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shusha [124]
2 years ago
14

HELPPPPP MEEE PLEASE

Mathematics
1 answer:
Brut [27]2 years ago
4 0
The answer is C I think
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Jared writes a multiplication expression with eight rational factors. Half of the factors are positive and half are negative. Is
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negative they are 4 negative factors

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Por favor denme una ecuación a esta pregunta:
Katena32 [7]

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x + 13 = 25

Step-by-step explanation:

Se puede representar "un número" con x.

x

"Adiciona" significa +.

x + 13

"El resultado" significa =.

x + 13 = 25

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Ms. Abrams receives a 4% commission plus a salary of $275.00 per week. If her total sales for one week are $2,880.00, how much d
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Which graph represents the function f(x) = 1/3 |x|
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Read 2 more answers
You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample
Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=160)  

And the distribution for \bar X is:

\bar X \sim N(\mu, \frac{160}{\sqrt{n}})  

We know that the margin of error for a confidence interval is given by:  

Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)  

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025  

Using the normal standard table, excel or a calculator we see that:  

z_{\alpha/2}=\pm 1.96  

If we solve for n from formula (1) we got:  

\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}  

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And we have everything to replace into the formula:  

n=(\frac{1.96(160)}{78.4})^2 =16  

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

n=(\frac{z_{\alpha/2} \sigma}{Me})^2  

And now we can replace the new value of Me and see what we got, like this:

n=(\frac{1.96*160}{60})^2 =27.32

And if we round up the answer we see that the value of n to ensure the margin of error required Me=\pm 60 $ is n=28.    

5 0
2 years ago
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