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netineya [11]
3 years ago
7

Which of the segments below is a secant?

Mathematics
2 answers:
mel-nik [20]3 years ago
5 0

Answer:

d.$\overline{BC}$

Step-by-step explanation :

We are given that a circle of center O.

We have to find the segment that is secant.

Secant : If a line which intersect the circle at two points.Then the line segment  is called  secant.

a.$\overline{OA}$

It is false because segment OA intersect the circle at one point A only.

b.$\overline{CD}$

It is false because segment CD intersect the circle at point D only.

c.$\overline{BC}$

It is true because it intersect the circle at two points .

d.$\overline{AB}$

It is false because it intersect the circle at one point A only.

Answer:c.$\overline{BC}$

Rainbow [258]3 years ago
3 0

Answer:

BC

Step-by-step explanation:

A secant is a line which intersects the circle at 2 points

In the given diagram this is BC

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Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o
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Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

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}.\pi^{x}.(1-\pi)^{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 \pi is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so \pi = 0.8.

(a) What is the probability she gets an A?

There are four problems, so n = 4

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find P(X \geq 3)

P(X \geq 3) = P(X = 3) + P(X = 4)

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

P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096

P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096

P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so n = 3

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find P(X \geq 2)

P(X \geq 3) = P(X = 2) + P(X = 3)

P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384

P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512

P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

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