Mathematics, 02.08.2019 02:00 Hrjohnson2004

Some people think it is unlucky if the 13th day of month falls on a friday. show that in that there year (non-leap or leap) there will always be least such unlucky friday but at one can be no more three. answer asap

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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.

Answer from: Quest

the direct variation says that:

$y \propto x$

then the equation is of the form is given by:

$y = kx$ where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

$3 = 2k$

divide both sides by 2 we get;

$k =\frac{3}{2} = 1.5$

⇒ $m(x) = 1.5x$

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

$m(x) = 1.5x$

$-1.5 = 1.5(-1)$

-1.5 = -1.5 true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

Answer from: Quest

chicken chicken chicken

step-by-step explanation:

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Some people think it is unlucky if the 13th day of month falls on a friday. show that in that there...