Subtract :
$$0.18$$ from $$0.6$$

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Subtract :
$$0.18$$ from $$0.6$$

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\frac{1}{\sqrt{0. ¯ ¯ ¯ 2} + \sqrt{0. ¯ ¯ ¯ 4}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 4} + \sqrt{0. ¯ ¯ ¯ 6}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 6} + \sqrt{0. ¯ ¯ ¯ 8}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 10} + \sqrt{0. ¯ ¯¯¯ ¯ 12}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 12} + \sqrt{0. ¯ ¯¯¯ ¯ 14}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 14} + \sqrt{0. ¯ ¯¯¯ ¯ 16}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 16} + \sqrt{0. ¯ ¯¯¯ ¯ 18}} + . . . . . .

. . . . . . . . . . . . . . . + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 38} + \sqrt{0. ¯ ¯¯¯ ¯ 40}} = \frac{3}{2} (\sqrt{a} + \sqrt{b}),

where a > b, a and b are rational numbers, then (a ÷ b) is

यदि

\frac{1}{\sqrt{0. ¯ ¯ ¯ 2} + \sqrt{0. ¯ ¯ ¯ 4}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 4} + \sqrt{0. ¯ ¯ ¯ 6}} + \frac{1}{\sqrt{0. ¯ ¯ ¯ 6} + \sqrt{0. ¯ ¯ ¯ 8}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 10} + \sqrt{0. ¯ ¯¯¯ ¯ 12}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 12} + \sqrt{0. ¯ ¯¯¯ ¯ 14}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 14} + \sqrt{0. ¯ ¯¯¯ ¯ 16}} + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 16} + \sqrt{0. ¯ ¯¯¯ ¯ 18}} + . . . . . .

. . . . . . . . . . . . . . . + \frac{1}{\sqrt{0. ¯ ¯¯¯ ¯ 38} + \sqrt{0. ¯ ¯¯¯ ¯ 40}} = \frac{3}{2} (\sqrt{a} + \sqrt{b}),

जहाँ a > b, a तथा b परिमेय संख्याएँ हैं, तब (a ÷ b ) का मान है

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If A and B are independent events, such that P(A)= 0.6, P(B) = 0.3, then P(A and not B) = ...

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Mark (✓) against the correct answer
What should be subtracted from .1 to get .04?

(a) 0.6
(b) 0.06
(c) 0.006
(d) none of these

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Study the table given below and find the value of '

A

' and '

B

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Subtract :$$0.18$$ from $$0.6$$