answer:The mode is a measure of central tendency that represents the most frequently occurring value in a dataset. The advantages of the mode include: It is easy to calculate and understand, making it a simple and intuitive measure of central tendency. It can be used with both numerical and categorical data, which is useful when working with datasets that contain non-numerical variables. It is not affected by extreme values or outliers, which can distort the mean and median. However, the mode also has some disadvantages: It may not be unique, as a dataset can have multiple modes if there are multiple values that occur with the same frequency. It may not accurately represent the typical value in the dataset, especially if the mode is not near the center of the distribution. It can be sensitive to the level of granularity in the data, as the mode can change depending on how the data is grouped or binned. It is not as useful for statistical calculations, such as calculating variance or standard deviation, as the mean is.

answer:Consider a dataset of exam scores for a group of students, with scores ranging from 0 to 100. The dataset contains the following scores: 60, 70, 70, 70, 80, 80, 80, 90, 90, 90 In this dataset, there are three modes: 70, 80, and 90, since each of these scores occurs with the same frequency (three times). This is an example of a multimodal distribution, where there are multiple modes that occur with the same frequency. This can make it more difficult to identify a single, representative value for the dataset.

answer:The range and interquartile range (IQR) are both measures of variability or dispersion in a dataset, but they differ in their calculation and interpretation. The range is the difference between the largest and smallest values in a dataset. It is calculated as: Range = Maximum value - Minimum value The range is sensitive to extreme values or outliers, which can greatly affect its value. For example, if a dataset contains one extremely high or low value, the range will be inflated, even if the rest of the data is clustered together. The interquartile range (IQR), on the other hand, is the difference between the 75th percentile (Q3) and the 25th percentile (Q1). It is calculated as: IQR = Q3 - Q1 The IQR is a more robust measure of variability, as it is less affected by extreme values or outliers. It provides a better representation of the spread of the middle 50% of the data. The IQR is often used in conjunction with the median to provide a more complete picture of the data's distribution. In general, the IQR is a more useful measure of variability than the range, as it provides a better representation of the data's spread and is less sensitive to extreme values.

answer:**Solution for (Ⅰ):** Given that a = sqrt{6}, b = 2c, and cos A = -frac{1}{4}, we apply the Law of Cosines: begin{align*} cos A &= frac{b^2 + c^2 - a^2}{2bc} -frac{1}{4} &= frac{(2c)^2 + c^2 - (sqrt{6})^2}{2 cdot 2c cdot c} -frac{1}{4} &= frac{4c^2 + c^2 - 6}{4c^2} -frac{1}{4} &= frac{5c^2 - 6}{4c^2}. end{align*} Multiplying both sides by 4c^2 and solving for c^2 gives: begin{align*} -1c^2 &= 5c^2 - 6 6c^2 &= 6 c^2 &= 1. end{align*} Taking the square root of both sides, we find c = 1 or c = -1. Since side lengths are positive, c = 1. boxed{c = 1} **Solution for (Ⅱ):** Given cos A = -frac{1}{4}, we find sin A using the Pythagorean identity: begin{align*} sin A &= sqrt{1 - cos^2 A} &= sqrt{1 - left(-frac{1}{4}right)^2} &= sqrt{1 - frac{1}{16}} &= sqrt{frac{15}{16}} &= frac{sqrt{15}}{4}. end{align*} Since b = 2c, we have sin B = 2sin C. Using the Law of Sines: begin{align*} frac{a}{sin A} &= frac{c}{sin C} frac{sqrt{6}}{frac{sqrt{15}}{4}} &= frac{1}{sin C} sin C &= frac{sqrt{15}}{4} cdot frac{1}{sqrt{6}} &= frac{sqrt{10}}{8}. end{align*} Therefore, sin B = 2sin C = 2 times frac{sqrt{10}}{8} = frac{sqrt{10}}{4}. boxed{sin B = frac{sqrt{10}}{4}} **Solution for (Ⅲ):** Given cos A = -frac{1}{4} and sin A = frac{sqrt{15}}{4}, we find sin 2A and cos 2A: begin{align*} sin 2A &= 2sin Acos A = 2 times frac{sqrt{15}}{4} times -frac{1}{4} = -frac{sqrt{15}}{8}, cos 2A &= 2cos^2 A - 1 = 2 times left(-frac{1}{4}right)^2 - 1 = -frac{7}{8}. end{align*} Given sin B = frac{sqrt{10}}{4}, we find cos B using the Pythagorean identity: begin{align*} cos B &= sqrt{1 - sin^2 B} = sqrt{1 - left(frac{sqrt{10}}{4}right)^2} = sqrt{1 - frac{10}{16}} = sqrt{frac{6}{16}} = frac{sqrt{6}}{4}. end{align*} Therefore, sin(2A - B) is calculated as: begin{align*} sin(2A - B) &= sin 2A cos B - cos 2A sin B &= -frac{sqrt{15}}{8} cdot frac{sqrt{6}}{4} - left(-frac{7}{8}right) cdot frac{sqrt{10}}{4} &= frac{sqrt{10}}{8}. end{align*} boxed{sin(2A - B) = frac{sqrt{10}}{8}}