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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
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1. Consider the trigonometric function f(t) = (a) What is the amplitude of f(t)? (b) What is the period of f(t)? (c) What are the maximum and minimum values attained by
Area Problem Now It is time to start second kind of integral: Definite Integrals. The area problem is to definite integrals what tangent & rate of change problems are to d
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
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Find the ratio in which the line segment joining A(6,5) and B(4,-3) is divided by the line y=2 (Ans:3:5) Ans : Let the ratio be k:1 x = 4 k + 6/ k +1 y
Example Sketch the graph of following f( x ) = 2x and g( x ) = ( 1 /2) x Solution Let's firstly make a table of values for these two functions. Following is
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Determine the slope following lines. Sketch the graph of line. The line which contains the two points (-2, -3) and (3, 1) . Solution we'll need to do is employ
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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