Answer:
Abput 12000 items were produced yes and i think it is nice that yk like yes
Step-by-step explanation:
Given:
![36q=18](https://tex.z-dn.net/?f=36q%3D18)
Required:
To solve the given equation.
Explanation:
Consider
![36q=18](https://tex.z-dn.net/?f=36q%3D18)
Divide 36 on both side, we get
![\begin{gathered} \frac{36q}{36}=\frac{18}{36} \\ \\ q=\frac{18}{36} \\ \\ q=\frac{1}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B36q%7D%7B36%7D%3D%5Cfrac%7B18%7D%7B36%7D%20%5C%5C%20%20%5C%5C%20q%3D%5Cfrac%7B18%7D%7B36%7D%20%5C%5C%20%20%5C%5C%20q%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Final Answer:
Answer:
18, 54
Step-by-step explanation:
6/2 = 3
2/(2/3) = 3
Each term is the previous term multiplied by 3.
6 * 3 = 18
18 * 3 = 54
Answer: 18, 54
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)
0.84: you would do .82/.98 (the probability of orchids and roses divided by probability of roses)