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Gaurav Seth 6 years, 4 months ago
Total number of coins tossed = 3.
Total number of outcomes n(S) = 8.
Let A be the event of getting at most 2 heads.
= > n(A) = {HHT,HTH,THH,TTH,THT,HTT,TTT} = 7
Required probability P(C) = 7/8
Posted by Jatashankar Sharma 6 years, 4 months ago
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Gaurav Seth 6 years, 4 months ago
Q: 6-4√3/6+4√3
Explanation is as follows:
On rationalizing,
=6-4√3×6-4√3/6+4√3×6-4√3
=36+48-48√3/36-48
=84-48√3/-12
=12(7-4√3)/-12
=-(7-4√3)
=4√3-7
Posted by Ritika Shaw 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
The constant polynomial whose coefficients are all equal to 0.
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Sia ? 6 years, 4 months ago
Here a = {tex}\frac {1} {3}{/tex}, b = {tex}\frac {1} {2}{/tex}, n = 3
{tex}\therefore{/tex} {tex}\frac{b-a}{n+1}=\frac{\frac{1}{2}-\frac{1}{3}}{3+1}{/tex}{tex}=\frac{\frac{3-2}{6}}{4}{/tex}{tex}=\frac{\frac{1}{6}}{4}=\frac{1}{24}{/tex}
{tex}\therefore{/tex} Three rational numbers between {tex}\frac {1} {3}{/tex} and {tex}\frac {1} {2}{/tex} are
{tex}\frac{1}{3}+\frac{1}{24}, \frac{1}{3}+2\left(\frac{1}{24}\right), \frac{1}{3}+3\left(\frac{1}{24}\right){/tex}
{tex}\frac{1}{3}+\frac{1}{24}{/tex}, {tex}\frac{1}{3}+\frac{1}{12}{/tex}, {tex}\frac{1}{3}+\frac{1}{8}{/tex}
{tex}\frac{3}{8}, \frac{5}{12}, \frac{11}{24}{/tex}
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Shakti Sarkar 6 years, 4 months ago
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