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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
Evaluate following. ∫ 0 ln (1 + π ) e x cos(1-e x )dx Solution The limits are little unusual in this case, however that will happen sometimes therefore don't get
#k1=f(Tn, Xn), k2=f (Tn + H.Y,Xn + H.Y.k1) Xn+1=Xn + H(a.k1+ b.k2) Find a relation between Y,a and b so that the method is second order consistent.
Consider the following proposal to deskew a skewed bitstream from a TRNG. Consider the bitstream to be a sequence of groups ot n bits for some n > 2. Take the first n bits, and o
Smith keeps track of poor work. Often on afternoon it is 5%. If he checks 300 of 7500 instruments what is probability he will find less than 20 substandard?
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Given the hypotenuse of a right triangle: Given that the hypotenuse of a right triangle is 18" and the length of one side is 11", what is the length of another side? a 2 +
Special Forms There are a number of nice special forms of some polynomials which can make factoring easier for us on occasion. Following are the special forms. a 2 + 2ab +
What is the value of tan? in terms of sin?. Ans: Tan ? = S i n ?/ C os ? Tan ? = S i n ? / √1 - S i n 2?
Find the Quadratic polynomial whose sum and product of zeros are √2 + 1, 1/ √2 + 1 Ans: sum = 2 √2 Product = 1 Q.P = X 2 - (sum) x + Product ∴ x 2 - (2 √2 )
Using the formulas and properties from above find out the value of the subsequent summation. c The first thing that we require to do here is square out the stuff being summe
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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