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Log problem_ISI 2010
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rongmon
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Apr 30, 2014; 11:00am
Log problem_ISI 2010
Given log(p)x = a and log(q)x = b, the value of log(p/q)x equals
(a) ab/(b - a)
(b) (b - a)/ab
(c) (a - b)/ab
(d) ab/(a - b)
log(p)x stands for log of x with base p. And similarly read log(q)x & log(p/q)x.
Dreyfus
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Apr 30, 2014; 11:14am
Re: Log problem_ISI 2010
rongmon
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Apr 30, 2014; 11:25am
Re: Log problem_ISI 2010
Thanks, great help.
rongmon
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Apr 30, 2014; 11:33am
Re: Log problem_ISI 2010
An unbiased coin is tossed until a head appears. What is the expected number of tosses required?
a. 1
b. 2
c. 4
d. infinity
Thank you.
Homer Simpson
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Apr 30, 2014; 3:44pm
Re: Log problem_ISI 2010
i think its infinity. not sure though!
“Operator! Give me the number for 911!”
Shefali
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Apr 30, 2014; 4:23pm
Re: Log problem_ISI 2010
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by rongmon
2tosses..
Dreyfus
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Apr 30, 2014; 4:57pm
Re: Log problem_ISI 2010
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by rongmon
Take summation of r*(1/2)^r from 1 to infinity.....u'll get 2 as expected no. of tosses
rongmon
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Apr 30, 2014; 5:07pm
Re: Log problem_ISI 2010
Could you expand that equation?
Thanks.
Homer Simpson
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Apr 30, 2014; 6:22pm
Re: Log problem_ISI 2010
thanks for clearing that up!
“Operator! Give me the number for 911!”
Dreyfus
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Apr 30, 2014; 6:24pm
Re: Log problem_ISI 2010
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by rongmon
Sum= 1/2 + 2*(1/2)^2 + 3*(1/2)^3+....
Multiply whole eq by 2
2Sum= 1+ 2*(1/2) + 3*(1/2)^2+.......
Subtract eq 1 from 2
Sum= 1+(1/2)+(1/2)^2+(1/2)^3+.......
dis series is a gp nd it sum converges to 2
rongmon
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Apr 30, 2014; 7:37pm
Re: Log problem_ISI 2010
Thank you!
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