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Firlakuza [10]
3 years ago
7

GEOMETRY QUESTION, HELP PLEASEEE.

Mathematics
1 answer:
choli [55]3 years ago
6 0

Answer:

The pre-image C is (2, 5)

Step-by-step explanation:

  • If the point (x, y) rotated about the origin by angle 90° clockwise, then its image is (y, -x)
  • If the point (x, y) rotated about the origin by angle 180° clockwise, then its image is (-x, -y)
  • If the point (x, y) rotated about the origin by angle 270° clockwise, then its image is (-y, x)

∵ Point C was rotated 270° clockwise around the origin

∵ C' = (-5, 2)

→ By using the 3rd rule above

∵ The image of the point (x, y) is (-y, x)

∴ (-y, x) = (-5, 2)

→ That means -y = -5 and x = 2

∴ x = 2

∴ -y = -5

→ Divide both sides by -1

∴ y = 5

∴ The pre-image C = (2, 5)

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Answer:

<h2>The Answer would be (0.25, -0.25)</h2>

Step-by-step explanation:

Since you started at the origin of the plane, it would be obvious that your place would be 0.25 on the x-axis. The template for this would be (x,y) x- right and left. y- up and down. On the y-axis, the bug moved 0.25 DOWN to point A. That would be a negative since on the y-axis, it counts up to positives and down to negative. Similar to the x-axis, it counts right to positives, left to negatives. Hope this helped!

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

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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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