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blagie [28]
3 years ago
10

Need help with stats!

Mathematics
1 answer:
Brut [27]3 years ago
8 0

Answer:

a) 1,440 ways

b) 59,280 or 64,000

Step-by-step explanation:

a) Aircraft boarding.

8 people, 2 in first class, boarding first, then 8 economy class.

The 2 people in first class board first, but they can board as AB or BA... so 2 ways here.

For the 6 economy class passengers, we have a permutation of 6 out of 6, so 720, as follows:

P(6,6) = \frac{6!}{(6 - 6)!} = 6! = 720

Since the two are independent, we multiply them to have a global number of ways: 2 * 720 = 1,440 different ways for the 8 passengers to board that plane.

b) combination lock.

Here we do have a little problem... the question doesn't specify if the 3 numbers are different numbers of not.  So, we'll calculate both:

Numbers go from 1 to 40 inclusively... so 40 possibilities.

Normally, in a combination lock, the numbers are different, so let's start with that one:

First number: 40 options available

Second number: 39 options available (cannot take the first one again)

Third number: 38 different options (can't take First or Second number again)

Overall, we then have 40 * 39 * 38 = 59,280 different lock combinations.

If we can pick pick the same number twice:

First number: 40 options available

Second number: 40 options available

Third number: 40 options available

Overall 40 * 40 * 40 = 64,000 different lock combinations

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En una fiesta se desea elegir un grupo de 3 personas de entre 6 matrimonios, si dos miembros de la misma pareja no pueden ser el
LenaWriter [7]

Answer:

Hay 160 maneras

Step-by-step explanation:

Para calcular de cuántas maneras se puede seleccionar x elementos de un grupo de n elementos podemos usar la siguiente fórmula:

nCx=\frac{n!}{x!(n-x)!}

Si vas a elegir 3 personas de entre 6 matrimonios y dos miembros de la misma pareja no pueden ser elegidos, entonces podemos decir que se está eligiendo 3 matrimonios y en cada matrimonio se está eligiendo un representante.

Entonces, podemos calcular de cuántas maneras se puede escoger 3 matrimonios de los 6, así:

6C3=\frac{6!}{3!(6-3)!}=20

Adicionalmente, para cada uno de los 3 matrimonios hay 2 opciones para ser representantes. Esto significa que hay 2^3 maneras de escoger representantes en cada una de las 20 formas calculadas anteriormente.

Por lo tanto se puede formar el grupo de 3 personas de 160 maneras diferentes:

20*2^3=160

6 0
3 years ago
Jose is working two summer jobs, making $10 per hour washing cars and $9 per hour walking dogs. Last week Jose worked a total of
Setler [38]

Answer:

x+y =13

10x + 9y =122

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations:

The sum of the hours he worked washing cars (x) and the hours he worked walking dogs(y) will be equal to the total hours worked (13)

  • x+y =13

For the second equation the product of the number of hours he worked washing cars(x) and the price per hour ($10) plus the product of the number of hours he worked walking dogs (y) and the price per hour ($9) will be equal to the total amount earned (122)-

  • 10x + 9y =122

So, the system of equations is:

  • x+y =13
  • 10x + 9y =122

Where:

x: number of hours Jose worked washing cars

y: number of hours Jose worked walking dogs  

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3 years ago
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7x +7y + 7 hope this helps as an expansion :)
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Complete the solution of the equation. Find the
forsale [732]

Answer:

y = -2

Step-by-step explanation:

Put x as -19 and solve for y.

2(-19) - 5y = -28

Multiply.

-38 - 5y = -28

Add 38 on both sides of the equation.

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Divide both sides by -5.

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I am Lyosha [343]

Answer:

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N___________________<u>M</u>_______O

               76                                 15

NO = MN + MO = 76 + 15 = 91

8 0
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