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Damm [24]
3 years ago
6

¿es cierto que si se suman los resultados de la tabla del con los de la tabla del 5 se obtienen los resultados del 7 ¿porque

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

Efectivamente, si se suman los resultados de la tabla del 2 con los de la tabla del 5 se obtienen los resultados de la tabla del 7.

Step-by-step explanation:

Esto es así porque 7 puede formarse a través de diversas sumas (7+0, 6+1, 5+2 o 4+3). Por ende, en el caso de las tablas de multiplicaciones, decir, por ejemplo, 7 x 3, es igual a decir 5 x 3 + 2 x3, dado que 5 + 2 es igual a 7.

Entonces, siguiendo el ejemplo, dado que 7 x 3 es igual a 21, la suma de 5 x 3 igual a 15 y 2 x 3 igual a 6 es 21 (15 + 6), con lo cual la afirmación es correcta.

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2 years ago
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5/6 of an estate goes to relatives. Of the remaining estate 1/2 goes to the American cancer society. What fraction of the estate
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6 0
3 years ago
Read 2 more answers
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Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

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Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0
3 years ago
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Bu using Pythagoras theorem
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7 0
3 years ago
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