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lozanna [386]
2 years ago
14

Researchers found that a person in a particular country spent an average of 5.5 hours a day watching TV in 2010. Assume the popu

lation standard deviation is 1.8 hours per day. A sample of 39 people averaged 6.3 hours of television viewing per day. Does this result support the findings of the​ study?
Mathematics
1 answer:
zmey [24]2 years ago
5 0

Answer:

Yes

Step-by-step explanation:

Because 6.3hours is still in the range 5.5+/-1.8 hours

Therefore the result support the findings of the study.

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Which of the following is not a Pythagorean triple?. . A.. 28, 45, 53. B.. 16, 63, 65. C.. 13, 84, 85. D.. 11, 61, 62.
Anna11 [10]
A set of numbers is said to be Pythagorean triple if the sum of the squares of the lesser numbers equal the square of the remaining number,

A. 28² + 45² = 2809    ;  53² = 2809     ; EQUAL
B. 16² + 63² = 4225    ;  65²= 4225      ; EQUAL
C. 13² + 84² = 7225    ;  85² = 7225     ; EQUAL
D. 11² + 61² = 3842    ;   62² = 3844     ; NOT EQUAL

The answer is letter D.
6 0
3 years ago
Read 2 more answers
If I had 10 cookies and promised my friend to give them 2 how many cookies will i have
Ksivusya [100]

Answer:

8 cookies

Step-by-step explanation:

edit one:the answer was 10 cookies, so please go give the person who said 10 cookies first some thanks/brainliest.

edit two:brainly.com/app/profile/31525083 is the user who got it first. Check comment if you are gonna say they didn't.

6 0
3 years ago
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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
2 years ago
What is 4 yards and 2 feet multiplied by 3
lesantik [10]

Answer:

12 yds and 6 feet, aka 42 feet.

Step-by-step explanation:

You multiply everything by 3. Then you convert yards to feet, then add everything together.

5 0
3 years ago
Which table represents a proportional relationships
Soloha48 [4]
I think The last one in right

y |2| 6 |10 |14
x |6 |18 |30 |42
6 0
3 years ago
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