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iogann1982 [59]
3 years ago
10

Write an equation of the circle with center (4,5) and radius 9.

Mathematics
2 answers:
mixer [17]3 years ago
6 0

Answer:

I'm going to write both of these because maybe you have a fill in the blank question. I don't know.

(x-4)^2+(y-5)^2=9^2

Simplify:

(x-4)^2+(y-5)^2=81

Step-by-step explanation:

(x-h)^2+(y-k)^2=r^2 is the equation of a circle with center (h,k) and radius r.

You are given (h,k)=(4,5) because that is the center.

You are given r=9 because it says radius 9.

Let's plug this in.

(x-4)^2+(y-5)^2=9^2

Simplify:

(x-4)^2+(y-5)^2=81

Inga [223]3 years ago
3 0

Answer:

(x-4)^2 + (y-5)^2 = 9^2

or

(x-4)^2 + (y-5)^2 =81

Step-by-step explanation:

The equation for a circle is (x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

(x-4)^2 + (y-5)^2 = 9^2

or

(x-4)^2 + (y-5)^2 =81

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Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three f
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Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are independent. 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.

He has a 3/4 probability of success.

This means that p = \frac{3}{4} = 0.75

Giannis shoots three free throws

This means that n = 3

What is the probability that he succeeds on at least two of them

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

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

P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219

P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438

84.38% probability that he succeeds on at least two of them

6 0
2 years ago
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