To find the class marks for each interval, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size of this data = 0.04. Taking 0.14 as assumed mean (a), dᵢ, uᵢ, fᵢuᵢ are calculated as follows.

From the table, we obtain
∑fᵢ = 30, ∑fᵢuᵢ = -31, a = 0.14 and h = 0.04
mean(X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 0.14 + (-31/30) × (0.04) = 0.14 – 0.04133 = 0.09867 = 0.099
Therefore, mean concentration of SO₂ in the air is 0.099 ppm.

See this Video explanation for better understanding😀✌

To find the class mark of each interval, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 17 aa assumed mean (a), dᵢ and fᵢdᵢ are calculated as follow.

From the table, we obtain
∑fᵢ = 40 + f, ∑fᵢdᵢ = -181 and a = 17
mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 17 + (-181/40) = 17 – 4.525 = 12.475 = 12.48
Therefore, the mean number of days is 12.48 days for which a student was absent.

(i) sin 60° cos 30° + sin 30° cos 60°
Putting the value of each trigonometric ratio, we get
(√3/2)(√3/2)+(1/2)(1/2) = 3/4 + 1/4 = 4/4 = 1

(ii) Putting the value of each trigonometric ratio, we get
2(1)²+(√3/2)²-(√3/2)² = 2 + 3/4 – 3/4 = 1

(iii) (cos 45°)/(sec 30°+cosec 30°)
Putting the value of each trigonometric ratios, we get
(1/√2)/(2/√3 + 2) = (1/√2)/((2+2√3)/√3) = √3/(√2(2+2√3)) = √3/(2√2 + 2√6)
= √3/(2√2 + 2√6) × (2√2 – 2√6)/(2√2 – 2√6) = (2√6 – 2√18)/((2√2)² – (2√6)²) = (2√6 – 6√2)/(8 – 24) = (-2(3√2 – √6))/(-16) = (3√2 – √6)/(8)

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
Putting the value of each trigonometric ratios, we get
(1/2 + 1 – 2√3)/(2√3 + 1/2 + 1) = ((√3 + 2√3 – 4)/(2√3))/((4 + √3 + 2√3)/(2√3)) = (3√3 – 4)/(3√3 + 4)
= (3√3 – 4)/(3√3 + 4) × (3√3 – 4)/(3√3 – 4) = (27 – 12√3 – 12√3 + 16)/((3√3)² – 4²)
= (43 – 24√3)/(27 – 16) = (43 – 24√3)/(11)

(v) (5 cos²60° + 4sec²30° – tan²45°)/(sin²30° + cos²30°)
Putting the value of each trigonometric ratios, we get
(5(1/2)² + 4(2/√3)² – (1)²)/((1/2)² + (√3/2)²) = (5/4 + 16/3 – 1)/(1/4 + 3/4) = ((15 + 64 -12)/(12))/(4/4) = 67/12

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Given that: sin A = 3/4
Let sin A = 3k/4/, where k is a real number.
In ∆ABC, by Pythagoras theorem, we have
AB² = AC²-BC²
= (4k)²-(3k)²
= 16k²-9k²
= 7k²
⇒ AB = √(7k²) = √7k
Hence, cos A = AB/AC = √7k/4k = √7/4 and tan A = BC/AB = 3k/√7k = 3/√7

Here is the video explanation of this question😃👇

Here, P(0,0) and Q(36, 15), using distance formula √((x₂-x₁)²+(y₂-y₁)²)
Distance between P(0,0) and Q(36, 15) is given by PQ
= √((36-0)²+(15- 0)²) = √(1296 + 225) = √1521 = 39
Using the distance formula: √((x₂-x₁)²+(y₂-y₁)² we can find the distance between the two towns.

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