A sample of water is taken and kept in a beaker in a freezer at a constant temperature of 0°C. If the system is at dynamic equilibrium, which of these statements is true?
First you need to multiply all the numbers to get 11548845 and then for example you need to do this (77*1515)+(77*99)+(99*1515) = 274263 to get and then you need you need to add 77,1515,99 to get 1691 and then you add 11548845, 274263, and 1691 to get 11824799 and that is the total number of possible outcomes of meals that you can get.
<em> hope this helps you and the the reason I explained is so you dont have to ask this type of question again.</em>
Answer:
![P(-2 \leq \bar X -\mu \leq 2)](https://tex.z-dn.net/?f=%20P%28-2%20%5Cleq%20%5Cbar%20X%20-%5Cmu%20%5Cleq%202%29)
If we divide both sides by
we got:
![P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})](https://tex.z-dn.net/?f=%20P%28%5Cfrac%7B-2%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B9%7D%7D%7D%5Cleq%20Z%20%5Cleq%20%5Cfrac%7B2%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B9%7D%7D%7D%29)
And we can use the normal distribution table or excel to find the probabilites and we got:
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=4](https://tex.z-dn.net/?f=%5Csigma%3D4)
We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean
is given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
And we want to find this probability:
![P(-2 \leq \bar X -\mu \leq 2)](https://tex.z-dn.net/?f=%20P%28-2%20%5Cleq%20%5Cbar%20X%20-%5Cmu%20%5Cleq%202%29)
If we divide both sides by
we got:
![P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})](https://tex.z-dn.net/?f=%20P%28%5Cfrac%7B-2%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B9%7D%7D%7D%5Cleq%20Z%20%5Cleq%20%5Cfrac%7B2%7D%7B%5Cfrac%7B4%7D%7B%5Csqrt%7B9%7D%7D%7D%29)
And we can use the normal distribution table or excel to find the probabilites and we got:
![P(-1.5 \leq Z \leq 1.5)= P(Z](https://tex.z-dn.net/?f=%20P%28-1.5%20%5Cleq%20Z%20%5Cleq%201.5%29%3D%20P%28Z%3C1.5%29%20-P%28Z%3C-1.5%29%20%3D%200.933-0.0668%3D0.866)
<em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>4</em><em> </em><em>ft</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Bro..
write down the question and take a pic and upload
it's not comprehensive enough like this