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3 years ago

Which is an equation of a circle with center (-10, 6) and radius 8?

1 answer:

8 0

An equation of a circle:

(x - a)² + (y - b)² = r²

(a; b) - a coordinates of a center
r - a radius

r = 8; (-10; 6) ⇒ a = -10 and b = 6

subtitute

(x - (-10))² + (y - 6)² = 8²

Answer: (x + 10)² + (y - 6)² = 64

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What is the product of 119 thousandths times 10?

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The answer is A becaues 119 thousand times 10 equals 1190000
then you divide it by 100 to get A.119 hunderths

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3 years ago

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Solve the linear system of equations using the linear combination method.

Mrrafil [7]

Answer:

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Step-by-step explanation:

We are going to eliminate the variable x. In order to do this, we are going to make the coefficients the same by multiplying the second equation by 2

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4 years ago

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dmitriy555 [2]

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3 years ago

A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the

Ne4ueva [31]

Answer:

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test positive

Event B: Has the disease

Probability of a positive test:

90% of 3%(has the disease).

1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So

$P(A) = 0.90*0.03 + 0.1*0.97 = 0.124$

Intersection of A and B:

Positive test and has the disease, so 90% of 3%

$P(A \cap B) = 0.9*0.03 = 0.027$

What is the conditional probability that she does, in fact, have the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177$

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

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3 years ago

When playing the roulette at a casino, you have exactly 1/35 chance of winning in each round if you play a single number. Your f

Vikentia [17]

Answer:

0.2275 = 22.75% probability that you actually won that round

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Fireworks going off

Event B: You won

Probability of fireworks going off.

100% of 1/35 = 0.0286(when you win)

10% of 34/35 = 0.9714(you lost). So

$P(A) = 0.0286 + 0.1*0.9714 = 0.12574$

Probability of you winning and fireworks going off:

100% of 1/35, so $P(A \cap B) = 0.0286$

If you failed to see the outcome of a round, but you see the fireworks going off, then what is the probability that you actually won that round?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0286}{0.12574} = 0.2275$

0.2275 = 22.75% probability that you actually won that round

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3 years ago