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True [87]
3 years ago
10

Graph the inequality y > 3x + 3

Mathematics
1 answer:
natita [175]3 years ago
7 0

Answer:

see below

Step-by-step explanation:

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2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac
Aleks04 [339]

Answer:

P(disease/positivetest) = 0.36116

Step-by-step explanation:

This is a conditional probability exercise.

Let's name the events :

I : ''A person is infected''

NI : ''A person is not infected''

PT : ''The test is positive''

NT : ''The test is negative''

The conditional probability equation is :

Given two events A and B :

P(A/B) = P(A ∩ B) / P(B)

P(B) >0

P(A/B) is the probability of the event A given that the event B happened

P(A ∩ B) is the probability of the event (A ∩ B)

(A ∩ B) is the event where A and B happened at the same time

In the exercise :

P(I)=0.025

P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975

P(PT/I)=0.904\\P(PT/NI)=0.041

We are looking for P(I/PT) :

P(I/PT)=P(I∩ PT)/ P(PT)

P(PT/I)=0.904

P(PT/I)=P(PT∩ I)/P(I)

0.904=P(PT∩ I)/0.025

P(PT∩ I)=0.904 x 0.025

P(PT∩ I) = 0.0226

P(PT/NI)=0.041

P(PT/NI)=P(PT∩ NI)/P(NI)

0.041=P(PT∩ NI)/0.975

P(PT∩ NI) = 0.041 x 0.975

P(PT∩ NI) = 0.039975

P(PT) = P(PT∩ I)+P(PT∩ NI)

P(PT)= 0.0226 + 0.039975

P(PT) = 0.062575

P(I/PT) = P(PT∩I)/P(PT)

P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116

8 0
3 years ago
What was the mistake? *
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Answer

Step-by-step explanation:

4 0
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tamaranim1 [39]

Answer:

slope= 3/2

Step-by-step explanation:

4 0
3 years ago
(BEST ANSWER GETS BRAINLIEST)(!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ksenya-84 [330]
<span>The initial cost is $12,950.

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4 0
3 years ago
Read 2 more answers
The box plot compares the number of events that Mrs. Cai’s and Mr. Roland’s classes won over the last 10 years on the school’s f
nata0808 [166]

Answer:

D. Mr. Roland’s class was more successful because his class’s lower quartile was the same as Mrs. Cai’s class’s upper quartile.

Step-by-step explanation:

Mrs. Cai's class's box ranges from 2 to 4, that is, her class’s upper quartile is 4.

Mr. Roland's class's box ranges from 4 to 7, that is, his class’s lower quartile 4.

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3 years ago
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