All categories

3 years ago

I need the answers for this assignment pls help :(

Mathematics

1 answer:

Kruka [31]3 years ago

3 0

Answer:

115

Step-by-step explanation:

You might be interested in

Tell me some irrational numbers

lana [24]

Answer:

square root of 2, pi, square root of 5, square root of 10, square root of 3, square root of 15

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Thanks so much for the help!!

saveliy_v [14]

Answer: Domain: x=-2. Range: -2<=y<1

Step-by-step explanation:

Domain: The line is only on one x-point and that is x=-2.

Range: The line is ranging form -2 to 1. However, y=1 is not included on the line but y=-2 is. Hence, the range is notated as -2<=y<1.

3 0

1 year ago

What is 2/3 plus 11/12 equal in a fraction

dlinn [17]

19/12

welcome :):):):)

6 0

3 years ago

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

3 years ago

Evaluate. 43 – 4:25 ο ο ο ο

meriva

This is an improper. Perhaps you can fix it, so that I can assist you with it? I apologise.

4 0

3 years ago