From the given data, it can be observed that the maximum class frequency is 20, belonging to 4000 – 5000 class interval. Therefore, modal class = 40 - 50 Lower limit () of modal class = 40 Frequency (f₁) of modal class = 20 Frequency (f₀) of class preceding modal class = 12 Frequency (f₂) of class sRead more
From the given data, it can be observed that the maximum class frequency is 20, belonging to 4000 – 5000 class interval.
Therefore, modal class = 40 – 50
Lower limit () of modal class = 40
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 12
Frequency (f₂) of class succeeding modal class = 11
Class size (h) = 10
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 40 + ((20 – 12)/(2 × 20 – 12 – 11)) × 10 = 40 + 8/17 × 10 = 40 + 4.7 = 44.7
Therefore, mode of the this data is 44.7 cars.
Area of sector DAFC = 90°/360° × πr² = 1/4 × π(8)² = 1/4 × 22/7 × 8 × 8 = 352/7 cm² Area of triangle ADC = 1/2 × DC × AD = 1/2 × 8 × 8 = 32 cm² Area of segment = Area of sector DAFC - Area of triangle ADC = (352/7 - 32) cm² = ((352 - 224)/7) cm² = (128/7) cm² Area of shaded region = Area of two segmRead more
Area of sector DAFC
= 90°/360° × πr² = 1/4 × π(8)²
= 1/4 × 22/7 × 8 × 8 = 352/7 cm²
Area of triangle ADC
= 1/2 × DC × AD
= 1/2 × 8 × 8
= 32 cm²
Area of segment = Area of sector DAFC – Area of triangle ADC
= (352/7 – 32) cm² = ((352 – 224)/7) cm² = (128/7) cm²
Area of shaded region = Area of two segments
= 2 × (128/7) cm² = 256/7 cm²
To find the class mark (xᵢ) for each interval, the following relation is used. Class mark xᵢ = (Upper tmit + Lower limit)/2 xᵢ and fᵢ can be calculated as follows. From the table, it can be observed that: ∑fᵢ = 20 and ∑fᵢxᵢ = 162
To find the class mark (xᵢ) for each interval, the following relation is used. Class mark xᵢ = (Upper tmit + Lower limit)/2 xᵢ and fᵢ can be calculated as follows.
From the table, it can be observed that:
∑fᵢ = 20 and ∑fᵢxᵢ = 162
To find the class mark for each interval, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Class size (h) of this data = 20.Taking 150 as assured mean (a), dᵢ, uᵢ, and fᵢuᵢ can be calculated as follows. From the table, it can be observed that ∑fᵢ = 50, ∑fᵢuᵢ = 12, a = 150 and h = 2Read more
To find the class mark for each interval, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Class size (h) of this data = 20.Taking 150 as assured mean (a), dᵢ, uᵢ, and fᵢuᵢ can be calculated as follows.
From the table, it can be observed that
∑fᵢ = 50, ∑fᵢuᵢ = 12, a = 150 and h = 20
mean(X̄) = a + (∑fᵢuᵢ/∑fᵢ)h = 150 + (-12/50) × 20 = 150 – 24/5 = 150 – 4.8 = 145.2
Therefore, the mean daily wage of the workers of the factory is ₹145.20
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below.
From the given data, it can be observed that the maximum class frequency is 20, belonging to 4000 – 5000 class interval. Therefore, modal class = 40 - 50 Lower limit () of modal class = 40 Frequency (f₁) of modal class = 20 Frequency (f₀) of class preceding modal class = 12 Frequency (f₂) of class sRead more
From the given data, it can be observed that the maximum class frequency is 20, belonging to 4000 – 5000 class interval.
See lessTherefore, modal class = 40 – 50
Lower limit () of modal class = 40
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 12
Frequency (f₂) of class succeeding modal class = 11
Class size (h) = 10
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 40 + ((20 – 12)/(2 × 20 – 12 – 11)) × 10 = 40 + 8/17 × 10 = 40 + 4.7 = 44.7
Therefore, mode of the this data is 44.7 cars.
Find the area of the shaded region.
Radius of sector 14 cm Area of sector = 90°/360° × πr² = 1/4 × πr² = 1/4 × 22/7 × 14 × 14 = 154 cm² In AABC, BC² AC² + AB² ⇒ BC² = (14)² + (14)² ⇒ BC² = 196 + 196 ⇒ BC² = 39² ⇒ BC = √392 ⇒ BC = 14√2 Therefore, the diameter of semicircle = BC = 14√2 Radius of semicircle = 7√2 cm Area of semicircle =Read more
Radius of sector 14 cm
See lessArea of sector
= 90°/360° × πr² = 1/4 × πr²
= 1/4 × 22/7 × 14 × 14 = 154 cm²
In AABC,
BC² AC² + AB² ⇒ BC² = (14)² + (14)² ⇒ BC² = 196 + 196
⇒ BC² = 39² ⇒ BC = √392 ⇒ BC = 14√2
Therefore, the diameter of semicircle = BC = 14√2
Radius of semicircle = 7√2 cm
Area of semicircle
= 1/2 × πr² = 1/2 × π(7√2)² = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm²
Area of triangle ABC
= 1/2 × AB × BC = 1/2 × 14 × 14 = 98 cm²
Area of shaded region
= Aera of triangle ABC + Area of semicircle – Area of quadrant
= (98 + 154 – 154) cm²
= 98 cm²
Calculate the area of the designed region in Figure common between the two quadrants of circles of radius 8 cm each.
Area of sector DAFC = 90°/360° × πr² = 1/4 × π(8)² = 1/4 × 22/7 × 8 × 8 = 352/7 cm² Area of triangle ADC = 1/2 × DC × AD = 1/2 × 8 × 8 = 32 cm² Area of segment = Area of sector DAFC - Area of triangle ADC = (352/7 - 32) cm² = ((352 - 224)/7) cm² = (128/7) cm² Area of shaded region = Area of two segmRead more
Area of sector DAFC
= 90°/360° × πr² = 1/4 × π(8)²
= 1/4 × 22/7 × 8 × 8 = 352/7 cm²
Area of triangle ADC
= 1/2 × DC × AD
= 1/2 × 8 × 8
= 32 cm²
Area of segment = Area of sector DAFC – Area of triangle ADC
= (352/7 – 32) cm² = ((352 – 224)/7) cm² = (128/7) cm²
Area of shaded region = Area of two segments
= 2 × (128/7) cm² = 256/7 cm²
Here is the video explanation 🤗👇
See lessA survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
To find the class mark (xᵢ) for each interval, the following relation is used. Class mark xᵢ = (Upper tmit + Lower limit)/2 xᵢ and fᵢ can be calculated as follows. From the table, it can be observed that: ∑fᵢ = 20 and ∑fᵢxᵢ = 162
To find the class mark (xᵢ) for each interval, the following relation is used. Class mark xᵢ = (Upper tmit + Lower limit)/2 xᵢ and fᵢ can be calculated as follows.
From the table, it can be observed that:
See less∑fᵢ = 20 and ∑fᵢxᵢ = 162
Consider the following distribution of daily wages of 50 workers of a factory.
To find the class mark for each interval, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Class size (h) of this data = 20.Taking 150 as assured mean (a), dᵢ, uᵢ, and fᵢuᵢ can be calculated as follows. From the table, it can be observed that ∑fᵢ = 50, ∑fᵢuᵢ = 12, a = 150 and h = 2Read more
To find the class mark for each interval, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Class size (h) of this data = 20.Taking 150 as assured mean (a), dᵢ, uᵢ, and fᵢuᵢ can be calculated as follows.
From the table, it can be observed that
See less∑fᵢ = 50, ∑fᵢuᵢ = 12, a = 150 and h = 20
mean(X̄) = a + (∑fᵢuᵢ/∑fᵢ)h = 150 + (-12/50) × 20 = 150 – 24/5 = 150 – 4.8 = 145.2
Therefore, the mean daily wage of the workers of the factory is ₹145.20