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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
How many homomorphism are there from z2 to z3. Zn is group modulo n
Mod(Z-25i) Sol) mod (Z-25i) means Z lies in the circumference of the circle with (0,25) at its centre and radius less then 15. so difference in the max and min value of arg Z is
If there are (2n+1)terms in an AP ,prove that the ratio of the sum of odd terms and the sum of even terms is (n+1):n Ans: Let a, d be the I term & Cd of the AP. ∴ ak =
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Polar Coordinates Till this point we've dealt completely with the Cartesian (or Rectangular, or x-y) coordinate system. Though, as we will see, this is not all time the easie
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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