A student organization wanted to study voting preferences in its student body during the 2012 presidential election. They selected 120 students at random from each class, freshmen through seniors. The sampling technique used is: O stratified random sampling. O volunteer sampling. multistage sampling. Osimple random sampling.

As per the inequality -0.4 < 0 which indicates that the low tide on Wednesday is also less than the typical low tide on Monday.

In this case, we can use the symbol < to indicate that the low tide on Tuesday morning is less than the low tide on Monday morning. This can be written as:

-0.8 < 0

The left-hand side of this inequality represents the low tide on Tuesday morning (-0.8 feet) and the right-hand side represents the typical low tide level on Monday morning (0 feet). The inequality indicates that the low tide on Tuesday is less than the typical low tide on Monday.

Similarly, we can compare the low tide on Wednesday morning to the low tide on Monday morning using the same inequality:

-0.4 < 0

In summary, we can use inequalities to compare the low tides on different days relative to a reference point. By using the typical low tide on Monday morning as the reference point, we can see that the low tides on Tuesday and Wednesday mornings are both lower than the typical low tide.

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The answer  is true. Inferential statistics use mathematical equations to generate probabilities, draw associations, and make predictions about a population, whereas descriptive statistics only provide summaries of data in the form of measures such as mean, median, and mode.

Inferential statistics involves the use of sample data to make inferences about a larger population. This involves using statistical methods such as hypothesis testing, confidence intervals, and regression analysis to draw conclusions about the population based on the sample data. These methods involve mathematical equations that allow for the calculation of probabilities and the estimation of parameters such as population means and variances.

In contrast, descriptive statistics simply involves summarizing and describing the characteristics of a dataset using measures such as central tendency, variability, and distribution. While descriptive statistics can provide useful information about a dataset, it does not involve making inferences or predictions about a larger population.

In summary, inferential statistics use mathematical equations to draw conclusions about a population based on sample data, while descriptive statistics simply provide summaries of data.

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Answer: (1,1)

Step-by-step explanation:

Our confidence interval is only representative of the sample that we selected, and there could be variation in the true population proportion.

To construct a confidence interval for the population proportion of all pregnant women who gave birth the week of their due date

the following steps can be taken:

1. Determine the sample size: The sample size can be determined based on the desired level of confidence and margin of error. Let's say we want a 95% confidence level and a margin of error of 5%, which means we want to be 95% confident that the true proportion falls within 5% of the sample proportion. Using a confidence interval calculator, the required sample size would be 385.

2. Select a random sample of women who have recently given birth: The sample should be selected randomly to ensure that it is representative of the population of all pregnant women.

3. Calculate the sample proportion: Determine the proportion of women in the sample who gave birth the week of their due date.

4. Calculate the standard error: The standard error can be calculated using the formula SE = √((p*(1-p))/n), where p is the sample proportion and n is the sample size.

5. Calculate the confidence interval: Using a confidence interval calculator, the confidence interval for the population proportion can be calculated. For our example, the confidence interval would be 0.404 to 0.556.

The confidence interval we constructed (0.404 to 0.556) does not contradict the study's claim that a certain proportion of pregnant women give birth the week of their due date, as the interval includes the possibility of this proportion being within the range of 0.40 to 0.56.

Our confidence interval is only representative of the sample that we selected, and there could be variation in the true population proportion.

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Since r satisfies all three properties of an equivalence relation (reflexivity, symmetry, and transitivity), we can conclude that r is an equivalence relation on Z.

If and only if a relation R on a set A is reflexive, symmetric, and transitive, then it qualifies as an equivalence relation. On the set, the equivalence relation is a relationship that is typically denoted by the symbol " ∼".

To prove that the relation r on Z is an equivalence relation, we need to show that it satisfies the three properties of an equivalence relation:

1. Reflexivity: For any integer a, we have a² = a², so a is related to itself under r. Therefore, r is reflexive.

2. Symmetry: For any integers a and b, if a is related to b under r, then a² = b². By taking the square root of both sides, we get |a| = |b|. Therefore, b is also related to a under r. Hence, r is symmetric.

3. Transitivity: For any integers a, b, and c, if a is related to b under r and b is related to c under r, then a² = b² and b² = c². By transitivity of equality, we have a² = c². Therefore, a is related to c under r. Hence, r is transitive.

Since r satisfies all three properties of an equivalence relation (reflexivity, symmetry, and transitivity), we can conclude that r is an equivalence relation on Z.

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The number of hours worked by each person is given as follows:

The variables for the system of equations are given as follows:

Daniel worked 6 more hours than Justin, hence:

y = x + 6.

They ironed 360 shirts between them, hence, considering the rates, we have that:

40x + 20y = 360.

From the graph given at the end of the answer, the intersection point of the two equations is given as follows:

(4, 10).

Hence the number of hours worked by each person is given as follows:

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$13.50 - $8 =

$5.50

...............

Answer: We can use a one-sample t-test to determine whether the sample mean pulse rate of 67.7 bpm is significantly different from the hypothesized population mean of 69 bpm. The test statistic can be calculated as:

t = (sample mean - hypothesized population mean) / (standard deviation / sqrt(sample size))

Substituting the given values, we get:

t = (67.7 - 69) / (10.6 / sqrt(165))

t = -1.3 / 0.823

t ≈ -1.58

Therefore, the value of the test statistic is -1.58.

we can estimate that the proportion of defectives being produced by the machine is around 0.316.

What is proportion?

A comparison between the size, number, or amount of one thing or group with that of another. In our class, there are three boys for everyone lady.

If the random sample of size 2 yields 2 defects, that means both items in the sample were defective. Let p be the proportion of defectives being produced by the machine.

The probability of selecting a defective item on the first draw is p, and the probability of selecting a defective item on the second draw is also p (assuming sampling without replacement).

Since both items were defective, the probability of this happening is p * p = p².

So,

p² = (number of samples with 2 defects) / (total number of samples)

We don't know the values of these numbers, but we can use them to estimate p. For example, if we had a total of 100 samples and 10 of them had 2 defects, then:

p² = 10/100 = 0.1

p ≈ √(0.1) ≈ 0.316

Hence, we can estimate that the proportion of defectives being produced by the machine is around 0.316.

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The probability that at most 2 accounting majors are on the committee is 60.6%.

To solve this problem, we can use the binomial probability formula:

P(X ≤ 2) = ΣP(X = i), where i = 0, 1, or 2

P(X = i) = (n choose i) * p^i * (1-p)^(n-i)

where n is the total number of available majors (18), p is the probability of selecting an accounting major (10/18), and (n choose i) is the binomial coefficient which gives the number of ways to select i accounting majors from n total majors.

So, to find the probability that at most 2 accounting majors are on the committee, we need to sum the probabilities of selecting 0, 1, or 2 accounting majors.

P(X = 0) = (8 choose 5) * (10/18)^0 * (8/18)^5 = 0.018
P(X = 1) = (10 choose 1) * (10/18)^1 * (8/18)^4 = 0.219
P(X = 2) = (10 choose 2) * (10/18)^2 * (8/18)^3 = 0.369

Therefore, P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.018 + 0.219 + 0.369 = 0.606 or 60.6%

So the probability that at most 2 accounting majors are on the committee is 60.6%.

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There are a total of six people in the group. To select two individuals at random, we can use the formula for combinations, which is nCr = n!/r!(n-r)!, where n is the total number of individuals in the group and r is the number of individuals to be selected. Therefore, the number of selections possible is:

6C2 = 6!/(2!(6-2)!) = 15

The formula for combinations is used to calculate the number of ways we can select r individuals from a total of n individuals. This formula is nCr = n!/r!(n-r)!. In this case, n is 6 (the total number of individuals in the group) and r is 2 (the number of individuals to be selected). Therefore, we substitute n=6 and r=2 into the formula to obtain the number of selections possible.

The number of selections possible from a group of six people, where two individuals are to be selected at random, is 15. This is obtained by using the formula for combinations, which is nCr = n!/r!(n-r)!, where n is the total number of individuals in the group and r is the number of individuals to be selected.

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The current yield on this bond is 6.49%.

A bond's yearly interest payment, market price, and par value must be known in order to determine the bond's current yield.

In this question, we are given that the bond pays an annual interest of $60, has a par value of $1,000, and is selling at a $75.25 discount from par value.

So, the market price of the bond is the par value minus the discount, which is [tex]$ 1,000[/tex] - [tex]$75.25[/tex] [tex]= $924.75.[/tex]

To calculate the current yield, we use the formula:

Current Yield [tex]=[/tex] (Annual Interest Payment / Current Market Price) x 100%

With our current values substituted, we obtain:

Current Yield [tex]= ($60 / $924.75) × 100%[/tex]

Current Yield [tex]= 0.0649 × 100%[/tex]

Current Yield [tex]= 6.49%[/tex]%

Therefore, (B) 6.49% is the right response.

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Since we weighed 15 German Shepherds, the degrees of freedom for the t-statistic would be: Degrees of freedom = Sample size - 1 = 15 - 1 = 14. So, the t-statistic has 14 degrees of freedom.

To determine the degrees of freedom for this t-statistic, we need to know the sample size. Since we weighed 15 German Shepherds, the degrees of freedom would be 15 - 1, which equals 14. Therefore, the t-statistic has 14 degrees of freedom.

To determine the degrees of freedom for the t-statistic in this case, you'll need to consider the sample size of German Shepherds you've weighed. Since you weighed 15 German Shepherds, the degrees of freedom for the t-statistic would be:
Degrees of freedom = Sample size - 1 = 15 - 1 = 14

So, the t-statistic has 14 degrees of freedom.

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Each of the three heat transfer equations is used under different circumstances.

The first equation, q = n x delta H, is used to calculate the amount of heat transferred during a chemical reaction or a physical change where the number of moles of the substances involved changes. This equation uses the enthalpy change (delta H) of the reaction and the number of moles (n) of the substance that undergoes the reaction to calculate the amount of heat (q) transferred.

The second equation, q = C x delta T, is used to calculate the amount of heat transferred during a temperature change. This equation uses the specific heat capacity (C) of the substance and the change in temperature (delta T) to calculate the amount of heat (q) transferred.

The third equation, q = m x cP x delta T, is used to calculate the amount of heat transferred during a temperature change of a substance with a constant mass. This equation uses the mass (m) of the substance, the specific heat capacity at constant pressure (cP), and the change in temperature (delta T) to calculate the amount of heat (q) transferred.

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The given confidence interval for the difference in mean number of hours spent playing two types of video games is (1.5, 3.8) at 90% confidence level.

The maker of one of the games claims that there is a significant difference in the population mean number of hours spent playing the two games. To determine if the claim is supported by the interval, we check if the interval contains 0. Since 0 is not in the interval, we can reject the null hypothesis of no difference in population means at 90% confidence level.

Therefore, the claim is supported by the interval and we can conclude that there is a significant difference in the mean number of hours spent playing the two types of video games.

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Answer:

D. The distance around a circle

a. To determine the maximum wait time that would result in at most 1.5% of customers receiving a free meal, we need to find the z-score associated with the 1.5% probability. Using a standard normal distribution table or calculator, we find that the z-score is approximately -2.33.

From the formula for a z-score:

z = (x - mu) / sigma

where x is the wait time we want to find, mu is the mean wait time (19 minutes), and sigma is the standard deviation (3.3 minutes), we can solve for x:

-2.33 = (x - 19) / 3.3

x - 19 = -7.689

x = 11.311

Therefore, we should advertise a maximum wait time of 11.3 minutes to ensure that no more than 1.5% of customers receive a free meal.

b. See attached graph.

The normal distribution is a continuous probability distribution that describes the probability of a random variable taking on a range of values. It is characterized by its mean and standard deviation and has a bell-shaped curve.

The normal distribution might apply to this situation because it is a common model for many real-world phenomena that exhibit random variation. In this case, the waiting times of customers are likely to be influenced by a variety of factors, including the time of day, the number of customers, and the efficiency of the staff. These factors may produce a random variation in waiting times that can be modeled using a normal distribution.

The specific normal distribution for this situation has a mean of 19 minutes and a standard deviation of 3.3 minutes. This means that the majority of customers can expect to wait around 19 minutes for their meal, with some variability around this average.

The graph created in part b represents the waiting times of the customers by showing the distribution of waiting times and highlighting the area corresponding to the 1.5% of customers who would receive a refund. The graph is centered at the mean of 19 minutes and has a spread of 3.3 minutes, which reflects the variability in waiting times.

The answer to part a suggests that we should advertise a maximum wait time of 11.3 minutes to ensure that no more than 1.5% of customers receive a free meal. For customers who do not receive a free meal, this wait time represents a reasonable expectation for their wait time. For customers who do receive a free meal, the wait time would be longer than expected but would still be within the realm of reasonable expectations. In practice, the company may choose to round this number up to a "nice" wait time, such as 10 minutes or 15 minutes, to make it more appealing to customers.

The proposal will not result in a loss of profit for the company because the maximum wait time advertised is well within the expected range of waiting times. By offering a refund for customers who wait longer than this time, the company is incentivizing its staff to work more efficiently and reduce waiting times overall. The cost of a few free meals is likely to be offset by increased customer satisfaction and loyalty, as well as reduced negative reviews and complaints. Additionally, the company can control the maximum refund rate by adjusting the advertised wait time as needed.

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Full Question ;

Suppose that you are working for a chain restaurant and wish to design a promotion to disabuse the public of notions that the service is slow. You decide to institute a policy that any customer that waits too long will receive their meal for free.

a. Determine an estimate of an advertised maximum wait time so that 1.5% of the customers would receive a free meal. Round to one decimal place.

b. Include a graph illustrating the solution. For the graph do NOT make an empirical rule graph, just include the mean and the mark off the area that corresponds to the 1.5% who would receive the refund. There is a Normal Distribution Graph generator linked in the resources area. Combine the above into as single file and upload using the link below.

c. Write a response to the vice president explaining your prescribed maximum wait time. Structure your essay as follows:

1. An advanced explanation of the normal distribution

2. Why the normal distribution might apply to this situation

3. Describe the specific normal distribution for this situation (give the mean and standard deviation)

4. Explain how the graph created in part b. represents the waiting times of the customers.

5. Explain the answer to part a. in terms of both the customers who get a free meal and those who do not. Feel free to use the accurate answer in part a to determine a "nice" wait time to be used in the actual advertising campaign.

6. Use the answers to parts a. and b. to explain how the proposal will not result in a loss of profit for the company.

The model with the higher PRE value is the one that has a greater proportional reduction in error, indicating that it explains more of the variance in the dependent variable and is therefore a better fit for the data.

To compare the pre (proportional reduction in error) for the two models, we need to calculate the pre value for each model and then compare them. The pre value measures the reduction in error achieved by using a model compared to using no model.

In the two-group model, pre can be calculated as the difference between the error rate of using no model and the error rate of using the two-group model, divided by the error rate of using no model. In the regression model, pre can be calculated as the difference between the error rate of using no model and the error rate of using the regression model, divided by the error rate of using no model.

Assuming that the two models are equally well-fitted to the data, the model with the higher pre value is the one that achieves a greater reduction in error. So, we need to compare the pre values for the two models.

If the two-group model achieves a higher pre value than the regression model, then it has a higher pre. Conversely, if the regression model achieves a higher pre value than the two-group model, then it has a higher pre. Without further information or analysis, it is not possible to say which model has a higher pre.

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The vectors [tex]v1[/tex], [tex]v2, v3, v4,[/tex] and [tex]v5[/tex] are linearly independent. The basis for the area covered by [tex]v1, v2, v3, v4,[/tex] and [tex]v5[/tex].

We can construct a matrix with the vectors as its columns and rows reduce it using Gaussian elimination to find out which of the five vectors is linearly independent. The vectors are linearly independent if the row-reduced matrix has a pivot in each row. If not, some of the vectors rely linearly.

The five-vectors will be indicated as follows:

[tex]v1 = [1 2 3 4][/tex]

[tex]v2 = [0 1 2 3][/tex]

[tex]v3 = [1 1 1 1][/tex]

[tex]v4 = [1 0 -1 0][/tex]

[tex]v5 = [0 1 0 -1][/tex]

They can be set up as a matrix's columns:

[tex]A = [1 0 1 1 0; 2 1 1 0 1; 3 2 1 -1 0; 4 3 1 0 -1][/tex]

To get this matrix's reduced row echelon form, we can row reduce it as follows:

[tex]R = [1 0 0 -1/2 1/2; 0 1 0 1/2 -1/2; 0 0 1 -1 2; 0 0 0 0 0][/tex]

The centre point in the row-reduced matrix indicates that the vectors [tex]v1[/tex], [tex]v2, v3, v4,[/tex] and [tex]v5[/tex] are linearly independent.

We can utilize the vectors themselves since they constitute a set that is linearly independent to identify a basis for the space that these vectors cover. The basis for the area covered by [tex]v1, v2, v3, v4,[/tex] and [tex]v5[/tex] is therefore only [tex]v1, v2, v3, v4[/tex], and [tex]v5.[/tex]

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A numerical value used as a summary measure for a sample is known as a statistic.

A statistic is a quantity that is calculated from a sample of data and is used to estimate an unknown quantity, such as a population parameter. It is important to note that a statistic is only an estimate of the true population parameter and may vary from sample to sample. Examples of statistics include the sample mean, sample standard deviation, and sample proportion.

what is quantity?

quantity refers to an amount or number of something that can be measured or counted. It can be expressed in numerical or non-numerical terms, and can be used to describe the size, volume, length, weight, or other physical attributes of an object or substance. In mathematics, quantity is often used in conjunction with variables and equations to represent mathematical relationships and solve problems.

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Karl scored 70 points on the Stats midterm.

We can use the formula for calculating z-scores:

z = (x - mu) / sigma

where:

x is the raw score

mu is the population mean

sigma is the population standard deviation

We know that Karl's z-score was -1, which means his score was 1 standard deviation below the mean. We also know that the mean was 76 points and the standard deviation was 6 points. Therefore:

-1 = (x - 76) / 6

Multiplying both sides by 6, we get:

-6 = x - 76

Adding 76 to both sides, we get:

x = 70

Therefore, Karl scored 70 points on the Stats midterm.

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The answer is 50 grams of salt in the tank at all times.

To find the number of grams of salt in the tank at time t, we need to use the formula:

a(t) = a(0) + (r - p) * t

where a(0) is the initial amount of salt in the tank, r is the rate at which brine is being pumped into the tank, p is the rate at which the solution is being pumped out of the tank, and t is the time elapsed.

In this case, a(0) = 50 grams, r = 3 grams per minute (since the brine contains 1 gram of salt per liter and 3 liters are being pumped in per minute, so 3 grams of salt are being added per minute), and p = 3 grams per minute (since the solution is being pumped out at the same rate as it is being pumped in). Therefore, we can simplify the formula to:

a(t) = 50 + 0 * t
a(t) = 50 grams

This means that the amount of salt in the tank remains constant over time, since the rate at which it is being pumped in and out is equal. Therefore, the answer is 50 grams of salt in the tank at all times.

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VII (line over top) CCLIV is equivalent to the Hindu-Arabic numeral 7,254.

The Roman numeral VII (line over top) CCLIV can be rewritten as follows:

VII (line over top) CCLIV = 7,254

The line over the Roman numeral VII signifies that its value should be multiplied by 1,000. Therefore, VII (line over top) represents 7,000. The Roman numerals CCLIV represent 254 in base 10.

Putting these values together, we get:

VII (line over top) CCLIV = 7,000 + 254 = 7,254

Therefore, VII (line over top) CCLIV is equivalent to the Hindu-Arabic numeral 7,254.

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The appropriate statistical technique to compare money spent at the store between two groups in a marketing research study is the independent-samples t-test.

The independent-samples t-test is used to compare the means of two independent groups. In this case, the groups are the ones who received the coupon and the ones who did not, and they are independent because the participants were randomly assigned to each group.

The null hypothesis for the independent-samples t-test is that there is no difference between the means of the two groups. The alternative hypothesis is that there is a significant difference between the means of the two groups.

By conducting an independent-samples t-test on the data, the researcher can determine whether the difference in money spent at the store between the two groups is statistically significant or simply due to chance. If the p-value is less than the significance level (usually 0.05), the researcher can reject the null hypothesis and conclude that there is a significant difference between the two groups in terms of the money spent at the store.

Therefore, the independent-samples t-test is an appropriate statistical technique to compare the means of two independent groups and test for significant differences between them.

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Morrison Inc. is using an R-Chart to monitor the variability of their 44.00-pound steel bars. They have taken 18 samples of 7 steel bars each.

Step 1: The Center Line of the control chart is the average of the sample means. Therefore, the Center Line for this R-chart would be the average of the average weights of the 7 steel bars over the 18 successive time periods. The Center Line can be calculated as follows:
Center Line = (44.04 + 43.96 + 44.01 + 44.02 + 43.95 + 44 + 44.04)/7 = 44.00

Step 2: The Upper Control Limit (UCL) can be calculated as follows:
UCL = Center Line + A2*R-bar
Where R-bar is the average range of the 18 samples and A2 is a constant based on the sample size (n = 7) and the desired level of significance (alpha = 0.05). From the table of constants, A2 = 0.482. The average range can be calculated as follows:
R-bar = (0.09 + 0.17)/2 = 0.13
Therefore, the UCL is:
UCL = 44.00 + 0.482*0.13 = 44.06

Step 3: The Lower Control Limit (LCL) can be calculated as follows:
LCL = Center Line - A2*R-bar
Therefore, the LCL is:
LCL = 44.00 - 0.482*0.13 = 43.94

Step 4: Using the sample data of the first time period, we can calculate the sample mean and sample range as follows:
Sample Mean = (44.04 + 43.96 + 44.01 + 44.02 + 43.95 + 44 + 44.04)/7 = 44.00
Sample Range = 44.04 - 43.95 = 0.09
The sample mean is within the control limits, so the process is in control.

Step 5: Using the sample data of the second time period, we can calculate the sample mean and sample range as follows:
Sample Mean = (44.04 + 43.96 + 44.01 + 44.02 + 43.95 + 44 + 44.04)/7 = 44.00
Sample Range = 44.04 - 43.95 = 0.09
The sample mean is within the control limits, so the process is in control.

Step 6: Using the sample data of the third time period, we can calculate the sample mean and sample range as follows:
Sample Mean = (43.99 + 43.98 + 44.04 + 44.13 + 43.96 + 44.04 + 43.98)/7 = 44.00
Sample Range = 44.13 - 43.96 = 0.17
The sample mean is within the control limits, but the sample range is above the UCL. Therefore, the process is out of control.

Step 7: The probability of making a Type I Error is the level of significance (alpha = 0.05) which represents the probability of rejecting the null hypothesis (process is in control) when it is actually true. Therefore, the probability of making a Type I Error is 0.05 or 5%. The probability that the process is really in control can be calculated using the concept of process capability indices, but it cannot be determined from the information given in this question.
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For a sample with margin of error 0.011 and confidence level 92%, the minimum sample size we should use to assure that your estimate of [tex] \hat p \: = 0.5[/tex] is 6328. So, option (a) is right one.

The sample size is a sample attribute and it is determined by the formula of margin of error with a confidence level. The size of the sample must be appropriate so that sample can estimate the population parameter with a small sampling error.

We have a sample with the following details, Margin of error, MOE = 0.011

Confidence level = 92%

The population proportion= p

Sample proportion [tex] \hat p \: = 0.5[/tex]

Now, using the table value of z score for 92% of confidence interval is equals to 1.75. So, [tex]Z_{ \frac{0.08}{2}} = 1.405[/tex]. The margin of error at the 93% confidence coefficient is [tex]ME = Z_{ \frac{0.08}{2}} × \sqrt{\frac{\hat p(1 - \hat p)}{n}}[/tex]

Substitute all known values in above formula, [tex]0.011= 1.750 × \sqrt{\frac{0.5(1 - 0.5)}{n}}[/tex]

Squaring both sides,

[tex]0.011² = ( 1.75)² (\frac{ 0.25}{n})[/tex]

=> [tex]n = \frac{ (1.75)² × 0.25}{0.011²}[/tex]

=> n = 6328

Hence, required value is 6328.

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Rounding up, the minimum sample size required is 6328

The minimum sample size required to ensure that the estimated proportion (p^) is within a desired margin of error with a certain level of confidence can be determined using a formula.

n = (z-score)^2 * p^(1-p^) / (margin of error)^2

Where:

The z-score is the standard normal distribution's critical value for the specified confidence level (92% with this case).

p^ is the estimated proportion of the population with the characteristic of interest (unknown in this case, so we can use 0.5 as a conservative estimate)

(1-p^) is the complementary proportion to p^

margin of error is the maximum allowed difference between the sample proportion and the true population proportion.

Inputting the values provided yields:

n = (1.751)^2 * 0.5 * 0.5 / (0.011)^2 n ≈ 6327.98

The formula considers the margin of error, confidence level, and estimated population proportion.

In this case, the margin of error is given as 0.011, the confidence level is 92%, and the population proportion is unknown. Using a conservative estimate of 0.5 for the population proportion, the minimum sample size is calculated as 6327.98, which is rounded up to 6328.

Therefore, a sample size of at least 6328 is required to ensure that the estimated proportion is within 0.011 of the true population proportion with a confidence level of 92%. Adequate sample size is important to obtain accurate estimates of population parameters and minimize sampling errors.

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Answer: In simple terms

Step-by-step explanation:

At 12ft from the lamppost:

Let's call the length of the shadow S.

We can see part of a right triangle formed by the player's height (7ft), the distance to the lamppost (12ft), and the hypotenuse which is the length of the shadow (S ft).

Using the Pythagorean theorem:

72 + 122 = S2

49 + 144 = 193

Therefore, at 12ft from the lamppost:

The length of the shadow (S) = 13ft

To find the rate at the shadow is changing:

As the player walks closer at 4ft/sec, the distance to the lamppost decreases by 4ft each second.

For each 4ft closer, the shadow length changes by:

Shadow length (13ft) x (4ft/12ft distance) = 2ft

So the shadow length changes by 2ft for each 4ft the player walks closer.

Therefore, the rate at the shadow length is changing at 12ft from the lamppost is:

2ft / 4ft walked closer = 0.5 ft/sec

The measures of the angles of cyclic quadrilaterals will be -

Blank 1: 98 degrees

Blank 2: 82 degrees

The sum of the opposite angles of a cyclic quadrilateral is 180 degrees.

Given that, the angles of the cyclic quadrilateral are

angle BAD = 14x

angle ADC = 10x + 5

angle ABC = 15x

Clearly angle ABC and angle ADC are opposite angles of a cyclic quadrilateral. So,

angle ABC + angle ADC = 180

15x + 10x + 5 = 180

25x = 180 - 5

25x = 175

x = 175/25

x = 7

So angle BAD = 14x = 14*7 = 98 degrees

angle ABC = 15x = 15*7 = 105 degrees

angle ADC = 10x + 5 = 10*7 + 5 = 75 degrees

Since we know that the sum of all angles of a quadrilateral is 360 degrees. So, the angle BCD = 360 - (98 + 105 + 75) = 82 degrees.

So Blank 1 will be '98 degrees' and Blank 2 will be '82 degrees'.

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Answer:

Parallel lines which are intersected by a transversal form congruent corresponding angles.

9x - 14 = 6x + 22

3x = 36, so x = 12