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3 years ago

The distance between Earth to Neptune is approximately 4.3514 x 10^9

Mathematics

1 answer:

Scrat [10]3 years ago

7 0

The answer would be D)Kilometers.

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1.Encuentra de dos formas diferentes el área de las figuras mostradas en cada literal:

MA_775_DIABLO [31]

Answer:

Cuadrilátero A: $A = y^{2}+4\cdot y +4$

Cuadrilátero B: $A = a^{2}+6\cdot a + 9$

Step-by-step explanation:

Existen dos formas distintas de determinar las áreas de cada cuadrilátero:

(i) <em>Obtener el área de cada cuadrado y sumar los resultados.</em>

(ii) <em>Calcular los lados del cuadrilátero grande y determinar el área. </em>

Cuadrilátero A

Método (i)

$A = y^{2}+2\cdot y +4+2\cdot y$

$A = y^{2}+4\cdot y +4$

Método (ii)

$A = (y+2)\cdot (y+2)$

$A = y^{2}+4\cdot y + 4$

Cuadrilátero B

Método (i)

$A = a^{2}+3\cdot a +9 + 3\cdot a$

$A = a^{2}+6\cdot a +9$

Método (ii)

$A = (a+3)\cdot (a+3)$

$A = a^{2}+6\cdot a + 9$

8 0

2 years ago

There are 45 students in Lisa's class. 35 of the students are boys and rest are girls

geniusboy [140]

If the question is how many students are girls, than it's ten(10).

8 0

3 years ago

Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.

Natalka [10]

Answer:

what lable i dont see nothing

Step-by-step explanation:

7 0

2 years ago

You consider yourself a bit of an expert at playing rock-paper-scissors and estimate that the probability that you win any given

cestrela7 [59]

Answer:

a) 0.37

b) 0.421

c) 0.25

Step-by-step explanation:

Since the probability of winning a game is binomial (P = 0.5) the expected value for number of winning when you play 60 games is

$\mu = 60*0.5 = 30$

And the standard deviation:

$\sigma = 60*0.5*0.5 = 15$

a) the cumulative probability of winning at least 35 games is

$P(X \geq 35) = 1 - P(X < 35) = 1 - 0.631 = 0.37$

b) $P(X < 27) = 0.421$

c) $P(30 \leq X \leq 40) = P(X \leq 40) - P(X \leq 30) = 0.75 - 0.5 = 0.25$

7 0

3 years ago

A certain rectangle is 5 times as long as it is wide. Suppose the length and width are both tripled. The perimeter of the second

astraxan [27]

It is tripled as well. You are tripling every distance and thus tripling the sum of the distances as well.

3 0

3 years ago