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slega [8]
3 years ago
10

Trisha needs to make at least 50 gift bags for an event. Each gift bag will contain at least 1 thumb drive or 1 key chain. She w

ants to use at least 5 times as many key chains as thumb drives. She has 25 thumb drives and 200 key chains.
Let x represent the number of key chains. Let y represent the number of thumb drives.

Which inequalities are among the constraints for this situation?

Select each correct answer.




x+y≥50

x≥5y

x≤5y

x+5y≤50

y≤25

Mathematics
2 answers:
mel-nik [20]3 years ago
8 0

Answer:

y ≤ 25

x ≥ 5y

x + y ≥ 50

Step-by-step explanation:

Archy [21]3 years ago
6 0

Answer:I Believe the Correct answers are:

x+5y ≤ 50

and

x+y≥50



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the drama club has sold 942 tickets to their play if their theater seats 57 each row how many full rows were there be how many e
Alona [7]

Since 57x17=969 I did 969-942 and got 27 so there is 16 full rows and then 27 left over so that means it won’t be a full row but 27 extra seats will be filled. (If your wondering where the 17 came from that’s how many rows can be made that’s over 942 because if I did 57x16 it would equal 912 and I would not have had a full row of chairs)

3 0
3 years ago
The mode of 1.5, 1.6, 2, 2, 3.4, 5.6, 5.9 is 3.14.<br><br> True<br><br> False
inessss [21]
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6 0
3 years ago
√50 as a mixed radical!
Hoochie [10]

Answer:

5sqrt(2)

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5 0
3 years ago
In how many ways can 3person study groups beselected from a class of 25students?Note: nrn!r!(n-r)!nEnter
nlexa [21]

Answer:

2,300

Explanation:

This is given as:

\begin{gathered} 25\text{ combination 3 represented as:} \\ ^nC_r=\frac{n!}{r!(n-r)!} \\ n=25 \\ r=3 \\ ^{25}C_3=\frac{25!}{3!(25-3)!} \\ ^{25}C_3=\frac{25\times24\times23\times22!}{3!\times22!} \\ ^{25}C_3=\frac{25\times24\times23}{3\times2\times1} \\ ^{25}C_3=2300 \end{gathered}

Therefore, there are 2,300 ways that 3 persons can be selected from 25 people

8 0
1 year ago
What is the answer to this question?
ElenaW [278]

sorry but i need the points thx

7 0
3 years ago
Read 2 more answers
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