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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
Strategy for Series Now that we have got all of our tests out of the way it's time to think regarding to the organizing all of them into a general set of strategy to help us
A box contains 12 balls out of which x are black .if one ball is drawn at random from the box what is the probability that it will be a black ball ? If 6 more black balls are pu
1. Consider the relation on A = {1, 2, 3, 4} with relation matrix: Assume that the rows and columns of the matrix refer to the elements of A in the order 1, 2, 3, 4. (a)
Help me how to solve equation by Quadratic Formula.
Example Find the Highest Common Factor of 54, 72 and 150. First we consider 54 and 72. The HCF for these two quantities is calculated as follows:
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G2=5.12 and G5=80 Find, r, G1, and S6
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Submit solutions for all of the following questions. Remember to set out your answers showing all steps completely and explicitly justify your steps. 1. Provide, in no more than
what is the domain of the function f(x)= 2x^2/x^2-9
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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