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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
More Optimization Problems Example A window is being built in which the bottom is rectangle and the top is a semicircle. If there framing materials is 12 meters what have
In a parallelogram ABCD AB=20cm and AD=12cm.The bisector of angle A meets DC at E and BC produced at F.Find the length of CF.
63*789
1. Solve the given differential equation, subject to the initial conditions: . x2y''-3xy'+4y = 0 . y(1) = 5, y'(1) = 3 2. Find two linearly independent power series soluti
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how do you round to the nearest dollars?
Given So calculate AB. Solution The new matrix will contain size 2 x 4. The entry in row 1 and column 1 of the new matrix will be determined by multiplying row 1 of
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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