I think it is 4 cause I love u
Wow that’s a lot of money for getting a B,if I don’t get 95 percent A in any of my classes I won’t be able to walk after my mom finds out.
Answer:
If it is a six-sided die then it has the probability of 1/6 or .16
Step-by-step explanation:
Answer:
The average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802
Step-by-step explanation:
We are given that
Standard deviation,
ounces
We have to find the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.
![P(x\geq 10)=0.99](https://tex.z-dn.net/?f=P%28x%5Cgeq%2010%29%3D0.99)
Assume the bag weight distribution is bell-shaped
Therefore,
![P(\frac{x-\mu}{\sigma}\geq 10)=0.99](https://tex.z-dn.net/?f=P%28%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cgeq%2010%29%3D0.99)
We know that
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
Using the value of z
Now,
![\frac{10-\mu}{0.2}=0.99](https://tex.z-dn.net/?f=%5Cfrac%7B10-%5Cmu%7D%7B0.2%7D%3D0.99)
![10-\mu=0.99\times 0.2](https://tex.z-dn.net/?f=10-%5Cmu%3D0.99%5Ctimes%200.2)
![\mu=10-0.99\times 0.2](https://tex.z-dn.net/?f=%5Cmu%3D10-0.99%5Ctimes%200.2)
![\mu=9.802](https://tex.z-dn.net/?f=%5Cmu%3D9.802)
Hence, the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802
Answer:
5, 7, 9, 11
Step-by-step explanation:
Solve the inequality
w - 4 < 8 ( add 4 to both sides )
w < 12
w ∈ { 5, 7, 9, 11 }