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Bess [88]
4 years ago
8

Juana tiene en su tienda un costal con 28 libras de azucar.Hizo seis paquetes de 2,5 kg cada uno,de los cuales vendio cinco.¿Cua

nta azucar quedo en el costal?
Mathematics
1 answer:
Likurg_2 [28]4 years ago
6 0

Answer:

Al comienzo Juana tiene 28 libras de azúcar.

Ella hace 6 paquetes de 2.5kg cada paquete.

Entonces, entre los 6 paquetes, hay:

6*2.5kg = 15kg de azúcar.

Ahora el problema es que tenemos dos unidades distintas, kilogramos y libras.

Aca podemos usar la relación:

1k = 2.2lb.

Entonces 15kg de azúcar son 15 veces 2.2 libras:

15kg = 15*2.2lb = 33lb.

Esto nos da mas de lo que hay originalmente en el costal.

Ahora si lo pensamos como:

Solo vendió 5 paquetes de 2.5kg, entonces el sobrante queda en el costal inicial:

5*2.5kg = 12.5kg

De vuelta, 1kg = 2.2lb

12.5kg = 12.5*2.2lb = 27.5lb

En este caso hay un sobrante de:

28lb - 27.5lb = 0.5lb

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95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}

E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}

\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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(a)

\begin{gathered} P(Z

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The Magazine Mass Marketing Company has received 10 entries in its latest sweepstakes. They know that the probability of receivi
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Answer:

P(X>4)= 0.624

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