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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
what is the graph of 2 sin x
6x^7-2x^3+4x-16 / 3x^2-7x+9
Before we look at this, let us learn what a multiple is. Take any number say 3. Multiply this number with natural numbers. We obtain 3, 6, 9, 12, 15, 18,.........
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OQRS IS A QUADRILATERAL SUCH THAT OQ= -6,3 OR= -3,7 AND OS= 1,5. T IS ON OQ SUCH THAT OT: TQ= 1:2 PROVE THAT QRST IS AA PARALLEGRRAM
What are the characteristics of a queuing system? (i) The input pattern (ii) The queue discipline (iii) The service mechanism
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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