Find the value of x.

576 x 895?with the shown work!!I know the answer is 515,520.

stiks02 [169]

Answer:

Correct, the answer is 515,520

Step-by-step explanation:

576

x895

----------

515,520

5 0

3 years ago

Evaluate x^1-x^-1+x^0 for x=2

Feliz [49]

Let p (x) = x^1 - x^-1 +x^0
p (2) = 2 -1/2 +1
=( 4 -1+2)/2
= 5/2

5 0

3 years ago

Read 2 more answers

Hey Siri found that 78% of students do their homework before 8:00 PM predict how many students out of 975 do their homework befo

cestrela7 [59]

Multiply 975 by .78 and you will get 760.50.

4 0

3 years ago

A Common Measure of Student Achievement, The National Assessment of Educational Progress (NAEP) is the only assessment that meas

vlada-n [284]

Answer:

a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

b) $270.283\leq \mu \leq 279.717$

c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

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".

$\bar X=275$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=60$ represent the population standard deviation

n=1077 represent the sample size

Part a

By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$