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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
Need help, please anybody solve this: Consider the universal set T and its subsets A, B and C underneath as: T = {a, b, c, d e, f} A = {a, d} B = {b, c, f} C = {a, c
Find the GCF of 70 and 112
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i need help in discrete mathematics on sets, relations, and functions.
Consider the Van der Pol oscillator x′′- µ(1 - x 2 )x′ + x = 0 (a) Write this equation as a system of first order equations (b) Taking µ = 2, use MatLab's routine ode45 to
Mike sells on the average 15 newspapers per week (Monday – Friday). Find the probability that 2.1 In a given week he will sell all the newspapers [7] 2.2 In a given day he will sel
1. In 1900, a certain country's population was 77,977,459 and it's area was 2,821,924 square miles, In 2000, the country's population was 283,575,229 and its area was 3,551,003 sq
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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