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Gnoma [55]
3 years ago
6

One stack has 6 cups, and its height is 15 cm. The other one has 12 cups, and its height is 23 cm. How many cups are needed for

a stack with a height of 50 cm?
Mathematics
1 answer:
Ronch [10]3 years ago
7 0

Answer:

Given

Number of stacks = 2

Stack 1 = 6 cups; h1 = 15cm

Stack 2 = 12 cups; h2 = 23cm

Let's first find the average:

With an average of 4/3, to obtain the number of cups needed to obtain a height of 50m, we have:

50 / (4/3)

= 50 * 3/4

= 150/4

= 37.5

From the answer, we can see that the number of cups is not really proportional to the height of the stack, because the average of stack one and stack 2 are different.

Step-by-step explanation:

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Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer
QveST [7]

Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

7 0
2 years ago
Find the radius of convergence, then determine the interval of convergence
galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

  • If β < 1, then the series converges
  • If β > 1, then the series diverges
  • If β = 1, then the series inconclusive

Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

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- 1 - 2 < x < 1 - 2

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We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

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Answer:

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Step-by-step explanation:

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