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Elodia [21]
3 years ago
9

Does anyone know the answer to this problem 8q-3=4q+9

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0
Add 3 to both sides gives
8q = 4q + 12minus 4q from both sides gives
4q = 12
divide both sides by 4 gives
q = 3
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Solve for x please help ! (show work)
Alina [70]

Answer:

x = -5

Step-by-step explanation:

-(5x-2) = 27

Distribute the minus sign

-5x +2 = 27

Subtract 2 from each side

-5x +2-2 = 27-2

-5x = 25

Divide by -5

-5x/-5 = 25/-5

x = -5

5 0
3 years ago
Read 2 more answers
lydia and annette are standing next to each other. lydia is 40 inches tall , and her shadow is 60 inches long. if annettes shado
Tasya [4]
75-60=15
40+15=55 so Annette is 55 inches tall
6 0
3 years ago
Which linear inequality is represented by the graph?
Virty [35]
The answer is the third one down because y moves up 1/3 and it’s 2 away from each even number
7 0
3 years ago
What is the probability that a resident reports high satisfaction, given that the resident is a renter?
dem82 [27]

Answer:

The probability that a resident reports high satisfaction while the resident is a renter is Option(a) 13.7\%

Step-by-step explanation:

Given:

                               Levels of satisfaction

                      High      Medium     Low       Total

Owners          26             34               18           78

Renters           15            12                5            22

Total               41             46               23          110

The Resident is a renter.

Step 1:

To find the probability that a resident reports high satisfaction.

P(Renter- higher satisfaction)= Higher Satisfaction by render ÷ Total level of satisfaction

=\frac{15}{110}

=0.137

13.7\% is the probability that a resident reports high satisfaction.

Therefore, Option (a) is a correct answer.

Learn more about Probability, refer:

  • brainly.com/question/13604758
  • brainly.com/question/10734660
7 0
2 years ago
Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho
oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0
3 years ago
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