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Klio2033 [76]
3 years ago
9

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) ther

e will always be least such unlucky Friday but at one can be no more three.
Please answer ASAP
Mathematics
1 answer:
Vlad1618 [11]3 years ago
5 0
<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>
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3 years ago
Solve for X, please show work!
Molodets [167]

Answer:

the solution set consists of {x < -2 ∪ x > 10/3}

Step-by-step explanation:

|3x - 2| > 8 is equivalent to the following set of inequalities:

1) 3x - 2 > 8

and

2) -(3x - 2) > 8

In case 1, above, add 2 to both sides, obtaining 3x > 10, or x > 10/3.

In case 2, above, carry out the indicated multiplication first:

-3x + 2 > 8.  Next, subtract 2 from both sides:  -3x > 6.

Next, divide both sides by -3, remembering to reverse the direction of the inequality sign:  x < -2.

Thus, the solution set consists of {x < -2 ∪ x > 10/3}

5 0
3 years ago
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3 3/5 + 9 3/5 answer with a mixed number in simplest form
Romashka [77]

Both fractions have the same denominator so it's going to be easy. First set up your equation:

3 \frac{3}{5} + 9 \frac{3}{5}

Add the whole numbers:

3 + 9 = 12

Now add the fractions:

\frac{3}{5} + \frac{3}{5} = \frac{6}{5}

add both together:

12 + \frac{6}{5} = 12 \frac{6}{5}

\frac{6}{5} is an improper fraction so change it:

12 \frac{6}{5} = 13 \frac{1}{5}

Since 6 is one more than 5, add 1 to the whole number and subtract the numerator and denominator(6 - 5 = 1) and make the remaining the new numerator. That leaves you with 13 \frac{1}{5}

Your answer is 13 \frac{1}{5}

3 0
3 years ago
PLEASE HELP ITS A TIMED QUIZ IM DESPERATE ILL GIVE U BRAINLIEST AND A THANKS PLEASE
den301095 [7]

Answer:

2000 = 600 + 2/5x ; 3500 ft

How To Solve:

2000 = 600 + 2/5x

Combine multiplied terms into a single fraction

2000 = 600 + 2/5x

2000 = 600 + 2x/5

Subtract 600 from both sides of the equation

2000 = 600 + 2x/5

2000 - 600 = 600 + 2x/5 - 600

Simplify

Subtract the numbers

1400 = 600 + 2x/5 − 600

Subtract the numbers

1400 = 2x/5

Solution

X = 3500

7 0
3 years ago
Se necesitan conductores voluntarios para llevar a 80 estudiantes al juego de béisbol de campeonato. Los conductores tienen auto
NemiM [27]

Esta pregunta esta incompleta

Pregunta completa

Se necesitan conductores voluntarios para llevar a 80 estudiantes al juego de béisbol del campeonato. Los conductores cuentan con automóviles, con capacidad para 4 estudiantes, o camionetas, con capacidad para 6 estudiantes. La ecuación describe la relación entre la cantidad de automóviles y la cantidad de camiones que pueden transportar exactamente 80 estudiantes.

Seleccione todas las afirmaciones que sean verdaderas sobre la situación.

a) Si van 12 coches, se necesitan 2 furgonetas.

b) c = 14 y v = 4 son un par de soluciones de la ecuación.

c) Si van 6 coches y van 11 furgonetas, habrá espacio adicional.

d) 10 coches y 8 furgonetas no son suficientes para transportar a todos los estudiantes.

e) Si pasan 20 autos, no se necesitan camionetas.

f) 8 furgonetas y 8 coches son números que cumplen con las limitaciones en esta situación.

Answer:

La opción b, e y f.

Step-by-step explanation:

Pasando por las opciones dadas:

c = coches que pueden llevar a 4 estudiantes

v = furgonetas con capacidad para 6 estudiantes

a) Si van 12 coches, se necesitan 2 furgonetas.

= 12 (4) + 2 (6)

= 48 + 12

= 60

60 ≠ 80

La opción a es incorrecta

b) c = 14 y v = 4 son un par de soluciones de la ecuación.

14 (4) + 4 (6)

56 + 24 = 80 estudiantes

La opción b es correcta

c) Si van 6 coches y van 11 furgonetas, habrá espacio adicional.

Por lo tanto:

5 (4) + 11 (6)

= 20 + 66

= 86.

86 es mayor que 80, la opción c es correcta, habría un espacio adicional para 6 estudiantes.

d) 10 coches y 8 furgonetas no son suficientes para transportar a todos los estudiantes.

Por lo tanto:

10 (4) + 8 (6)

40 + 48

= 88

e) Si pasan 20 autos, no se necesitan camionetas.

20 (4) = 80 estudiantes

La opción e es correcta

f) 8 furgonetas y 8 coches son números que cumplen con las limitaciones en esta situación.

8 (6) + 8 (4)

= 48 + 32

= 80 estudiantes.

La opción b es correcta

Por lo tanto, la ecuacióna que describe la relación entre la cantidad de automóviles y la cantidad de camiones que pueden transportar exactamente 80 estudiantes es la opción b, e y f.

7 0
3 years ago
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