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Bond [772]
3 years ago
13

hypothesis that there is a relationship between parents’ and children’s party identification. Would we be correct in inferring t

hat such a relationship also exists in the population? Explain your answer. What is the probability that any relationship we found is due to pure chance?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
3 0

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.

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Simplify the expression 6 + 3(2x - 4(3x - 2)].​
I am Lyosha [343]

Answer:

-30x + 30

Step-by-step explanation:

6 + 3 × [2x - 12x + 8]

6 + (-30x) + 24

-30x + 24

hope its correct if not sorry

8 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
(1 point) In an experiment, a ball is drawn from an urn containing 8 orange balls and 12 yellow balls. If the ball is orange, th
klemol [59]

Answer:

a) 8

b) 12

Step-by-step explanation:

With orange: 2×2×2 = 8

With yellow: 2×2 = 4

Total = 8+4 = 12

3 0
4 years ago
SEE PICTURE FOR QUESTION​
castortr0y [4]
I think the answer is B
7 0
3 years ago
How do i solve this?
bixtya [17]

Answer:

5.x^8.y^3

Step-by-step explanation:

15.x^10.y^5.z^2 ÷ 3.x^2.y^2.z^2

= 15/3. x^(10-2). y^(5-2). z^(2-2)

= 5.x^8.y^3.z^0

= 5.x^8.y^3

7 0
3 years ago
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