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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 Var(x) = 4, find Var (3x+8), where X is a random variable. Var (ax+b) = a 2 Var x Var (3x+8) = 3 2 Var x = 36
x/15=50/20
Our objective is solve the following fourth-order BVP: (a(x)u'' )'' = f (x) u(0) = u(1)=0 u(0)' = u(1)'=0 (a) Give the variational formulation of the above BVP. (b) Describe the
what is the value of zero to the power raised to zero?
how do you solve expressions
how do you write this polynomial in standerd form 5x3 + x5 - 8 + 4x ?
?[1,99] x^5+2x^4+x^3+5x^2+6x+2÷x^2+2x
Euler Equations - Series Solutions to Differential Equations In this section we require to look for solutions to, ax 2 y′′ + bxy′ + cy = 0 around x0 = 0. These ki
A firm is manufacturing 45,000 units of nuts. The probability of having a defective nut is 0.15 Compute the given i. The expected no. of defective nuts ii. The standard an
Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. (Ans:999720) Ans: LCM of 24, 15, 36 LCM = 3 × 2 × 2 × 2 × 3 × 5 = 360 Now, the greatest six digit
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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