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ikadub [295]
2 years ago
15

HELP HELP HELP TODAY IS THE DEADLINE

Mathematics
1 answer:
disa [49]2 years ago
5 0

C, 3 units translated to the left due to the +3 inside the bracket, vertically reflected on the x-axis as it is negative outside the bracket and translation 1 unit up because of the +1 outside the brackets.

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jeka57 [31]

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2 years ago
D+0.5=0.75<br><br> what is the answer put numbers
11111nata11111 [884]

Answer:

d = 0.25

Step-by-step explanation:

d + 0.5 = 0.75

    -0.5     -0.5

d = 0.25

Hope this helps!!!

4 0
3 years ago
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An expression is shown.<br>2 3/4 x 2/3<br>What is the value of the expression?​
Novay_Z [31]

Answer:

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Step-by-step explanation:

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3 years ago
Simplify the following expression.<br><br> 0.82 / 0.2
AlekseyPX

Answer:

0.41

Step-by-step explanation:

0.82 = 0.2 x 0.41

0.82 / 0.2

= ( 0.2 x 0.41 ) / 0.2

= 0.41

5 0
2 years ago
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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
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

3 0
3 years ago
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