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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
[3+tan20+tan80]/tan20+tan80
in a veggie mix the ratio of cups of carrots to cups of broccolie is 4 to 5 if you made this party mix larger how many cups of carrots would be needed to mix with fo cups of brocco
We know that a factor is a quantity which divides the given quantity without leaving any remainder. Similar to LCM above we can find a highest common factor (HCF)
Consider a database whose universe is a finite set of vertices V and whose unique relation .E is binary and encodes the edges of an undirected (resp., directed) graph G: (V, E). Ea
4/x+4-3/x+3=2/x+2-1/x+1
what Is the common denominator for 1/2 and 1/4
Tetracycline 500 mg capsules Sig: 1 cap po bid for 14 days. Refills: 2 What is the dose of this medication:____________________ (0.5 point) How many doses are given per day:______
Solve the inequality |x - 1| + |x - 2|≤ 3. Working Rule: First of all measure the expression to zero whose modulus happens in the given inequation and from this search the va
if an object weighed 11 pounds how many ounces would it weigh
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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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