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1 year ago

En una caja hay el doble de monedas que en otra. Si se pasan 7 monedas de la

1 answer:

8 0

De acuerdo con las ecuaciones lineales, la caja de mayor cantidad tuvo 28 monedas y la caja de menor cantidad tuvo 14 monedas.

<h3>Cuántas monedas hay en cada caja?</h3>

De acuerdo con el enunciado, existen dos cajas y una caja tiene el doble de monedas que la otra. Luego, se traslada 7 monedas a la caja de menor cantidad y se obtiene igual cantidad en ambas cajas. En consecuencia, la situación queda representada por la siguiente ecuación lineal:

2 · x - 7 = x + 7

(2 · x - x) - 7 = 7

2 · x - x = 7 + 7

(2 - 1) · x = 14

x = 14

La caja de menor cantidad tiene 14 monedas y la cantidad de monedas en la otra caja es:

y = 2 · x

y = 2 · 14

y = 28

La caja con mayor cantidad tiene 28 monedas.

Para aprender más sobre ecuaciones lineales: brainly.com/question/28371342

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(b) The confidence level is 68%.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

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The general form of a (1 - α)% confidence interval is:

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SD = standard deviation

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Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

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Compute the confidence interval as follows:

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The percentage of z-distribution between -1 and 1 is, 68%.

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The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

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For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

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*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

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