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Strike441 [17]
3 years ago
15

The minimum monthly payment for Anita's credit card is 2% of her balance or $10, whichever is greater. If Anita's balance at the

end of her last billing cycle was $360, what is her minimum monthly payment?
A. $17.20

B. $2.80

C. $10.00

D. $7.20
Mathematics
1 answer:
sveticcg [70]3 years ago
4 0

Answer:

C. $10.00

Step-by-step explanation:

She has to pay which ever is greater.

To get whichever is greater you need to check what 2% of her last billing cycle. That would be done by multiplying .02 and 360. That equals $7.20.

10>7.20 and therefore the answer would be 10 dollars aka C.

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The following frequency table summarizes the number of children that dads in Dads Club have.
attashe74 [19]

Answer:

C

Step-by-step explanation:

Now there are 6+4+8+1+1=20 dads in the Dads Club.

Fewer than 3 children have

  • 1 child - 6 dads
  • 2 children - 4 dads.

So, 6+4=10 dads have fewer than 3 children.

The probability that the dad of Dads Club has fewer than 3 children is

Pr=\dfrac{10}{20}=0.5 \text{ or } 50\%.

The probability that the next dad to join Dads Club has fewer than 3 children is reasonable to be 50%.

8 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
Plz help me<br><br> 56+(96x3)-34=?
Veseljchak [2.6K]
The answer to it is 310
7 0
3 years ago
Read 2 more answers
Is 0.1304347826 a rational number?
VLD [36.1K]
Yes it is a rational number
4 0
3 years ago
A coin is tossed 600times with the frequencies as:
brilliants [131]

Answer:

  • see below

Step-by-step explanation:

Total number of trials =600.

Number of heads = 342.

Number of tails = 258

On tossing a coin,

let  \: E_1  \: and  \: E_2  \: be  \: the  \: events \:  of \:  getting  \: a  \: head

and of getting a tail respectively.

then,

1)  \: P \:  (getting \:  a  \: head) = \: P \:  (E_1)

\longrightarrow \:  \frac{number \: of \: heads \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{342}{600}  =  \frac{57}{100}

\longrightarrow \: 0.57

\red{ \rule{150pt}{3pt}} \:

2) P (getting \: a\: tail) =P (E_2)

\longrightarrow \:  \frac{number \: of \: tails \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{258}{600}  =  \frac{43}{100}

\longrightarrow \: 0.43

6 0
3 years ago
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