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3 years ago

Ok so I legit just want to make internet friends which is stupid but here is info (sorta)

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

5 0

Answer:

Look, i get it, but this is brainly, no offense. but uh

Step-by-step explanation:

so basically this is brainly please do not use it thiss is badddddddddddddddddddddddddd ok? 1. because this is for asking SCHOOL QUESTIONS 2. S C H O O L Q U E S T I O N S

ok bye , sorry :)

-random person

uranmaximum [27]3 years ago

3 0

Answer:

So you want to make internet friends? You might want to consider the fact that Brainly isn't the best place to make friends. It's more of a study website. Social media is better for making friends. And I don't know why you are doing this.

Step-by-step explanation:

1. Get social media.

2. Friend people there.

3. Now you have internet friends.

4. Nominate me Brainliest answer.

AND THERE YOU GO! Hope this helps.

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Why do certificates of deposit tend to offer better interest rates than money market accounts?

victus00 [196]

Answer:

Step-by-step explanation:

Certificates of deposit are only available to the bank's most proven and valuable customers, so the bank can afford to give a greater interest rate on them.

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3 years ago

An amount of $25,000 is borrowed for 9 years at 9% interest, compounded annually. If the loan is paid in full at the end of that

Aleks04 [339]

Answer:

$36,566

Step-by-step explanation:

"The calculator provided" will tell you the monthly payment is $338.57, so the 108 payments made over the period of 9 years will total ...

108 × $338.57 = $36,565.56

Rounded to the nearest dollar, the amount is $36,566.

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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:

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.

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.