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WITCHER [35]
2 years ago
15

A grade school class is nearing the final round of musical chairs. 5 students are left and there are 3 chairs available. How man

y combinations could be made of these 5 students in the 3 chairs?
Mathematics
1 answer:
Sunny_sXe [5.5K]2 years ago
3 0

Answer:

15

Step-by-step explanation:

its as simple as the numbers time each other

5x3=15

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Which sentence would be a good counter example to this statement ?
ArbitrLikvidat [17]

Answer:

The answer is "Option D".

Step-by-step explanation:

A line is a horizontal 1D-dimensional representation without any thickness and extends in every way. It sometimes is also known as the straight line. The line, which connects two planes lies simultaneously on both planes, that's why in this question only "option D" is correct.

8 0
3 years ago
A clock radio phone cost 83.95. Tax is $4.49 for each radio phone how much do 3 clock radio phones cost
zimovet [89]

Step-by-step explanation:

If each clock radio costs $83.95 + taxes ($4.49), each clock radio has the real cost of $88.44

Therefore if you multiply the real cost of the 3 pieces, you'll get the result:

$88.44 x 3 = $265.32

Answer:

$265.32

5 0
3 years ago
What did I do wrong. I thought it was 15
CaHeK987 [17]
Its not 15 isnt it 10
3 0
3 years ago
Read 2 more answers
30, 16, 49, 25 find the median
Ksivusya [100]

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em>

5 0
3 years ago
Read 2 more answers
Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?
-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>
  • In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

  1. U = {x | x is a real number}
  2. A = {x | x is an odd integer}
  3. R = {x | x = 3, 7, 11, 27}

So,

  • A = 1, 3, 5, 7, 9, 11... are the elements of set A.
  • R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
  • Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(C) no because each element in set A is not represented in set R

(D) no, because each element in set R is not represented in set A

8 0
1 year ago
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