The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
Median class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
(i) (2 tan 30°)/(1 - tan²30°) Putting the value of each trigonometric ratios, we get (2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2 We know that sin 60° =Y hence the option (A) is correct. (ii)(1 - tan² 45°)/(1 + tan² 45°) Putting the value of each trigonometric ratios, wRead more
(i) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2
We know that sin 60° =Y hence the option (A) is correct.
(ii)(1 – tan² 45°)/(1 + tan² 45°)
Putting the value of each trigonometric ratios, we get
(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0
Hence, the option (D) is correct.
(iii) sin 2A = 2 sin A
We know that sin 0 = 0, hence, the option (A) is correct.
(iv) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3
We know that tan 60° = √3, hence, the option (C) is correct.
Here is the video explanation of the above question😁🤗
(i) tan 48° tan 23° tan 42° tan 67° = 1 LHS = tan 48° tan 23° tan 42° tan 67° = tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ] = tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° × 1/tan 48° × 1/tan 23° = 1 = RHS (ii) cos 38° cos 52° - sin 38° sin 52° = 0 LHS = cos 38° cos 52° - sinRead more
(i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ]
= tan 48° tan 23° cot 48° cot 23°
= tan 48° tan 23° × 1/tan 48° × 1/tan 23°
= 1 = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos 52° – cos (90°-38°) cos (90°-52) [∵ cos (90°-0°) = sinθ]
= cos 38° cos 52° – cos52° – cos 38°
= 0 = RHS
(3/2)x + (5/3)y = 7 9x - 10y = 14 Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6 ⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent. For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/ See here for video Solution 👀👇
(3/2)x + (5/3)y = 7
9x – 10y = 14
Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.
The following table gives the distribution of the life time of 400 neon lamps:
The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
See lessMedian class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
See less∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
Choose the correct option and justify your choice:
(i) (2 tan 30°)/(1 - tan²30°) Putting the value of each trigonometric ratios, we get (2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2 We know that sin 60° =Y hence the option (A) is correct. (ii)(1 - tan² 45°)/(1 + tan² 45°) Putting the value of each trigonometric ratios, wRead more
(i) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2
We know that sin 60° =Y hence the option (A) is correct.
(ii)(1 – tan² 45°)/(1 + tan² 45°)
Putting the value of each trigonometric ratios, we get
(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0
Hence, the option (D) is correct.
(iii) sin 2A = 2 sin A
We know that sin 0 = 0, hence, the option (A) is correct.
(iv) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3
We know that tan 60° = √3, hence, the option (C) is correct.
Here is the video explanation of the above question😁🤗
See lessShow that the following trigonometric relation are true.
(i) tan 48° tan 23° tan 42° tan 67° = 1 LHS = tan 48° tan 23° tan 42° tan 67° = tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ] = tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° × 1/tan 48° × 1/tan 23° = 1 = RHS (ii) cos 38° cos 52° - sin 38° sin 52° = 0 LHS = cos 38° cos 52° - sinRead more
(i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° cot(90°-42°) cot(90°-67°) [∵ cot(90°-θ) = tanθ]
= tan 48° tan 23° cot 48° cot 23°
= tan 48° tan 23° × 1/tan 48° × 1/tan 23°
= 1 = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos 52° – cos (90°-38°) cos (90°-52) [∵ cos (90°-0°) = sinθ]
= cos 38° cos 52° – cos52° – cos 38°
= 0 = RHS
see this video explanation of the above question🧐
See lessOn comparing the ratios, a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the pair of linear equations are consistent, or inconsistent. (3/2)x + (5/3)y = 7; 9x – 10y = 14
(3/2)x + (5/3)y = 7 9x - 10y = 14 Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6 ⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent. For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/ See here for video Solution 👀👇
(3/2)x + (5/3)y = 7
9x – 10y = 14
Here, a₁/a₂ = (3/2)/9 = 1/6 and b₁/b₂ = (5/3)/-10 = 1/6
⇒ a₁/a₂ ≠ b₁/b₂, so, pair of linear equations are consistent.
For more details click here => https://www.tiwariacademy.com/ncert-solutions/class-10/maths/
See here for video Solution 👀👇
See less