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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
Scalar Equation of Plane A little more helpful form of the equations is as follows. Begin with the first form of the vector equation and write a vector for the difference. {
PROOF OF VARIOUS LIMIT PROPERTIES In this section we are going to prove several of the fundamental facts and properties about limits which we saw previously. Before proceeding
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#question.find the number of combinations of the letters a, b, c, and d taken 3 at a time.
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y'' + 2y = 2 - e-4t, y(0) = 1 use euler''s method with a step size of 0.2 to find and approximate values of y
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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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