Answer:
No
It could be purely due to chance.
Step-by-step explanation:
A population is defined as the whole group which has the same characteristics. For example a population of the college belongs to the same college . But a sample may be an element of a population.
So it is not necessary for a population to have the same characteristics as the sample.
But it is essential for the sample to have at least one same characteristics as the population.
So we would not be correct in inferring that such a relationship also exists in the population.
It is a hypothesis which can be true or false due to certain conditions or limitations as the case maybe.
For example in a population of smokers some may be in the habit of taking cocaine. But a sample of cocaine users does not mean the whole population uses it.
It could be purely due to chance if we find out that there is a relationship between parents’ and children’s party identification in the population.
<span>2/15 if drawn without replacement.
1/9 if drawn with replacement.
Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are:
1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5
Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15
If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are:
1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6
And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
Answer:
Step-by-step explanation:
Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve
mu =74 and sigma =8
A grade starts from 100-8 = 92nd percentile
Z score for 92nd percentile = 1.405
X score = 74+8(1.405) = 85.24
--------------------
B cut off is to next 16%
Hence C would start for scores below 100-(8+16) = 76%
76th percentile = 0.705*8+74 =79.64
The firts thig we are going to do is create tow triangles using the angles of elevation of Paul and Jose. Since the problem is not giving us their height we'll assume that the horizontal line of sight of both of them coincide with the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as

.
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:



Now we can use the law of sines to find the distance

between Paul and Jose:



Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:

We can conclude that Jose is 33.9m from the base of the tree.