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Kisachek [45]
3 years ago
9

Which of these match plz don’t say random ones more then one answer possibly

Mathematics
1 answer:
natali 33 [55]3 years ago
3 0
A and E
Just rearrange where the -5 is
You might be interested in
From a standard deck of cards, a 5-card hand is dealt. Calculate the number of hands that contain 4 diamonds and 1 black king
rjkz [21]

Answer:

1430

Step-by-step explanation:

Given that a standard deck of cards is 52 cards, we have 13 Diamonds and we need 4 diamonds.

Hence, we have 13C4.

Also, we have 4 kings in a standard deck of cards, and in which we have 2 black kings but we need 1 king.

Hence we have a 2C1

Therefore:

13C4 * 2C1

=> 13! ÷ [4! (13 - 4)!] * 2! ÷ [1! (2-1)!]

=> 13! ÷ [4! (9)!]

=> (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13) ÷ [(1 x 2 x 3 x 4) (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9)]

=> (10 x 11 x 12 x 13) ÷ 24

=> (17160 ÷ 24) * (24 ÷ 6)

=> 715

2! ÷ [1! (2-1)!]

=> 2

Hence, we have 715 * 2

=> 1430

Hence, in this case, the correct answer to the question is 1430

7 0
3 years ago
Una heladeria dispone de 20 frutas distintas para elaborar sus malteadas. Si los clientes pueden elegir tres sabores para mezcla
Dahasolnce [82]

Answer:

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

Step-by-step explanation:

En este caso, el cliente que adquiere una malteada de tres sabores distintos sigue el siguiente procedimiento:

1) El primer sabor sale de cualquiera de las 20 frutas disponibles.

2) El segundo sabor es distinto al primer sabor, es decir, que sale de las 19 frutas restantes.

3) El tercer sabor es distinto al primer sabor y al segundo sabor, es decir, que sale de las 18 frutas restantes.

Puesto que existe una doble conjunción y que puede importar el orden según la preferencia del cliente, se habla matemáticamente de una permutación, definida como:

n\mathbb{P}k = \frac{n!}{(n-k)!} (1)

Donde:

n - Número de sabores disponibles, adimensional.

k - Número de sabores escogidos, adimensional.

Si tenemos que n = 20 y k = 3, entonces la cantidad de malteadas de tres sabores distintos es:

n\mathbb{P}k = \frac{20!}{(20-3)!}

n\mathbb{P}k = \frac{20!}{17!}

n\mathbb{P}k = 20\cdot 19\cdot 18

n\mathbb{P}k = 6840

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

5 0
3 years ago
Ahahhajdockdndn help
Elina [12.6K]
Domain: 6,-5,-1
Range: 2, 0, 2
6 0
3 years ago
Factor: <img src="https://tex.z-dn.net/?f=%2025y%5E%7B2%7D%20" id="TexFormula1" title=" 25y^{2} " alt=" 25y^{2} " align="absmidd
chubhunter [2.5K]
(5y-2)to the power of 2
5 0
3 years ago
Read 2 more answers
You can draw a quadrilateral with two sets of parallel lines and no right angles. true false
yuradex [85]
False. When a quadrilateral has two sets of parallel sides ,like I drew below, all four corners form right angles.

5 0
3 years ago
Read 2 more answers
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