Arranging the numbers: 1, 2, 3, 4, 5, 6, 7, 8. The median is the average of the 4th and 5th numbers: (4 + 5) / 2 = 4.5.

Arrange the numbers: 19, 22, 24, 25, 28, 30. The median is the average of the 3rd and 4th values: (24 + 25) / 2 = 24.5.

Mean = 5. Variance = [(2-5)² + (4-5)² + (4-5)² + (4-5)² + (5-5)² + (5-5)² + (7-5)² + (9-5)²] / 8 = 4.

Mean = (3 + 7 + 7 + 19) / 4 = 9. Variance = [(3-9)² + (7-9)² + (7-9)² + (19-9)²] / 4 = 25.

Mean = (5 + 7 + 9 + 11) / 4 = 8. Variance = [(5-8)² + (7-8)² + (9-8)² + (11-8)²] / 4 = 3.5.

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.

The mode is 4, as it appears 3 times, more than any other number.

The mode is 8 and 9, but since we need to select one, we can choose 8 as it appears first.

The mode is 30, as it appears three times, which is more than any other number.

The number 8 appears 3 times, which is more than any other number, making it the mode.

The mode is 4, which appears most frequently (three times).

The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan⁻¹(|(m1 - m2) / (1 + m1*m2)|) = tan⁻¹(5/3), which is approximately 90 degrees.

The coefficient of x^5 in (3x - 4)^7 is C(7, 5) * (3)^5 * (-4)^2 = 21 * 243 * 16 = 68016.

The left limit as x approaches 1 is 1, the right limit is 2, and f(1) = 2. Since the left and right limits do not match, f(x) is not continuous at x = 1.

At x = 1, f(1) = 2(1) - 1 = 1 and lim x→1- f(x) = 1, lim x→1+ f(x) = 1. Thus, f(x) is continuous at x = 1.

The function f(x) = |x| is continuous at x = 0 since both the left-hand limit and right-hand limit equal f(0) = 0.

f'(x) = 2x - 4; Setting f'(x) = 0 gives x = 2 as the critical point.

Using the power rule, f'(x) = 3x^2 - 4.

Using the power rule, f'(x) = 5x^4 - 9x^2 + 2.

Using the distance formula: d = √[(2 - (-1))² + (2 - (-1))²] = √[(2 + 1)² + (2 + 1)²] = √[9 + 9] = √18 = 3√2 ≈ 4.24.

Using the distance formula: d = √[(0 - 0)² + (8 - 0)²] = √[0 + 64] = √64 = 8.

Using the distance formula: d = √[(4 - 1)² + (6 - 2)²] = √[9 + 16] = √25 = 5.

Using the distance formula: d = √[(2 - 2)² + (-1 - 3)²] = √[0 + 16] = √16 = 4.

Using the formula for distance from a point to a line, the distance is |2(1) + 3(2) - 6| / sqrt(2^2 + 3^2) = 1.

f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. f''(x) = 2 > 0 indicates a local minimum at x = 2.

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f''(x) = 12x^2 - 16. Minima at x = 0.

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 4.

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.

The vertex is at x = -4/(2*(-2)) = 1. The maximum value is f(1) = -2(1)^2 + 4(1) + 1 = 3.

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 8.