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Irina-Kira [14]
2 years ago
15

In a group of 31 pupils, 4 play the flute only. 14 play the piano only. 7 play neither instrument. A student is selected at rand

om. What is the probability the student plays both instruments?
Mathematics
2 answers:
astra-53 [7]2 years ago
8 0
<h3>Solution:</h3>

<u>Note that:</u>

  • Total pupils: 31
  • 4 pupils = flute only
  • 14 pupils = piano only
  • 7 pupils = neither
  • This means that {31 - (4 + 14 + 7)} pupils play both instruments.

<u>First, let's find out the number of pupils who play both instruments.</u>

  • => {31 - (4 + 14 + 7)} pupils
  • => {31 - (25)} pupils
  • => {31 - 25} pupils
  • => 6 pupils

<u>Probability form:</u><em> Student who plays both instruments/Total students</em>

  • => 6/31

The probability that this student plays both instruments is 6/31.

solmaris [256]2 years ago
7 0

Answer:

probability the student plays both instruments is [  \frac{6}{31}  ]

<h2>                         <u>Explanation</u>:</h2>

students who play both instruments = total students - flute players - piano players - students who play neither instruments

students who play both instruments

→ 31 - 4 - 14 - 7

→ 6 students play both instruments.

probability:

<h2>→  \boxed{\frac{students- who -play- both- instruments}{total -students} }</h2><h2>→                     \boxed{\frac{6}{31} }</h2>
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.EXPLANATION

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Answer:

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Step-by-step explanation:

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__

23. y = 27.8112·1.18832^t (see the second attachment)

__

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Answer:

C

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