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Marina86 [1]
3 years ago
6

A total of 100 students are taking a history exam. Each student is required to write an essay on one of six possible topics. Wha

t is the fewest possible number of students who could have chosen the topic selected by most students
Mathematics
1 answer:
amid [387]3 years ago
4 0

The fewest number of students who could have chosen the topic selected by most students is 17.

Since there are six essays and each student is required to write an essay on one of the six topics, the probability of choosing one of the six topic is

P(one topic) = one topic/total number of topics

= 1/6.

Since we have 100 students, the fewest number of students that could choose the topic selected by most students is

n = Probability of selecting one topic × total number of students

= P(one topic) × 100

= 1/6 × 100

= 16.67

≅ 17 students.

So, the fewest number of students who could have chosen the topic selected by most students is 17.

Learn more about probability here:

brainly.com/question/19916581

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Question 1:

Price per milliliter

Small $4.50 / 250ml

Medium $9.95 / 500 ml

Large $16.95 / 1000 ml

Question 2:

$9.95 / 500 ml > $4.50 / 250ml > $16.95 / 1000 ml

Question 3 - 4:

$4.50 / 250ml = ($4.50 * 6) / (250ml * 6 ) = $27 / 1500ml = 0.018

$9.95 / 500 ml = ($9.95 * 3) / (500 ml * 3) = $29.85 / 1500ml = 0.019

$16.95 / 1000 ml = ($16.95 * 3) / (1000 ml * 3) = $50.85 / 1500ml = 0.0339

( Use the first one for question 3 and the second for question 4)

$4.50 / 250ml would be the cheapest way to get 1500ml.

$9.95 / 500ml would be the most expensive way to get 1500ml.

I hope this helps you! Tell me if I'm wrong!

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3 years ago
Can you help me please?
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Answer:

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Step-by-step explanation:

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Write 26% as a decimal.
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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

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Step-by-step explanation:

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