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WITCHER [35]
2 years ago
8

Need answer ASAP will give brainliest check out the picture

Mathematics
1 answer:
valentinak56 [21]2 years ago
8 0

Answer:

i think... x=6

Step-by-step explanation:

3x+9=x+21

3x−x=12

2x=12

x=6

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Can you please help me ​
sergiy2304 [10]

Answer:

  x = 23

Step-by-step explanation:

The marked angles are "corresponding," hence congruent:

  8x -27 = 7x -4

  x = 23 . . . . . . . . . add 27-7x to both sides

5 0
3 years ago
Which of the following best describes the graph shown below?
Valentin [98]
A -------------------------------------
4 0
3 years ago
Read 2 more answers
Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let D^c be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,P(D)=\frac{1}{10000}

Only 0.4% of the people who don't have it test positive.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

P(D^c)=1-\frac{1}{10000}

Formula:P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}

P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

(B)

P(D^c|A^c)=probability that someone does not have disease given that he tests negative

P(A^c|D^c)=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0
3 years ago
4 days ago, Emily blew up a balloon (that is a perfect sphere) to a radius of 5cm. Since then, the balloon has lost air and decr
Elan Coil [88]

Answer:

27/125

Step-by-step explanation:

3 0
3 years ago
What is 6(m + 4) =_m +_
Zielflug [23.3K]

Answer:

6m+24

Step-by-step explanation:

basically you multiply m and 4 by six soooo

6 times m is 6x

6 times 4 is 24

soo

6m+24 is correct

can i have rainliest pleaseee:)b

5 0
3 years ago
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