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2 years ago

12

Under his cell phone plan, Elijah pays a flat cost of $64.50 per month and $5 per gigabyte. He wants to keep his bill at $87.50

per month

1 answer:

photoshop1234 [79]2 years ago

6 0

Gotta go stingy by adding 64.5 us 87.5

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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

I need help on thi and the first person who answer correctly gets a BRANLIST and show work

const2013 [10]

Answer:

w<6 THE SMALL LINE UNDER THE LESS THAN SIGN IS STILL THERE I JUST CANNOT ADD IT

Step-by-step explanation:

first solve the equation to the right.

-3(2w+1) = -6w-3

now solve the whole equation

-33-w< -6w-3

-w+6w < -3+33

5w < 30

divide the 5 from both sides to simplify

5/5w < 30/5

w < 6

THE ANSWER IS w < 6

3 0

2 years ago

Plz help I need help

Zarrin [17]

Answer:

0.65 is the correct answer

3 0

2 years ago

Read 2 more answers

I need help what is 30% of 31

Ainat [17]

Answer:

9.3

Step-by-step explanation:

Is means equals and of means multiply

W = 30% * 31

Change to decimal form

W = .30 *31

W = 9.3

4 0

2 years ago

Read 2 more answers

I’ll mark brainlest:( i need help!

enyata [817]

Answer: Choice A) $3 \sqrt{26}$

==================================================

Work Shown:

$\text{The two points are } (x_1,y_1) = (6,7) \text{ and } (x_2,y_2) = (3,-8)\\\\d = \text{Distance between } (x_1,y_1) \text{ and } (x_2,y_2)\\\\d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} = \text{Distance Formula}\\\\d = \sqrt{(6-3)^2+(7-(-8))^2}\\\\d = \sqrt{(6-3)^2+(7+8)^2}\\\\$

$d = \sqrt{(3)^2+(15)^2}\\\\d = \sqrt{9+225}\\\\d = \sqrt{234}\\\\d = \sqrt{9*26}\\\\d = \sqrt{9}*\sqrt{26}\\\\d = 3\sqrt{26}\\\\$

When going from 234 to 9*26, the idea here is to factor where one factor is the largest perfect square possible. That way we can use the rule $\sqrt{x*y} = \sqrt{x}*\sqrt{y}$ to break up the root and simplify.

7 0

3 years ago