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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
Solve the subsequent IVP. y′′ + 11y′ + 24 y = 0 y (0) =0 y′ (0)=-7 Solution The characteristic equation is as r 2 +11r + 24 = 0 ( r + 8) ( r + 3) = 0
Factoring quadratics of the form x 2 + bx + c ? This tutorial will help you factor quadratics that look something like this: x 2 + 7x + 12 (Positive coefficients; no lea
( 1/2)2 x 1/5+1/6=
sin(x)+cos(x)
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#given that the perimeter of the buildig is 108m and the area of the floor is 138m, find the volume of the screed in m3 if it is 30mm thick
If a school has lockers with 50 numbers on each combination lock, how many possible combinations using three numbers are there.
Three Dimensional geometry Intorduction In earlier classes we studied about the coordinates in two planes that is the XY plane. Here we are going to study in detail about th
find the angel between the vectors 4i-2j+k and 2i-4j online answer
Determine the domain of each of the following functions. f( x ) = x - 4 / x 2 - 2 x -15 Solution With this problem we have to avoid division by
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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