0.99999... Let x = 0.99999 ... ...(i) Multiplying equation (i) by 10 both sides ⇒ 10x = 9.99999... ⇒ 10x = 9 + 0.99999... ... ⇒ 10x = 9 + x [From equation (i)] ⇒ 10x - x = 9 ⇒ 9x = 9 ⇒ x = 9/9 = 1 The answer makes sense as 0.99999... is very close to 1, that is why we can say that 0.99999 = 1.
0.99999…
Let x = 0.99999 … …(i)
Multiplying equation (i) by 10 both sides
⇒ 10x = 9.99999…
⇒ 10x = 9 + 0.99999… …
⇒ 10x = 9 + x [From equation (i)]
⇒ 10x – x = 9
⇒ 9x = 9
⇒ x = 9/9 = 1
The answer makes sense as 0.99999… is very close to 1, that is why we can say that 0.99999 = 1.
The maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16 (less than 17). By performing the actual division, we get 1/17 = 0.058823529411764705882352941176470588235294117647... So, the maximum number of digits that can be in the repeating blockRead more
The maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16 (less than 17).
By performing the actual division, we get
1/17 = 0.058823529411764705882352941176470588235294117647…
So, the maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16.
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 - 5000. Therefore, modal class = 4000 5000 Lower limit () of modal class = 4000 Frequency (f₁) of modal class = 18B Frequency (f₀) of class preceding modal class = 4 Frequency (f₂) of claRead more
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 – 5000.
Therefore, modal class = 4000 5000
Lower limit () of modal class = 4000
Frequency (f₁) of modal class = 18B
Frequency (f₀) of class preceding modal class = 4
Frequency (f₂) of class succeeding modal class = 9
Class size (h) = 1000
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 4000 + ((18 – 4)/(2 × 18 – 4 – 9)) × 1000 = 4000 + 608.695 = 4608.695
Therefore, mode of the given data is 4608.7 runs
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
Express 0.99999 …. in the form p/q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
0.99999... Let x = 0.99999 ... ...(i) Multiplying equation (i) by 10 both sides ⇒ 10x = 9.99999... ⇒ 10x = 9 + 0.99999... ... ⇒ 10x = 9 + x [From equation (i)] ⇒ 10x - x = 9 ⇒ 9x = 9 ⇒ x = 9/9 = 1 The answer makes sense as 0.99999... is very close to 1, that is why we can say that 0.99999 = 1.
0.99999…
See lessLet x = 0.99999 … …(i)
Multiplying equation (i) by 10 both sides
⇒ 10x = 9.99999…
⇒ 10x = 9 + 0.99999… …
⇒ 10x = 9 + x [From equation (i)]
⇒ 10x – x = 9
⇒ 9x = 9
⇒ x = 9/9 = 1
The answer makes sense as 0.99999… is very close to 1, that is why we can say that 0.99999 = 1.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.
The maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16 (less than 17). By performing the actual division, we get 1/17 = 0.058823529411764705882352941176470588235294117647... So, the maximum number of digits that can be in the repeating blockRead more
The maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16 (less than 17).
See lessBy performing the actual division, we get
1/17 = 0.058823529411764705882352941176470588235294117647…
So, the maximum number of digits that can be in the repeating block of digits in the decimal expansion of 1/17 is 16.
Look at several examples of rational numbers in the form p/q (q ≠0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
2/5 = 0.4, 1/10 = 0.1, 3/2 = 1.5, 7/8 = 0.875 The denominator of all the rational numbers are in the form of 2ᵐ x 5ⁿ, where m and n are integers.
2/5 = 0.4, 1/10 = 0.1, 3/2 = 1.5, 7/8 = 0.875
See lessThe denominator of all the rational numbers are in the form of 2ᵐ x 5ⁿ, where m and n are integers.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Three non-terminating non-recurring decimals: 1) 0.414114111411114 .. 2) 2.01001000100001.. 3) π = 3.1416 ..
Three non-terminating non-recurring decimals:
See less1) 0.414114111411114 ..
2) 2.01001000100001..
3) π = 3.1416 ..
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 - 5000. Therefore, modal class = 4000 5000 Lower limit () of modal class = 4000 Frequency (f₁) of modal class = 18B Frequency (f₀) of class preceding modal class = 4 Frequency (f₂) of claRead more
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 – 5000.
See lessTherefore, modal class = 4000 5000
Lower limit () of modal class = 4000
Frequency (f₁) of modal class = 18B
Frequency (f₀) of class preceding modal class = 4
Frequency (f₂) of class succeeding modal class = 9
Class size (h) = 1000
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 4000 + ((18 – 4)/(2 × 18 – 4 – 9)) × 1000 = 4000 + 608.695 = 4608.695
Therefore, mode of the given data is 4608.7 runs
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