Answer:
Ok, here is the answers! brainliest pleaseee :D
Step-by-step explanation:
13. - 0.624 kg. (Explanation: 104 grams per one pack of pop-tarts. 104 g x 6 packs of pop-tarts equal 624 grams in total. 624 grams to kilograms is 0.624 kg.)
14. - No. The tablecloth is 155 cm long. It needs to cover a table that is 1.6 meters long. 155 cm to meters is 1.55 meters. The tablecloth is 0.05 to short to cover the table. 1.6 meters - 1.55 meters = 0.005.
15 - Alex's sister rode the farthest. She rode 1,000 meters farther than Alex. 2 kilometers in meters is 2,000 meters. His sister rode 1,000 meters more than that, 3,000.
Answer: 5/2
5/6 / 1/3 = 2.5
2.5 in fraction = 5/2
Answer:
There can be at most 
attendees in a corporate team-building event.
Step-by-step explanation:
Let 
denotes number of attendees in a corporate team-building event.
Fixed cost = $19
Cost charged per attendee = $1
Budget for the corporate team-building event = $31
Therefore,
So, there can be at most attendees in a corporate team-building event.
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:
- U = {x | x is a real number}
- A = {x | x is an odd integer}
- 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
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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
