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3 years ago

10

a group of students are booked to go on a sports tour. The total cost of theA tour is $18000. write an expression in x for the a

mount that each learner must pay

1 answer:

monitta3 years ago

3 0

Answer:

x+y=18000

Step-by-step explanation:

x= ammount of students

y= number of days or places going to

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100

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3 years ago

What are prime factors of 1000?

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1,2,4,5,8,10,20,25,40,50,10,125,200,250,500,1000

Step-by-step explanation:

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2 years ago

A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable

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Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.3$ .

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066$

$P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014$

$P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002$

$P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024$

$P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002$

$P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0$

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082$

0.0082 = 0.82% probability that he will pass

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