Answer:
The answer is 1/6
Step-by-step explanation:
So Think of a half and then divide it into 3 parts for a half it will be 1/3 but for a meter it will be 1/6 because there is 2 half's in 1 whole
Hope this helps!! :D
Please mark brainliest if this helped you!
Answer:
E(M) = 23/3
Var(M) = 53/9.
Step-by-step explanation:
There are three possible values for M: 4, 6, and 10.
![\begin{array}{|l|l|}\cline{1-2}\\[-1em]\text{Larger Number} & \text{Smaller Number}\\ \cline{1-2}\\[-1em]4 & 2 \\\cline{1-2}\\[-1em]6 & 2, ~4\\\cline{1-2}\\[-1em] 10 & 2, ~4, ~6\\\cline{1-2} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cl%7Cl%7C%7D%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%5Ctext%7BLarger%20Number%7D%20%26%20%5Ctext%7BSmaller%20Number%7D%5C%5C%20%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D4%20%26%202%20%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D6%20%26%202%2C%20~4%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%2010%20%26%202%2C%20~4%2C%20~6%5C%5C%5Ccline%7B1-2%7D%20%5Cend%7Barray%7D)
.
That's six unique combinations in total. If the balls are drawn randomly, the probability for getting each combination shall be equal. That is:
![\begin{array}{|c|c|}\cline{1-2}\\[-1em]m & P(\text{M} = m)\\\cline{1-2}\\[-1em] 2 & 0\\\cline{1-2}\\[-1em] 4 & 1/6 \\\cline{1-2}\\[-1em] 6 & 2/6\\\cline{1-2}\\[-1em] 10 & 3/6\\\cline{1-2}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7C%7D%5Ccline%7B1-2%7D%5C%5C%5B-1em%5Dm%20%26%20P%28%5Ctext%7BM%7D%20%3D%20m%29%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%202%20%26%200%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%204%20%26%201%2F6%20%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%206%20%26%202%2F6%5C%5C%5Ccline%7B1-2%7D%5C%5C%5B-1em%5D%2010%20%26%203%2F6%5C%5C%5Ccline%7B1-2%7D%5Cend%7Barray%7D)
.
Consider the formula for the expected value of a discrete random variable:
![\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 \times \frac{1}{6} + 6\times \frac{2}{6} + 10 \times \frac{3}{6}\\ &= \frac{23}{3}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20E%28%7B%5Crm%20M%7D%29%26%20%3D%20%5Csum%7B%5Bm%20%5Ccdot%20P%28%5Cmathrm%7BM%7D%20%3D%20m%29%5D%7D%5C%5C%20%26%3D%204%20%5Ctimes%20%5Cfrac%7B1%7D%7B6%7D%20%2B%206%5Ctimes%20%5Cfrac%7B2%7D%7B6%7D%20%2B%2010%20%5Ctimes%20%5Cfrac%7B3%7D%7B6%7D%5C%5C%20%26%3D%20%5Cfrac%7B23%7D%7B3%7D%5Cend%7Baligned%7D)
.
Formula for the variance of a discrete random variable (note that this formula can take many other forms):
![\begin{aligned}Var(\mathrm{M}) &= \sum{[m^{2}\cdot P(\mathrm{M}= m)]} - \mu^{2}\\&= 4^{2}\times \frac{1}{6} + 6^{2} \times \frac{2}{6} + 10^{2} \times \frac{3}{6} - \left(\frac{23}{3}\right)^{2}\\&= \frac{53}{9}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DVar%28%5Cmathrm%7BM%7D%29%20%26%3D%20%5Csum%7B%5Bm%5E%7B2%7D%5Ccdot%20P%28%5Cmathrm%7BM%7D%3D%20m%29%5D%7D%20-%20%5Cmu%5E%7B2%7D%5C%5C%26%3D%204%5E%7B2%7D%5Ctimes%20%5Cfrac%7B1%7D%7B6%7D%20%2B%206%5E%7B2%7D%20%5Ctimes%20%5Cfrac%7B2%7D%7B6%7D%20%2B%2010%5E%7B2%7D%20%5Ctimes%20%5Cfrac%7B3%7D%7B6%7D%20-%20%5Cleft%28%5Cfrac%7B23%7D%7B3%7D%5Cright%29%5E%7B2%7D%5C%5C%26%3D%20%5Cfrac%7B53%7D%7B9%7D%5Cend%7Baligned%7D)
.
Answer:
2 . True
Step-by-step explanation:
2. The two events are mutually exclusive events. This means that the probabilities on coins 1 and coins 2 are independent on each other. In other words, the probability of the first coin does not affect the probability of the second coin outcome.
Its B or D!
I hope this helps!
take a lucky guess!
Answer:
y=7x+14
Step-by-step explanation:
-y=14-7x
y=-14+7x
y=7x-14
