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kogti [31]
2 years ago
13

Eight identical slips of paper, each containing one number from one to eight, inclusive, are mixed up inside a bag. Subset A of

the sample space represents the complement of the event in which the number 6 is drawn from the bag.
Which shows subset A?

A = {6}
A = {1, 3, 5, 7}
A = {1, 2, 3, 4, 5, 7, 8}
A = {1, 2, 3, 4, 5, 6, 7, 8}
Mathematics
2 answers:
DochEvi [55]2 years ago
7 0

Answer:  The correct option is (C) A = {1, 2, 3, 4, 5, 7, 8}.

Step-by-step explanation: Given that eight identical slips of paper, each containing one number from one to eight, inclusive, are mixed up inside a bag.

The sample space, S will be

S = {1, 2, 3, 4, 5, 6, 7, 8}.

Subset A of the sample space represents the complement of the event in which the number 6 is drawn from the bag.

Let, 'B' represents the event in which the number 6 is drawn from the bag.

So, B = {6}.

Since 'A' is the complement of 'B', so the set 'A' will contain those element of the sample space 'S' that are not in the set 'B'.

Therefore,

A = B' = {1, 2, 3, 4, 5, 7, 8}

⇒ A = {1, 2, 3, 4, 5, 7, 8}.

Thus, (C) is the correct option.

d1i1m1o1n [39]2 years ago
6 0

I will assume that it is a single slip then subset A = {1,2,3,4,5,7,8} (i.e. "1 is drawn", "2 is drawn" etc.)

which is the complement of A^c = {6} ("6 is drawn")


The answer is A = {1,2,3,4,5,7,8}

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Kathy has $50,000 to invest today and would like to determine whether it is realistic for her to achieve her goal of buying a ho
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Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

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\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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