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.

    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.

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  2. 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.

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  3. 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
    The denominator of all the rational numbers are in the form of 2ᵐ x 5ⁿ, where m and n are integers.

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  4. Three non-terminating non-recurring decimals: 1) 0.414114111411114 .. 2) 2.01001000100001.. 3) π = 3.1416 ..

    Three non-terminating non-recurring decimals:
    1) 0.414114111411114 ..
    2) 2.01001000100001..
    3) π = 3.1416 ..

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  5. 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

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