Taking a final exam rn and im stuck on this question, pls help ;-;

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

3 years ago

Which box-and-whisker plot shows the high temperatures in Pittsburgh, Pennsylvania during the first two weeks of March: 33, 45,

nikitadnepr [17]

There is no picture so i am not sure what to tell u

6 0

2 years ago

What is negative 21/9 times negative 18/7

diamong [38]

Answer:

negative 21/9 times negative 18/7 is 6

Step-by-step explanation:

The phrase negative 21/9 times negative 18/7 can be written mathematically as

$-\frac{21}{9} \times -\frac{18}{7}$

Lets first convert the fractions into decimal values

$-\frac{21}{9}$ = -2.33

$-\frac{18}{7}$ = -2.57

Now on simplyfying

=> $-2.33 \times -2.57$

=> 5.988

=>6 (rounding off)

We will get a positive value because the multiplicative law states that the product of two negative number is positive

5 0

3 years ago

The question is asking to determine which statement is true?

Advocard [28]

Answer:

the answer is true which

7 0

2 years ago

Find all the real zeros of f(x)=3x^3-9x^2+3x-9 Can someone help me please

vaieri [72.5K]

Answer:

x=0,3

Step-by-step explanation:

start by dividing every term by 3

x^3-3x^2+x-3

group into 2 terms

(x^3-3x^2)+(x-3)

simplify as much as you can

x^2(x-3)+(x-3)

combine terms

(x^2)(x-3)

x=3, 0

4 0

3 years ago