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 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%
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.
(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 lessThe 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%
A 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.
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