/ml . Let y be the dose size administered, equate and solve for y:

[tex]10mg=1ml\\24.9475mg=y\\\\\thereforey= \frac{1ml\times24.9475mg}{10mg}\\\\y=2.4948[/tex]

Hence, the dose required is 2.4948ml

Answer:

2.5

Step-by-step explanation:

Answer:

See answers below

Step-by-step explanation:

# prior

mu0 <- 5

k0 <- 1

s20 <- 4

nu0 <- 2

# read in data

dat <- list()

dat[1] <- read.table("school1.dat")

dat[2] <- read.table("school2.dat")

dat[3] <- read.table("school3.dat")

n <- sapply(dat, length)

ybar <- sapply(dat, mean)

s2 <- sapply(dat, var)

# posterior

kn <- k0 + n

nun <- nu0 + n

mun <- (k0 * mu0 + n * ybar)/kn

s2n <- (nu0 * s20 + (n - 1) * s2 + k0 * n * (ybar - mu0)^2/kn)/nun

# produce a posterior sample for the parameters

s.postsample <- s2.postsample <- theta.postsample <- matrix(0, 10000, 3, dimnames = list(NULL,  

   c("school1", "school2", "school3")))

for (i in c(1, 2, 3)) {

   s2.postsample[, i] <- 1/rgamma(10000, nun[i]/2, s2n[i] * nun[i]/2)

   s.postsample[, i] <- sqrt(s2.postsample[, i])

   theta.postsample[, i] <- rnorm(10000, mun[i], s.postsample[, i]/sqrt(kn[i]))

}

# posterior mean and 95% CI for Thetas (the means)

colMeans(theta.postsample)

## school1 school2 school3  

##   9.294   6.943   7.814

apply(theta.postsample, 2, function(x) {

   quantile(x, c(0.025, 0.975))

})

##       school1 school2 school3

## 2.5%    7.769   5.163   6.152

## 97.5%  10.834   8.719   9.443

# posterior mean and 95% CI for the Standard Deviations

colMeans(s.postsample)

## school1 school2 school3  

##   3.904   4.397   3.757

apply(s.postsample, 2, function(x) {

   quantile(x, c(0.025, 0.975))

})

##            school1 school2 school3

## 2.5%    2.992   3.349   2.800

## 97.5%   5.136   5.871   5.139

The question calls for Bayesian inference with a normal model and a conjugate prior to derive posterior distributions and probabilities concerning means and standard deviations. Statistical software such as R or Python with libraries like PyMC3 is needed to perform the detailed computations, which are not provided here.

The provided question involves advanced statistical analysis, specifically Bayesian inference with prior distributions and the computation of posterior probabilities. Unfortunately, without software or computational methods to process the provided datasets for schools school1.dat, school2.dat, and school3.dat, a comprehensive answer cannot be provided. However, the general approach to tackle this problem would involve:

To complete the analysis, one would typically utilize statistical software capable of Bayesian computation, such as R or Python with appropriate libraries (e.g., PyMC3).

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Answer:

a) B(t)= 64.8e^(-0.01255t), k=-0.0125 b) 56.5 c)2150

Step-by-step explanation:

a) B(t)= Ae^(-kt)

at t=0, B(t)=64.8

A=64.8

at t=6, B(t)=60.1

60.1=64.8e^(-6k)

k=0.0125

b) B(t)=64.8e^(-0.0125×11)

B(t)= 56.5

c) 10=64.8e^(-0.0125t)

0.15432=e^(-0.0125t)

-0.0125t=ln(0.15432)

-0.0125t=-1.869

t=149.5 or 150

Year= 2150

a)k ≈ 0.01267 and the equation is [tex]B(t) = 64.8e^{-0.01267t[/tex]

b)B(11) ≈ 56.38 lb

c) t ≈ 143.86

A beef consumption model follows exponential decay where B(t), the annual beef consumption t years after 2000,

can be modeled by the equation: [tex]B(t) = B_0e^{-kt}.[/tex]

Given that [tex]B_0[/tex] = 64.8 lb in 2000 and B(6) = 60.1 lb in 2006,

we first need to find the decay constant k.

a) Find the value of k, and write the equation:

We use the given data points to solve for k.

[tex]B(6) = 64.8e^{-6k} = 60.1[/tex]

[tex]e^{-6k} = 60.1 / 64.8[/tex]
[tex]e^{-6k}[/tex] ≈ 0.9272
Taking the natural log of both sides:
-6k = ln(0.9272)
k ≈ -ln(0.9272) / 6
k ≈ 0.01267

Therefore, the equation for B(t) is:

[tex]B(t) = 64.8e^{-0.01267t[/tex]

b) Estimate the consumption of beef in 2011:

For t = 11 (since 2011 is 11 years after 2000):

[tex]B(11) = 64.8e^{-0.01267 * 11[/tex]
B(11) ≈ 64.8[tex]e^{-0.13937[/tex]
B(11) ≈ [tex]64.8 * 0.8699[/tex]
B(11) ≈ 56.38 lb

c) In what year (theoretically) will the consumption of beef be 10 lb?

We need to solve for t when B(t) = 10 lb:

[tex]10 = 64.8e^{-0.01267t[/tex]
[tex]e^{-0.01267t} = 10 / 64.8[/tex]
[tex]e^{-0.01267t[/tex] ≈ 0.1543
Taking the natural log of both sides:
-0.01267t = ln(0.1543)
t ≈ -ln(0.1543) / 0.01267
t ≈ 143.86

Thus, the theoretical year is approximately 2000 + 144 = 2144.