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musickatia [10]
3 years ago
9

What are the qualification of? 0.3%

Mathematics
1 answer:
KIM [24]3 years ago
4 0

Answer:

30

Step-by-step explanation:

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Solve the equation.<br> 5.16 = 4n<br> Show your work.
astra-53 [7]

Answer:

5.16=4n

Step 1: Flip the equation.

4n=5.16

Step 2: Divide both sides by 4.

4n/4= 5.16/4

anwers n=1.29

Step-by-step explanation:

I hope this help ;-)

8 0
3 years ago
Subtract. 225.37−41.46
Aleksandr-060686 [28]
The answer is 183.91
8 0
3 years ago
Read 2 more answers
explain why multiplying the numerator and the denominator of a fraction by the same number results in an equivalent fraction.​
alisha [4.7K]

Answer:

We also know that when you have the same numerator and denominator in a fraction, it always equals 1. For example: So as long as we multiply or divide both the top and the bottom of a fraction by the same number, it's just the same as multiplying or dividing by 1 and we won't change the value of the fraction.

Step-by-step explanation:

7 0
3 years ago
Help me out here???? I need help!!​
wel
Answer: x would equal 30
7 0
3 years ago
Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a
Nana76 [90]

Answer:

0.00002 = 0.002% probability of actually having 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: Positive test

Event B: Having the disease

Probability of having a positive test:

0.05 of 1 - 0.000001(false positive)

0.99 of 0.000001 positive. So

P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094

Probability of a positive test and having the disease:

0.99 of 0.000001. So

P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}

What is the probability of actually having the disease

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002

0.00002 = 0.002% probability of actually having the disease

6 0
3 years ago
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