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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 would I graph the equation 2x-5y=5?
different types of ellipse
maria has a slice of pizza that is 1/6 of the pizaa.Ben has a slice of pizza that is 1/3 of the pizza, marias slice is bigger.draw pizzas to show how this is possible.
Descrbe about Arithmetic and Geometric Sequences? When numbers are listed according to a particular pattern, we call the list a sequence. In a sequence, the numbers are separat
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The points A,B,C and D represent the numbers Z1,Z2,Z3 and Z4.ABCD is rhombus;AC=2BD.if Z2=2+i ,Z4=1-2i,find Z1 and Z3 Ans) POI of diagonals: (3-i)/2. Using concept of rotation:
Fermat's Theorem : If f ( x ) contain a relative extrema at x = c & f ′ (c ) exists then x = c is a critical point of f ( x ) . Actually, it will be a critical point such that f
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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