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Juliette [100K]
2 years ago
15

Expresa algebraicamente como varia el costo en cada uno de los hoteles al aumentar la estancia Confecciona, para cada uno de los

hoteles la tabla de valores relativa a los diez primeros días de estancia en el hotel hotel la laguna Precio por persona / día MP $70 primer día gratis Estancia mínima cinco días Hotel del Mar Precio por persona / día MP $60 Estancia mínima de dos días c Representa gráficamente las funciones obtenidas en el aparto anterior. d ¿ Cuanto pagará una persona al cabo de cinco días en cada uno de los hoteles? e ¿ Al cabo de cuantos días resulta mas económico el Hotel del mar ? f El costo de la estancia de una persona en el hotel es de $480 ¿En que el hotel se ha alojado? ¿Cuántos días ? Ayudenme porfa
Mathematics
1 answer:
Leokris [45]2 years ago
3 0

Answer:

I do not know how to read Spanish sorry........

Step-by-step explanation:

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Round 2.9856 to the nearest 100th.
Roman55 [17]

Answer:

2.98

Step-by-step explanation:

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3 years ago
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(a−2b)3 when a=−4 and b=−1/2
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Answer:

-9

Step-by-step explanation:

(-4-2*-1/2)3

(-4+1)3

-3*3

-9

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B-3±6-2b i need it know pls help me
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Answer : b=3

Hope this helps have a great day! :)
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3 years ago
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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=5 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)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

6 0
2 years ago
Craig went bowling with 25$ to spend. He rented shoes for $5.25 and paid $4.00 for each game. What was the greatest number of ga
lidiya [134]
25-5.25 gives you the amount of money Craig can spend for games. 25-5.25=19.75. If Craig can spend 19.75, he can play only 4 whole games.
5 0
3 years ago
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