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The probability that a leap year will have 53 sunday is ? and how please explain it ?(a)1/7 (b) 2/7 (c) 5/7 (d)6/7Sol)A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days. The remaining 2 days may be any of the following : (i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday (iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday (vii) Saturday and Sunday For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday. n(S) = 7 n(E) = 2 P(E) = n(E) / n(S) = 2 / 7
Here we will use the expansion method Firstly lim x-0 log a (1+x)/x firstly using log property we get: lim x-0 log a (1+x)-logx then we change the base of log i.e lim x-0 {l
the traffic light at three different road crossing change after every 48 seconds, 72 seconds and 108 seconds respectively. if they change simultaneously at 7 a.m., at what time wil
1,5,14,30,55 find the next three numbers and the rule
If p=10 when q=2,find p when q=5
3 1/2 x 1 4/7 x 1 1/3
A polynomial satisfies the following relation f(x).f(1/x)= f(x)+f(1/x). f(2) = 33. fIND f(3) Ans) The required polynomial is x^5 +1. This polynomial satisfies the condition state
We are here going to begin looking at nonlinear first order differential equations. The first type of nonlinear first order differential equations which we will see is separable di
solve: 4ydx+xdy=0
4into 5 is;
A reaction following first-order kinetics was studied by determining the reactant concentrations at equal time intervals. Each successive pair of concentrations, [A] o and [A] 1
A non - leap year contains 365 days 52 weeks and 1 day more.i) We get 53 Sundays when the remaining day is Sunday.Number of days in week = 7∴ n(S) = 7Number of ways getting 53 Sundays.n(E) = 1n E 1n S 7=∴ Probability of getting 53 Sundays =1/7
A leap year consits of 366 days in those 364 days are completely 52 weeks so it contains 52 sundays remaining 2 days there are 2/7 probabilities are there.
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