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2 years ago

Interpret the following Sigma Notation n=152n

1 answer:

8 0

Okay so basically the answer would be
-8305
And in this form it would be
-151 + -302 + -435 -604 -755 + -906 + 1057 +-1208 + -1359 + 1510

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Amount of money you earn per hour if you make d dollars in 15 hours??

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The answer to the question is d/15

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3 years ago

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3 years ago

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Hunter-Best [27]

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2 years ago

Write the absolute value equation given the transformations

Evgesh-ka [11]

Answer:

y = - 0.5 | x - 5 | + 2 

Step-by-step explanation:

g(x) = |x|

$1^{st}$ step - shifts right 5 units: g(x + 5) = | x - 5 |

$2^{nd}$ step - shifts up 2 units: g(x + 5) + 2 = | x - 5 | + 2

$3^{rd}$ step - reflected: - g(x + 5) + 2 = - | x - 5 | + 2

$4^{th}$ step - stretched by 0.5: y = - 0.5 | x - 5 | + 2

y = - 0.5 | x - 5 | + 2

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3 years ago

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