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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
If we "break up" the root into the total of two pieces clearly we get different answers. Simplified radical form: We will simplify radicals shortly so we have to next
The picture frame given below has outer dimensions of 8 in by 10 in and inner dimensions of 6 in by 8 in. Find the area of section A of the frame. a. 18 in 2 b. 14 in 2
what is -6.4 as a fraction?
(3x+2)^2 d^2y/dx^2+3(3x+2)dy/dx-36y=3x^2+4x+1
DIFFERENCE BETWEEN RIGHT ANGLE AND SCALENE
nC6:n-3C3=91:4
if the numerator of a fraction is decreased by 40% and the denominator is increased by 100% the new value is 1. what was the original factor
Cross Product In this last section we will look at the cross product of two vectors. We must note that the cross product needs both of the vectors to be three dimensional (3D
An advertising project manager developed the network diagram shown below for a new advertising campagign. In addition, the manager gathered the time information for each activity,
a) Let n = (abc) 7 . Prove that n ≡ a + b + c (mod 6). b) Use congruences to show that 4|3 2n - 1 for all integers n ≥ 0.
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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