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Nat2105 [25]
3 years ago
7

HELP ASAPPPP !!!!!!!!!!!......

Mathematics
1 answer:
irina [24]3 years ago
8 0

Answer:

just use a calculator

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A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3. We may conclude
valentina_108 [34]

Answer:

A. there is a 99% probability that μ is between 3 and 9.

Step-by-step explanation:

From a random sample, we build a confidence interval, with a confidence level of x%.

The interpretation is that we are x% sure that the interval contains the true mean of the population.

In this problem:

99% confidence interval.

6 ± 3.

So between 6-3 = 3 and 6 + 3 = 9.

So we are 99% sure that the true population mean is between 3 and 9.

So the correct answer is:

A. there is a 99% probability that μ is between 3 and 9.

4 0
3 years ago
Read 2 more answers
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
2 years ago
X squared +20 equals 2X
Ivahew [28]

x={1+i√19 , 1-i√19}

solve this by using quadratic formula

5 0
3 years ago
Is x= -2 a solution to 3x-6-12?
Ann [662]

Answer:

Actually x would be 6

4 0
2 years ago
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What was the name of the small group of Jews that was especially opposed to Hellenism and Roman rule? A. Zealots B. Martyrs C. G
Klio2033 [76]
The answer is A. Zealots

4 0
3 years ago
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