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3 years ago

PLEASE HELP ME!! BRAINLIEST!!

Mathematics

1 answer:

-Dominant- [34]3 years ago

8 0

Answer:

200

Step-by-step explanation:

Given the first term 39 and the last term 11 then the sum of the 8 layers can be calculated as

$S_{n}$ = $\frac{n}{2}$ ( f + l) ← f is the bottom layer anf f the top layer, thus

$S_{8}$ = $\frac{8}{2}$ (39 + 11) = 4 × 50 = 200

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ticket sales for a local concert totaled $101,244 yesterday. After the ticket window closed today, the cashiers counted 968 tick

STALIN [3.7K]

The cost of one concert ticket is approximately $148.59.

The tickets sales for the local sales is totalled $101, 244 yesterday.

ticket sales yesterday = $101244

He sold 968 tickets with a two-days total of $143, 836 (both yesterday and today)

The cost of one ticket can be calculated below:

cost of one ticket = 143,836 / 968

cost of one ticket = 148.590909091

cost of one ticket ≈ $ 148.59

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2 years ago

The temperature at noon at a ski resort is 3 F . The temperature drops 8 F by midnight what is the temperature at midnight

DIA [1.3K]

Answer:

-5

Step-by-step explanation:

3 minus 8 equals -5

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4 0

3 years ago

Read 2 more answers

Solve. 4x + 10= 2x – 14

hammer [34]

Answer:

x = -12

Step-by-step explanation:

Step 1: Subtract 2x from both sides

2x + 10 = -14

Step 2: Subtract 10 from both sides

2x = -24

Step 3: Divide both sides by 2

x = -12

8 0

3 years ago

Read 2 more answers

A student answer 86 problems on a test correctly and received a grade of 98% how many problems were on the test if all the probl

adoni [48]

Answer:

87 or 88

Step-by-step explanation:

86 is 98% of 87.755102

So, the number of problems must have been 87 or 88

6 0

2 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.

5 0

3 years ago