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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
You are going on a road trip and you buy snack packs and three different kind of beverages. You buy 7 Cokes, 5 Pepsis and 4 Dr. Peppers. You pull out two beverages at random. An
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assigenment of b.sc.1sem
every rational nmber is expressible either as a_________or as a____________decimal.
from 0->1: Int sqrt(1-x^2) Solution) I=∫sqrt(1-x 2 )dx = sqrt(1-x 2 )∫dx - ∫{(-2x)/2sqrt(1-x 2 )}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x 2 ) - ∫-x 2 /√(1-x 2 ) Let
Diagonals of a trapezium divide each other proportionally: Given : In trapezium ABCD , AB// DC R.T.P :AO/OC = BO/OD Construction: Draw the line PQ; parallel to AB or C
The math equation is written exactly this way: 0+50x1-60-60x0+10=??? The answer I get is 10 and others say 0 0+50=50 50x1=50 50-60=-10 -10-60=-70 -70x0=0 0+10=10
I need help solving this question...You have to design a rectangular flyer. The top and bottom must have 5" margins and the left and right sides must have 2" margins. If you must
limit x APProaches infinity (1+1/x)x=e
A circle touches the sides of a quadrilateral ABCD at P, Q, R and S respectively. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.
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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