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vladimir2022 [97]
3 years ago
10

Translate algebraically and simplify.

Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

Step-by-step explanation:

Im not answering the question but omg ur and army my bias is suga and im also in middle school

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There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0
4 years ago
How much do a dozen cookies weigh?
yaroslaw [1]
0.43 oz

                   Hope this helps...
7 0
3 years ago
Read 2 more answers
Please help me with this!!!
krok68 [10]

Answer:

the answer is A because if it is raised to the power greater than one and is added it will be greater than1

3 0
3 years ago
In March, there will be two conferences: one for math and one for history. So far, 7 people have signed up for the math conferen
otez555 [7]

Answer:

After 4 days, the number of people attending both conferences be the same.      

Step-by-step explanation:

We are given the following in the question:

Maths conference:

Number of people already signed = 7

Number of people who sign up each day = 2

Thus, the number of people who will sign up for maths conference in x days will be given by the linear function:

 f(x) = 7+2x

History conference:

Number of people already signed =11

Number of people who sign up each day = 1

Thus, the number of people who will sign up for maths conference in x days will be given by the linear function:

 g(x) =11+x

Both conference will have same number of people when

f(x) = g(x)\\\Rightarrow 7 + 2x = 11+x\\\Rightarrow x = 4

Thus, after 4 days, the number of people attending both conferences be the same.

5 0
3 years ago
Question 5
DochEvi [55]
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