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Elenna [48]
3 years ago
14

In a random sample of 80 people, 52 consider themselves as baseball fans. Compute a 92% confidence interval for the true proport

ion of people consider themselves as baseball fans and fill in the blanks appropriately. We are 92% confident that the true proportion of people consider themselves as baseball fans is between__________ and___________ .
Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

k0020

Step-by-step explanation:

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Find the difference of(ab+8a+1)-(-6ab+4)
vagabundo [1.1K]
Hey there!

When finding the difference of expressions like these, there's a few things you should keep in mind. 

1. When finding the difference of two polynomials, multiply the polynomial that implied to be "negative" by –1 to expand the problem and make it easier to solve. 

2. You combine terms based on their unknown terms, like x, y, x², ab, etc. You cannot combine the terms x and x², but you can combine the terms 6b and 8b by adding them. 

3. You can rearrange your terms however you see fit. If a polynomial is implied as "positive", you can multiply that entire polynomial by +1 to get rid of the parentheses. 

With all that in mind, you can go ahead and solve for your difference, like so:

(ab+8a+1)-(-6ab+4)

(ab+8a+1) (-1(-6ab+4))

(ab+8a+1) + (6ab-4)

ab+6ab+8a+1-4

7ab+8a-3

7ab+8a-3 will be your difference. 

Hope this helped you out! :-)
8 0
3 years ago
Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are
Svetlanka [38]

Answer:

(a) Number of sets B given that

  • A⊆B⊆C: 2¹⁰.  (That is: A is a subset of B, B is a subset of C. B might be equal to C)
  • A⊂B⊂C: 2¹⁰ - 2.  (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let x_1, x_2, \cdots, x_{20} denote the 20 elements of set X.

Let x_1, x_2, \cdots, x_{10} denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain x_1, x_2, \cdots, x_{10}.

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from x_1, x_2, \cdots, x_{20}.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 2^{10} -2 = 1024 - 2 = 1022 possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's 2^{20} = 1048576 possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 2^{20} - 2^{10} possibilities left.

5 0
3 years ago
Sam owes the bank a little less then $40 what could be his balance
bija089 [108]
D 39.25 which isn't far from 40

8 0
3 years ago
Calculate produce of 1/6 and 12/13
Sergeu [11.5K]

Answer:

1/6×12/13

2/13

Step-by-step explanation:

thats the answer

8 0
4 years ago
PLEASE HELP!! DUE SOON!!
ryzh [129]

Answer:

Do you still need help??

Step-by-step explanation:

5 0
3 years ago
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