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3 years ago

Need help on num 15

Mathematics

1 answer:

Yakvenalex [24]3 years ago

8 0

Answer:

Step-by-step explanation:

D's equation shows y equals 7.8 times x. 7.8 is the unit rate.

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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

6 0

2 years ago

What is the simplest form of 64/96

Delicious77 [7]

The answer is 2/3
64/96 to 32/48 to 16/24 to 8/12 to 4/6 to 2/3..........(64/96 in simplest form becomes 2/3)

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3 years ago

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

Olin [163]

If either $t=0$ or $t=9$ , the ODE has the trivial solution $y=0$ , but $y(1)=1\neq0$ . So any solution to the ODE that satisfies the initial condition can only exist on the interval $0$ .