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pshichka [43]
2 years ago
6

Into how many equal parts can the same cake be cut if the cuts can only be made along the gridlines? cake can be cut into equal

parts.

Mathematics
1 answer:
avanturin [10]2 years ago
3 0

Into how many equal parts can the same cake be cut if the cuts can only be made along the gridlines, cake can be cut into 20 equal parts.

This is further explained below.

<h3>What are gridlines?</h3>

Generally, In spreadsheet software, gridlines are the light gray lines that divide the cells, rows, and columns. Gridlines are often used to maintain track of data. Gridlines are widely used in popular spreadsheet programs like Spreadsheet.

In conclusion, If the cuts can only be done along the gridlines, the cake may be divided into 20 equal pieces when cutting along the lines.

Read more about gridlines

brainly.com/question/2773987

#SPJ1

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Simplify.<br> 1/4(3x-7)+3/4x
Ivan

Answer:

C. 1\frac{1}{2} x - 1\frac{3}{4}

Step-by-step explanation:

c

7 0
1 year ago
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Which of the following can be used as "reasons" in a two-column proof?
Fudgin [204]

Answer:

A definition and a theorem can be used as a reason in a two-column proof. A two column proof is assembled into statement and reason columns, where each statement should have verified reason.

Step-by-step explanation:

3 0
3 years ago
Countering a cost $9.40 per sales tax. After sales tax, it cost $9.87. What is the sales tax rate?
Nady [450]
STEP 1:
find the sales tax (decimal form)

x= sales tax

Cost + (Cost * Sales Tax)= Total
plug in known numbers

$9.40 + ($9.40 * x)= $9.87

9.40 + 9.40x= 9.87
subtract 9.40 from both sides

9.40x= 0.47
divide both sides by 9.40

x= 0.47/9.40

x= 0.05 sales tax decimal form


STEP 2:
find sales tax percentage

= 0.05 * 100
or move decimal to the right two decimal places

= 5% sales tax percent form


ANSWER: The sales tax is 5% (or 0.05 in decimal form)

Hope this helps! :)
6 0
3 years ago
Read 2 more answers
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
Complete the following statement. The conditional probability of B given A can be found by​ _______. a. adding​ P(A) and​ P(B).b
Ksju [112]

Answer: Assuming that event A has occurred

Step-by-step explanation:

7 0
3 years ago
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