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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
Suppose that at some future time every telephone in the world is assigned a number that contains a country code, 1 to 3 digits long, that is, of the form X, XX , XXX or followed
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#questionShow that the system oscillates in simple harmonic motion demonstrated by; , for which the general solution where X = (x – x0)..
1. A point P(a,b) becomes (3,c) after reflection in x - axis, and (d,6) after reflection in the origin. Show that a = 3, b = - 6, c = 6, d = 2 2. If the pair of lines ax² + 2pxy
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1) At a midway game at the state fair, the probability of winning an individual game is advertised to be 30% ( p = . 3). Suppose 50 people played the game (assume all 50 outcomes
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Ratio Test In this part we are going to take a look at a test that we can make use to see if a series is absolutely convergent or not. Remind that if a series is absolutely c
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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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