Type one is genetic and you are born with it. type two is acquired through an unhealthy lifestyle such as eating junk food and not exercising regularly
Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
recall that a cube has all equal sides, check the picture below.
![\bf \textit{volume of a cube}\\\\ V=x^3~~ \begin{cases} x=side's~length\\[-0.5em] \hrulefill\\ V=5.12 \end{cases}\implies 5.12=x^3 \\\\\\ \sqrt[3]{5.12}=x\implies 1.72354775\approx x](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5C%0AV%3Dx%5E3~~%0A%5Cbegin%7Bcases%7D%0Ax%3Dside%27s~length%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0AV%3D5.12%0A%5Cend%7Bcases%7D%5Cimplies%205.12%3Dx%5E3%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B5.12%7D%3Dx%5Cimplies%201.72354775%5Capprox%20x)
Answer:
12%
Step-by-step explanation:
We have to find percentage of employee both manager and has MBA degree.
Percentage of employee both manager and has MBA degree=P(MBA and manager)*100
We are given that
P(MBA)=0.25
P(Manager)=0.20
P(MBA/ manager)=0.60
P(MBA and manager)=?
P(MBA/ manager)=P(MBA and manager)/P(Manager)
P(MBA/ manager)*P(Manager)=P(MBA and manager)
P(MBA and manager)=0.6*0.2
P(MBA and manager)=0.12
Percentage of employee both manager and has MBA degree=0.12*100
Percentage of employee both manager and has MBA degree=12%.
Thus, the percentage of the employees is both manager and has MBA degree is 12%
3 is 25% of 12, so if you apply the same proportion to 180, a quarter of 180 is 45!!
