All categories

2 years ago

What is the distance between the points (5, -1) and (5, 3)

Mathematics

1 answer:

Diano4ka-milaya [45]2 years ago

7 0

By using distance formula (which is attached below), you can solve this problem:

√(x2-x1)^2+(y2-y1)^2

(5, -1) and (5, 3)

√(5-5)^2+(3-(-1))^2

√0^2+4^2

√0+16

√16

The distance between the two points is 4.

You might be interested in

And the last question is:

Leya [2.2K]

It should be " b " That should be right but u may want to wait for others to respond to this to make sure

6 0

3 years ago

Read 2 more answers

A country's population in 1994 was 184 million. in 1997 it was a 190 million. estimate the population in 2015 using the exponent

cestrela7 [59]

Answer:

107.82

Step-by-step explanation:

5 0

3 years ago

All of the following statements about learning are true EXCEPT __

Rasek [7]

Craming last minute can lead to unecessary confusion however most examinars dont allow asking questions during exam

So if asking questions for clarification is allowed. A) is the odd one out

And these options are rather for test taking than learning

3 0

3 years ago

Bicyclist mike hall rode around the world in 91 days.what fraction is the simplest form represents the part of a year in weeks h

victus00 [196]

52 weeks
365 days
7days a week
91 days divided by 7 equals
13 weeks

8 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