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3 years ago

Shawna has $435.15 in her savings

Mathematics

1 answer:

galina1969 [7]3 years ago

4 0

Answer:

H. $360.71

Step-by-step explanation:

The 1 in $435.15 is in the tenths place. It has a value of 1/10. One tenth that value is 1/100. To have that value, a 1 would need to be in the hundredths place: 0.01.

Of the offered choices, the only one with a 1 in the hundredths place is ...

H. $360.71

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If we were to do this equation how would I solve it

s344n2d4d5 [400]

Answer:

$- 6 \frac{1}{3}$

Step-by-step explanation:

$\frac{10}{3} - \frac{1}{3} - \frac{29}{3} - ( - \frac{1}{3} ) \\ \\ = \frac{10}{3} - \frac{1}{3} - \frac{29}{3} + \frac{1}{3} \\ \\ = \frac{10}{3} - \frac{29}{3} - \frac{1}{3} + \frac{1}{3} \\ \\ = \frac{10 - 29}{3} \\ \\ = \frac{ - 19}{3} \\ \\ = - 6 \frac{1}{3}$

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4 years ago

How are corresponding angles of two similar figures related

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Answer:

the corresponding angles are equal in similar figures.

Step-by-step explanation:

When two figures are similar then the corresponding angles of the two figures are equal. the side of the figures are in proportional.

for example, when triangle ABC is similar to triangle DEF

then the corresponding angles are equal

<A=<D

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so the corresponding angles are equal in similar figures.

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3 years ago

The process of digging a site includes a. carefully recording the location of artifacts. b. accurate descriptions of the artifac

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Answer:

The process of digging a site includes

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4 years ago

Laura finished her history assignment in 1/4 hours. Then she completed her English assignment in 3/5 hours. What was the total a

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Step-by-step explanation:

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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.

5 0

3 years ago