<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
P = 2(l + w)
p = 356
l = 96
356 = 2(96 + w)
356 = 192 + 2w
356 - 192 = 2w
164 = 2w
164/2 = w
82 = w
check....
p = 2(l + w)
356 = 2(96 + 82)
356 = 2(178)
356 = 356 (correct)
so the width (w) = 82 cm
I’d like to answer the but What angle ??
1. Introduction. This paper discusses a special form of positive dependence.
Positive dependence may refer to two random variables that have
a positive covariance, but other definitions of positive dependence have
been proposed as well; see [24] for an overview. Random variables X =
(X1, . . . , Xd) are said to be associated if cov{f(X), g(X)} ≥ 0 for any
two non-decreasing functions f and g for which E|f(X)|, E|g(X)|, and
E|f(X)g(X)| all exist [13]. This notion has important applications in probability
theory and statistical physics; see, for example, [28, 29].
However, association may be difficult to verify in a specific context. The
celebrated FKG theorem, formulated by Fortuin, Kasteleyn, and Ginibre in
[14], introduces an alternative notion and establishes that X are associated if
∗
SF was supported in part by an NSERC Discovery Research Grant, KS by grant
#FA9550-12-1-0392 from the U.S. Air Force Office of Scientific Research (AFOSR) and
the Defense Advanced Research Projects Agency (DARPA), CU by the Austrian Science
Fund (FWF) Y 903-N35, and PZ by the European Union Seventh Framework Programme
PIOF-GA-2011-300975.
MSC 2010 subject classifications: Primary 60E15, 62H99; secondary 15B48
Keywords and phrases: Association, concentration graph, conditional Gaussian distribution,
faithfulness, graphical models, log-linear interactions, Markov property, positive
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