Solve the questions.Kindly answer in English I don't understand swahili

(a) Differentiate between each of the following types of sets. (i) Finite set and infinite set; • A finite set is a set whose elements can be counted, and the counting process eventually ends. It has a definite, limited number of elements. • An infinite set is a set whose elements cannot be counted, as the counting process would never end. It has an unlimited number of elements. (ii) Empty set and universal set. • An empty set (or null set) is a set that contains no elements. It is denoted by or . • A universal set is a set that contains all elements relevant to a particular context or problem. It is usually denoted by U. (b) Calculate the harmonic mean, mode, and median for the given data. First, we prepare a table with midpoints (x), frequencies (f), 1/x, f/x, and cumulative frequencies (cf). | Marks | Number of Students (f) | Midpoint (x) | 1/x | f/x | Cumulative Frequency (cf) | | :---- | :----------------------- | :------------- | :---- | :---- | :-------------------------- | | 0 - 10 | 2 | 5 | 0.2 | 0.4 | 2 | | 10 - 20 | 7 | 15 | 0.066667 | 0.466669 | 9 | | 20 - 30 | 15 | 25 | 0.04 | 0.6 | 24 | | 30 - 40 | 25 | 35 | 0.028571 | 0.714275 | 49 | | 40 - 50 | 45 | 45 | 0.022222 | 1.0 | 94 | | 50 - 60 | 32 | 55 | 0.018182 | 0.581824 | 126 | | 60 - 70 | 15 | 65 | 0.015385 | 0.230775 | 141 | | 70 - 80 | 9 | 75 | 0.013333 | 0.12 | 150 | | Total | f = N = 150 | | | (f/x) ≈ 4.173543 | | (i) Harmonic Mean (HM) The formula for the harmonic mean for grouped data is: HM = (N)/( f)x Step 1: Sum the frequencies (N) and the values of f/x. N = 150 (f)/(x) = 0.4 + 0.466669 + 0.6 + 0.714275 + 1.0 + 0.581824 + 0.230775 + 0.12 ≈ 4.173543 Step 2: Calculate the harmonic mean. HM = (150)/(4.173543) HM ≈ 35.939 The harmonic mean is 35.939. (ii) Mode The formula for the mode for grouped data is: Mode = L + ( (f_m - f_1)/(2f_m - f_1 - f_2) ) × h The modal class is 40 - 50 (highest frequency, f_m = 45). L = 40 (lower boundary of the modal class) f_m = 45 (frequency of the modal class) f_1 = 25 (frequency of the preceding class, 30 - 40) f_2 = 32 (frequency of the succeeding class, 50 - 60) h = 10 (class width) Step 1: Substitute the values into the formula. Mode = 40 + ( (45 - 25)/(2(45) - 25 - 32) ) × 10 Step 2: Simplify the expression. Mode = 40 + ( (20)/(90 - 57) ) × 10 Mode = 40 + ( (20)/(33) ) × 10 Mode = 40 + 0.60606 × 10 Mode = 40 + 6.0606 Mode ≈ 46.061 The mode is 46.061. (iii) Median The formula for the median for grouped data is: Median = L + ( (N)/(2) - cf_bf_m ) × h First, find the median position: (N)/(2) = (150)/(2) = 75. The median class is 40 - 50 (where the 75th observation falls, as cf goes from 49 to 94). L = 40 (lower boundary of the median class) N = 150 (total frequency) cf_b = 49 (cumulative frequency of the class before the median class) f_m = 45 (frequency of the median class) h = 10 (class width) Step 1: Substitute the values into the formula. Median = 40 + ( (75 - 49)/(45) ) × 10 Step 2: Simplify the expression. Median = 40 + ( (26)/(45) ) × 10 Median = 40 + 0.57777 × 10 Median = 40 + 5.7777 Median ≈ 45.778 The median is 45.778. Send me the next one 📸