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kotegsom [21]
3 years ago
6

Can anyone hknow number 2?

Mathematics
1 answer:
fomenos3 years ago
3 0

Answer:

n = y/(r+s)

Step-by-step explanation:

(rn+sn) / y = 1

Multiply each side by y

(rn+sn) / y  *y= 1*y

rn+sn = y

Factor out n

n(r+s) = y

Divide each side by (r+s)

n(r+s) / (r+s) = y/(r+s)

n = y/(r+s)

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A test has 20 true/false questions. What is the probability that a student passes the test if they guess the answers? Passing me
Minchanka [31]

Using the binomial distribution, it is found that:

The probability that the student will get 15 correct questions in this test by guessing is 0.0207 = 2.07%.

For each question, there are only two possible outcomes, either the guess is correct, or it is not. The guess on a question is independent of any other question, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • There are 20 questions, hence n = 20.
  • Each question has 2 options, one of which is correct, hence p = \frac{1}{2} = 0.5

The probability is:

P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

In which:

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

P(X = 15) = C_{20,15}.(0.5)^{15}.(0.5)^{5} = 0.0148

P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046

P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011

P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002

P(X = 16) = C_{20,19}.(0.5)^{19}.(0.5)^{1} = 0

P(X = 17) = C_{20,20}.(0.5)^{20}.(0.5)^{0} = 0

Then:

P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0 + 0 = 0.0207

The probability that the student will get 15 correct questions in this test by guessing is 0.0207 = 2.07%.

You can learn more about the binomial distribution at brainly.com/question/24863377

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