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3 years ago

How to solve the inequality

Mathematics

1 answer:

GaryK [48]3 years ago

4 0

The problem is not shown clearly therefore i am not sure if my answer will be correct, but, this inequality is already solved. all this means is that y is more than or equal to 4.5 (meaning that y can be 4.5, 5, 6, 7, 8 and so own but never a number less than 4.5), the "I I" only mean that y will never be a negative.

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Can someone help please?!! Major Help Pic above

Mila [183]

I believe it’s c. I hope that this is helpful!

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3 years ago

Nina ran 12 4/5 laps in 1/4 of an hour. At this rate, how many laps could she run in an hour? (Simplify your answer completely)

kozerog [31]

i think the answer is 42 but i don’t know how to get in into its smallest form but i hope this helps

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3 years ago

You roll a six-sided die. Find the probability of each of the following scenarios. (a) Rolling a 5 or a number greater than 3.

sertanlavr [38]

Answer:

Step-by-step explanation:

let A= number 5

B=number >3

P(A)=1/6 {5}

P(B)=3/6 {4,5,6}

p(A∩B)=1/6 {5}

P(AUB)=P(A)+P(B)-P(A∩B)=1/6+3/6-1/6=3/6=1/2

let event C=number<5

D=even number

C=[1,2,3,4}

D={2,4,6}

C∩D={2,4}

P(C)=4/6

P(D)=3/6

P(C∩D)=2/6

P(C∪D)=P(C)+P(D)-P(C∩D)=4/6+3/6-2/6=5/6

let event E=number 2

F= an odd number.

E={2}

F={1,3,5}

E∩F={}

P(E)=1/6

P(F)=3/6

P(E∩F)=0

P(E∪F)=P(E)+P(F)-P(E∩F)=1/6+3/6-0=4/6=2/3

7 0

2 years ago

What is. <br> 3q + 2 (q+1)

Fudgin [204]

Answer:

5q + 2

Explanation:

3q + 2(q + 1)

Distribute:

= 3q + (2)(q) + (2)(1)

= 3q + 2q + 2

Combine Like Terms:

= 3q + 2q + 2

= (3q + 2q) + (2)

= 5q + 2

5 0

2 years ago

Read 2 more answers

In Prof. Leahy's class, if a student does homework most days, the chance of passing the course is 90%. On the other hand, if a s

34kurt

Answer:

a) p(the student did not do homework and he/she passed the course) = 0.09

b) p(the student did not do homework given that she/he passed the course) = 0.125.

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.

a. p(the student did not do homework and he/she passed the course)

If a student does not do homework most days, the chance of passing the course is only 30%. 100 - 70 = 30% don't do homework on a regular basis.

0.3*0.3 = 0.09

p(the student did not do homework and he/she passed the course) = 0.09

b. p(the student did not do homework given that she/he passed the course)

Conditional probability.

Event A: Passed the course

Event B: Did not do homework.

p(the student did not do homework and he/she passed the course) = 0.09

This means that $P(A \cap B) = 0.09$

Probability that the student passes the course:

90% of 70%(do homework)

30% of 30%(do not do homework).

This means that:

$P(A) = 0.9*0.7 + 0.3*0.3 = 0.72$

p(the student did not do homework given that she/he passed the course)

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09}{0.72} = 0.125$

p(the student did not do homework given that she/he passed the course) = 0.125.

4 0

3 years ago