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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
If α, β are the zeros of the polynomial x 2 +8x +6 frame a Quadratic polynomial whose zeros are a) 1/α and 1/β b) 1+ β/α , 1+ α/β. Ans. P(x) = x 2 +8x +6 α + β = -8
UA and DU are preparing for the NCAA basketball game championship. They are setting up their strategies for the championship game. Assessing the strength of their "benches", each c
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when is the trnscribing process of data preparation irrelevant ? a)CAPI b) mall panel c) in home interview d) all of them
Above we have seen that (2x 2 - x + 3) and (3x 3 + x 2 - 2x - 5) are the factors of 6x 5 - x 4 + 4x 3 - 5x 2 - x - 15. In this case we are able to find one facto
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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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