Answer:
The sample of sizes 2 and their mean are given below.
Step-by-step explanation:
The population consist of 5 values, S = {1, 3, 4, 4, 6}.
The number of samples of size 2 (without replacement) that can be formed from these 5 values is:

Th formula to compute the mean is:

List the 10 samples and their mean as follows:
<u>Sample</u> <u>Mean</u>
(1, 3) ![\bar x=\frac{1}{2}[1+3]=\frac{4}{2}=2.0](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B1%2B3%5D%3D%5Cfrac%7B4%7D%7B2%7D%3D2.0)
(1, 4) ![\bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B1%2B4%5D%3D%5Cfrac%7B5%7D%7B2%7D%3D2.5)
(1, 4) ![\bar x=\frac{1}{2}[1+4]=\frac{5}{2}=2.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B1%2B4%5D%3D%5Cfrac%7B5%7D%7B2%7D%3D2.5)
(1, 6) ![\bar x=\frac{1}{2}[1+6]=\frac{7}{2}=3.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B1%2B6%5D%3D%5Cfrac%7B7%7D%7B2%7D%3D3.5)
(3, 4) ![\bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B3%2B4%5D%3D%5Cfrac%7B7%7D%7B2%7D%3D3.5)
(3, 4) ![\bar x=\frac{1}{2}[3+4]=\frac{7}{2}=3.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B3%2B4%5D%3D%5Cfrac%7B7%7D%7B2%7D%3D3.5)
(3, 6) ![\bar x=\frac{1}{2}[3+6]=\frac{9}{2}=4.5](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B3%2B6%5D%3D%5Cfrac%7B9%7D%7B2%7D%3D4.5)
(4, 4) ![\bar x=\frac{1}{2}[4+4]=\frac{8}{2}=4.0](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B4%2B4%5D%3D%5Cfrac%7B8%7D%7B2%7D%3D4.0)
(4, 6) ![\bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B4%2B6%5D%3D%5Cfrac%7B10%7D%7B2%7D%3D5.0)
(4, 6) ![\bar x=\frac{1}{2}[4+6]=\frac{10}{2}=5.0](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D%5B4%2B6%5D%3D%5Cfrac%7B10%7D%7B2%7D%3D5.0)
The initial population is
P₀ = 94 million in 1993
The growth formula is

where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.
Because P(5) = 99 million (in 1999),

In the year 2005, t = 12 years, and

Answer: 106 million (nearest million)
Answer : y>0
f(x) = 9*2^x
f(x) is an exponential function

When we plug in positive value for x , the value of y is positive
When we plug in negative value for x , the value y is also positive
So for any value of x, the y value is positive always.
Range is the set of y values for which the function is defined
y values are positive , so range is y >0
Answer:
0.94
Step-by-step explanation:
The question after this basically is:
<em>"If the applicant passes the "aptitude test for managers", what is the probability that the applicant will succeed in the management position?"</em>
<em />
So,
P(successful if hired) = 60% = 0.6 [let it be P(x)]
P(success at passing the test) = 85% = 0.85 [let it be P(y)]
P(successful and pass the test) = P(x) + P(y) -[P(x)*P(y)]
So,
P(successful and pass the test) = 0.6 + 0.85 - (0.6*0.85) = 0.94 (94%)