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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
Use the graph of y = x2 - 6x to answer the following: a) Without solving the equation (or factoring), determine the solutions to the equation x 2 - 6x = 0 usi
integrate sin(x) dx
In the view below of the robot type of Cartesian Coordinates, is not the "Z" and "Y" coordinates reversed? http://www.expertsmind.com/topic/robot-types/cartesian-coordinates-91038
We desire to construct a box whose base length is three times the base width. The material utilized to build the top & bottom cost $10/ft 2 and the material utilized to build the
About Zeros in the Denominator of Rational Expressions One thing that you must be careful about when working with rational expressions is that the denominator can never be zero
Consider a database whose universe is a finite set of vertices V and whose unique relation .E is binary and encodes the edges of an undirected (resp., directed) graph G: (V, E). Ea
Hi I have a maths question related to construction as its a construction management course...i could send some example sheets too...could it be done?
Exercise 12c question number 24
Find the length of the second diagonal of a rhombus, whose side is 5cm and one of the diagonals is 6cm.
A boy standing on a horizontal plane finds a bird flying at a distance of 100m from him at an elevation of 300. A girl standing on the roof of 20 meter high building finds the angl
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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