sab goemetry ke piche pade ho ye loTen tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. what is the prob that largest no. appearing on the selected ticket is 7 is ??
bhai is ques ka logic zaroor dena ..ye ques to phle wale thread pe kafi chaya hua tha..
10^6 total ab max 7 chahiye to 1-7 hi select kar sakte hain , so 7^6 par isme cases aaenge jisme 7 ek bhi bar select nai hua ho, aise max. cases tab honge jab 1-6 mein hi select kar lia ho - 6^6 so (7^6-6^6)/10^6
10^6 totalab max 7 chahiye to 1-7 hi select kar sakte hain , so 7^6par isme cases aaenge jisme 7 ek bhi bar select nai hua ho, aise max. cases tab honge jab 1-6 mein hi select kar lia ho - 6^6so (7^6-6^6)/10^6
nai hua ho, aise max. cases tab honge jab 1-6 mein hi select kar lia ho - 6^6
Is the answer c)R(n-1) = Rn/ ( 3+2*root(2)) So, infinite GP with r = 1/(3 + 2root(2)) -> Rn*(1 - r^n) / ( 1-r) n approaches infinity = > Rn*(1/(1-r)) => option C
kya baat hai.bhai..phla post aur bang on.......cheers..
nai hua ho, aise max. cases tab honge jab 1-6 mein hi select kar lia ho - 6^6Isko aur ek baar aur samjha diyo..
yar max. 7 chahiye hi par 7^6 includes cases like 1,1,1,1,1,1; 6,6,6,6,6,6; 1,2,4,5,1,6; i.e u are not selecting 7th, so condition is no fulfilled remove all such cases where 7 is not included =>7^6-6^6
@chandrakant.k ...lo bhai..apna gussa thoda shaant kar lo....algebra se khelte hai chalo..
Q.) a, b, c, d be real numbers in G.P. If u, v,w satisfy the equations u + 2v + 3w = 6; 4u + 5v + 6w = 12; 6u + 9v = 4 then roots of the equations (1/u + 1/v + 1/w)x2 + [(b − c)2 + (c − a)2 + (d − b)2]x + u + v + w = 0 and 20x2 + [10(a − d)2]x − 9 = 0 are
It is whole square right ? I mean the co-efficient 2 written inside brackets.Solving we get u = -1/3 , v = 2/3, w = 5/3 .. We see, that coefficient of x^2 of one is same as constant terms of the other and the middle term gets adjusted ... => reciprocal ? Option b) ?
well yes..they are powers..but which coef of x?^2 u r lukng at...and comparing with which const term...n hw u adjusted middle term????
@chandrakant.k ...lo bhai..apna gussa thoda shaant kar lo....algebra se khelte hai chalo..Q.) a, b, c, d be real numbers in G.P. If u, v,w satisfy the equations u +2v + 3w = 6; 4u + 5v + 6w = 12; 6u + 9v = 4 then roots of the equations(1/u + 1/v + 1/w)x2 + [(b − c)2 + (c − a)2 + (d − b)2]x + u + v + w = 0 and20x2 + [10(a − d)2]x − 9 = 0 area) equal to each other b) reciprocal to each other c) at least one of the foregoing d) all of the foregoing
u=-1/3,v=2/3,w=5/3
(1/u + 1/v + 1/w)x2 + [(b − c)2 + (c − a)2 + (d − b)2]x + u + v + w = 0 reduces to
Well, I didn't adjust the middle term but my sixth sense tells me that it will get adjusted I used this property : f(x) = ax^2 + bx + c = 0 Reciprocal roots will satisfy the quadratic, g(x) = c(x^2) + b(x) + a = 0So, you can see that the constant term and x^2 got interchanged.
In a certain town, every girl is acquainted with more boys then girls. Furthermore, all the girls acquainted with a given boy are also acquainted with each other.Then
a) The number of boys is always greater than the number of girls. b) The number of girls is always greater than the number of boys . c) There are at least as many boys as girls in the town . d) There are at least as many girls as boys in the town.
b) 1 ?p^2 + 3p - 1 = N All primes other than 2,3 are of the form 6k+1/ 6k-1so, for p = 2, 3 , we need to check manually -> p = 2 => p^2 + 3p - 1 = 9 (rejected)p = 3 => p^2 + 3p -1 = 17 Now for p = 6k+ 1 or 6k - 1N = (p-1)*(p+1) +3pThis N is divisible by 3 => No other value of p exists.
...cheers..
..algebra se khelte hai chalo..