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Gennadij [26K]
3 years ago
7

The choir sold boxes of candy and teddy bears near Valentine’s Day to raise money. They sold 57 items altogether. Bears sold for

$8.00 each and each box of candy sold for $6.00. They collected $380. How much of each item did they sell
Mathematics
1 answer:
nevsk [136]3 years ago
6 0

Answer:

19 bears, 38 boxes of candy.

Step-by-step explanation:

We can use a system of equations (b = # of bears, c = # of candy boxes):

<em>b + c = 57, 8b + 6c = 380</em>.

We can use substitution by isolating b in the first eq.: <em>b = 57 - c</em>, then substitution b with 57 - c in the second eq.: <em>8(57 - c) + 6c = 380</em>.

Then, we solve: 456 - 8c + 6c = 380, 76 = 2c, c = 38 candy boxes.

Finally we plug in 38 for c in the first eq.: b + 38 = 57, b = 19 bears.

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sashaice [31]

Answer: B

Step-by-step explanation:

The converse means X --> Y , so Y --> X.

In other terms, you just need to reverse the terms and review that it turns out correct. In answer B, it does that just right.

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My question got deleted for some reason??? anyways
lakkis [162]

Answer:

5.5 + 8.5 = 14

14/2=7

The average is 7

7 x 6.5 = 45.5

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7 x 6.5 = 45.5

Step-by-step explanation:

6 0
3 years ago
Phyllis invested 32000 dollars, a portion earing a simple interest rate of 5 percent per year and the rest earning a rate of 6 p
il63 [147K]

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Answer:

  • $10,000 at 5%
  • $22,000 at 6%

Step-by-step explanation:

Let x represent the amount invested at 6%. Then 32000-x is the amount invested at 5%, and the total interest earned is ...

  0.06x +0.05(32000 -x) = 1820

  0.01x +1600 = 1820 . . . . simplify

  0.01x = 220 . . . . . . . . . .  subtract 1600

  x = 22,000 . . . . . . . . . . . multiply by 100

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Phyllis invested $22,000 at 6% and $10,000 at 5%.

6 0
3 years ago
For the given set, first calculate the number of subsets for the set, then calculate the
vodomira [7]

Answer:

\fbox{\begin{minipage}{14em}Number of subsets: 16\\Number of proper subsets: 15\end{minipage}}

Step-by-step explanation:

<em>Given:</em>

The set A = {5, 13, 17, 20}

<em>Question: </em>

Find the number of subsets of A

Find the number of proper subsets of A

<em>Simple solution by counting:</em>

Subset of A that has 0 element:

{∅} - 1 set

Subset of A that has 1 element:

{5}, {13}, {17}, {20} - 4 sets

Subset of A that has 2 elements:

{5, 13}, {5, 17}, {5, 20}, {13, 17}, {13, 20}, {17, 20} - 6 sets

Subset of A that has 3 elements:

{5, 13, 17}, {5, 13, 20}, {5, 17, 20}, {13, 17, 20} - 4 sets

Subset of A that has 4 elements:

{5, 13, 17, 20} - 1 set

In total, the number of subsets of A: N = 1 + 4 + 6 + 4 + 1 = 16

The number of proper subsets (all of subsets, except subset which is equal to original set A): N = 16 - 1 = 15

<u><em>Key-point:</em></u>

The counting method might be used for finding the number of subsets when the original set contains few elements.

The question is that, for a set that contains many elements, how to find out the number of subsets?

The answer is that: there is a fix formula to calculate the total number (N) of subsets of a set containing n elements: N = 2^{n}

With original set A = {5, 13, 17, 20}, there are 4 elements belonged to A.

=> Number of subsets of A: N = 2^{4} = 16

(same result as using counting method)

<em>Brief proof of formula: N = </em>2^{n}<em />

Each element of original set is considered in 2 status: existed or not.

If existed => fill that element in.

If not => leave empty.

For i.e.: empty subset means  that all elements are selected as not existed, subset with 1 element means that all elements are selected as not existed, except 1 element, ... and so on.

=> From the point of view of a permutation problem, for each element in original set, there are 2 ways to select: existed or not. There are n elements in total. => There are 2^n} ways to select, or in other words, there are 2^{n} subsets.

Hope this helps!

:)

8 0
3 years ago
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