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
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