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dimaraw [331]
3 years ago
5

La mama de maximo le pidio que fuera a la tienda dijo que comprara 0.5 litros de leche, el dueño de la tienda le dijo que solo t

enia leche en envases de ½ litros ¿Es mas, menos o igual a la capacidad que le pidió su mama?
Mathematics
1 answer:
Agata [3.3K]3 years ago
8 0

Answer:

1/2 litros es igual a la capacidad que le pidió su mamá.

Step-by-step explanation:

Para encontrar la respuesta debes escribir 0.5 como fracción. Para hacerlo, el numerador es el número original sin el punto decimal y para definir el denominador debes determinar cuántos digitos decimales tienes. En este caso, el número tiene un dígito decimal, por lo cual el denominador es 10. De acuerdo a esto:

0.5 se escribiría como \frac{5}{10} y esta fracción se puede simplificar porque los dos números son divisibles entre 5:

5/5= 1

10/5= 2

Por esto, 0.5= \frac{1}{2}

Esto significa que 1/2 litros es igual a la capacidad que le pidió su mama.

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Answer:

2.618-2.98\frac{0.078}{\sqrt{15}}=2.56    

2.618+2.98\frac{0.078}{\sqrt{15}}=2.68    

So on this case the 99% confidence interval would be given by (2.56;2.68)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=2.618

The sample deviation calculated s=0.078

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that t_{\alpha/2}=2.98

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2.618-2.98\frac{0.078}{\sqrt{15}}=2.56    

2.618+2.98\frac{0.078}{\sqrt{15}}=2.68    

So on this case the 99% confidence interval would be given by (2.56;2.68)    

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