For f(x)=x4−x3, which gives an output of 54?

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

6 0

3 years ago

What is the center and radius of the circle with equation (x - 2)^2 + (y - 5)^2 = 100?

ololo11 [35]

Answer:
Center = (2,5)
Radius = 10
Choice A

To find this answer, first write the equation
(x-2)^2 + (y-5)^2 = 100
into
(x-2)^2 + (y-5)^2 = 10^2

Note how the second equation is in the form
(x-h)^2 + (y-k)^2 = r^2

We see that (h,k) = (2,5) is the center
and r = 10 is the radius

5 0

3 years ago

Read 2 more answers

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago