Let X have the geometric distribution. X is the number of trials until the first success. X ~ Geom(p) where p is the success probability the probability
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This shows you that around 0, looks like , and so is positive for x>0. As you can easily see that exp win for big x, I think that the easier way to conclude for all x
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Ex 8.4 : Q.5(ix) : Prove the following identities, where Ch 8 | Math for Class X CBSE :
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Ex 1.3: Q.2 : Prove that sum of 3 and product of 2 and square root Ch 1 | Class Xth Math for CBSE :
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Let X have the geometric distribution. X is the number of trials until the first success. X ~ Geom(p) where p is the success probability the probability
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Then I decided to take the natural log of both sides and try my hand at the other two cases (which I think should be 0
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This shows you that around 0, looks like , and so is positive for x>0. As you can easily see that exp win for big x, I think that the easier way to conclude for all x
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Ex 1.3: Q.3(i) : Prove that quotient of 1 and square root of 2 is Ch 1 | Class Xth Math for CBSE :
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Prove the Limit as x Approaches 0 of (e^x-1)/x :
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Hindi : Ex 8.4 : Q.5(ix) : Prove the following identities, where Ch 8 | Math for Class X CBSE :
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NOTE: e^x^ means that x is an exponent to the number e proof can be done only by using calculus methods!
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Proof: d/dx(e^x) = e^x :
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(x0 = e^x. Similarly we can prove f(x) = e^x, (...) Thus e^x= 1+x/1!+x^2/2!+x^3/3!+(...)x^n/n!+x^(n+1)/(n+1)!+(...).. or. e^x >= 1+x/1!+x^2/2!+x^3/3!+(...)..+x…
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Ex 1.3: Q.1 : Prove that square root of 5 is irrational - Ch 1 | Class Xth Math for CBSE Students :
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