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 +Read more
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 - 4.525Read more
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.
To find the class marks, the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Class size (h) for this data = 10. Tking 70 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 35, ∑fᵢuᵢ = -2, a = 70 and h = 10 mean (X̄) = a + (∑fᵢuᵢ /∑fᵢ)h = 70Read more
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) for this data = 10. Tking 70 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 35, ∑fᵢuᵢ = -2, a = 70 and h = 10
mean (X̄) = a + (∑fᵢuᵢ /∑fᵢ)h = 70 + (-2/35) × 10 = 70 – 4/7 = 70 – 0.57 = 69.43
Therefore, mean literacy rate is 69.43%
(i) False, Because, tan A = (Opposite of angle A)/(Adjecent side of angle A), if opposite side > adjacent side, then value of tan A is greater than 1. (ii) True, Because, sec A = (Hypotenuse)/(Adjecent side of angle A) and we know that hypotenuse is always greater than adjacent side. (iii) False,Read more
(i) False,
Because, tan A = (Opposite of angle A)/(Adjecent side of angle A), if opposite side > adjacent side, then value of tan A is greater than 1.
(ii) True,
Because, sec A = (Hypotenuse)/(Adjecent side of angle A) and we know that hypotenuse is always greater than adjacent side.
(iii) False,
Because, cos A is used for cosine of angle A.
(iv) False,
Because, cot A is used for cotangent of angle A.
(v) False,
Because, sin 0 = (Opposite side of angle A)/(Hypotenuse), we know that hypotenuse is always greater than opposite side.
(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 eachRead more
(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)
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
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 +Read more
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😀✌
See lessA class teacher has the following absentee record of 40 students of a class for the whole term.
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 - 4.525Read more
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
See less∑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.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
To find the class marks, the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Class size (h) for this data = 10. Tking 70 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 35, ∑fᵢuᵢ = -2, a = 70 and h = 10 mean (X̄) = a + (∑fᵢuᵢ /∑fᵢ)h = 70Read more
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) for this data = 10. Tking 70 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
See less∑fᵢ = 35, ∑fᵢuᵢ = -2, a = 70 and h = 10
mean (X̄) = a + (∑fᵢuᵢ /∑fᵢ)h = 70 + (-2/35) × 10 = 70 – 4/7 = 70 – 0.57 = 69.43
Therefore, mean literacy rate is 69.43%
State whether the following are true or false. Justify your answer.
(i) False, Because, tan A = (Opposite of angle A)/(Adjecent side of angle A), if opposite side > adjacent side, then value of tan A is greater than 1. (ii) True, Because, sec A = (Hypotenuse)/(Adjecent side of angle A) and we know that hypotenuse is always greater than adjacent side. (iii) False,Read more
(i) False,
Because, tan A = (Opposite of angle A)/(Adjecent side of angle A), if opposite side > adjacent side, then value of tan A is greater than 1.
(ii) True,
Because, sec A = (Hypotenuse)/(Adjecent side of angle A) and we know that hypotenuse is always greater than adjacent side.
(iii) False,
Because, cos A is used for cosine of angle A.
(iv) False,
Because, cot A is used for cotangent of angle A.
(v) False,
Because, sin 0 = (Opposite side of angle A)/(Hypotenuse), we know that hypotenuse is always greater than opposite side.
Here is the explanation video of this question🙌😀
See lessEvaluate the following trigonometric equations:
(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 eachRead more
(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
See Video Explanation for better understanding😀🙌
See less