Answer with Step-by-step explanation:
The number of marbles are as under
3 red , 3 green , 1 Lavender total = 7
Now to select five marbles from a total of 7 marbles such that at least 2 marbles are included are the sum of the following cases:
1) We select 2 exactly 2 red marbles from 3 reds and the remaining 3 marbles are selected from 4 of other colours
Thus ![n_{1}=\binom{3}{2}\binom{4}{3}=\frac{3!}{2!}\cdot \frac{4!}{3!}=12](https://tex.z-dn.net/?f=n_%7B1%7D%3D%5Cbinom%7B3%7D%7B2%7D%5Cbinom%7B4%7D%7B3%7D%3D%5Cfrac%7B3%21%7D%7B2%21%7D%5Ccdot%20%5Cfrac%7B4%21%7D%7B3%21%7D%3D12)
2)We select all the 3 red marbles and the remaining 2 are selected from the remaining 4 marbles
![n_2=\binom{4}{2}=\frac{4!}{2!\cdot 2!}=6](https://tex.z-dn.net/?f=n_2%3D%5Cbinom%7B4%7D%7B2%7D%3D%5Cfrac%7B4%21%7D%7B2%21%5Ccdot%202%21%7D%3D6)
Thus the total number of ways are ![n_1+n_2=12+6=18](https://tex.z-dn.net/?f=n_1%2Bn_2%3D12%2B6%3D18)
Answer:
A= 115.625
Step-by-step explanation:
3,800x2-200=64A
Combine all like terms.
7,400=64A
Your going to have to divide 64 on bothe sides to leave 'A' alone one one side of the equal sign.
![\frac{7,400=64A}{64}](https://tex.z-dn.net/?f=%5Cfrac%7B7%2C400%3D64A%7D%7B64%7D)
Your final answer will be <u>115.626=A</u>
It would be 42m+12 and that would be the expanded version
So that the means of the sample falls into the dangerous region then ![{\displaystyle {\overline {Y}}}-\mu_{{\displaystyle {\overline {Y}}}} > 3](https://tex.z-dn.net/?f=%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D-%5Cmu_%7B%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D%7D%20%3E%203)
Where
is normally distributed with mean
and standar desviation ![\sigma_{{\displaystyle {\overline {Y}}}} =\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D%7D%20%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Then ![n = 49](https://tex.z-dn.net/?f=n%20%3D%2049)
![\mu_{{\displaystyle {\overline {Y}}}}=\mu= 10](https://tex.z-dn.net/?f=%5Cmu_%7B%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D%7D%3D%5Cmu%3D%2010)
![\sigma_{{\displaystyle {\overline {Y}}}} = \frac{1}{\sqrt{49} }](https://tex.z-dn.net/?f=%5Csigma_%7B%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B49%7D%20%7D)
So:
![P(\frac{{\displaystyle {\overline {Y}}}-\mu}{\frac{\sigma}{\sqrt{n}}})> \frac{3}{\frac{1}{\sqrt{49}}}](https://tex.z-dn.net/?f=P%28%5Cfrac%7B%7B%5Cdisplaystyle%20%7B%5Coverline%20%7BY%7D%7D%7D-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%29%3E%20%5Cfrac%7B3%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B49%7D%7D%7D)
Finally
.
The probability that the sample mean falls in the dangerous region is approximately zero.
This makes sense, since as the sample size increases, it is less likely to deviate more than 2 standard deviations from the mu population mean.