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expeople1 [14]
3 years ago
9

What is the quotient of -18 divided by [-1/6]

Mathematics
2 answers:
Levart [38]3 years ago
7 0

First step is to flip the fraction 1/6 to 6/1 then the next step is to multiply 1/18 x 6/1 that gives you 6/18 so then simflifly it then that would give you 1/3

mark me as the brainlest plzzz and most inportantly hope i helped 

Sloan [31]3 years ago
4 0
18?
-18 divided by [-1/6] 
is 18
hopefully that's what you were looking for.?!!
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Mattie uses the discriminant to determine the number of zeros the quadratic equation 0 = 3x2 – 7x + 4 has. Which best describes
r-ruslan [8.4K]

Answer:

The equation has two zeros because the discriminant is greater than 0.

Step-by-step explanation:

3x^2 – 7x + 4

a=3   b = -7   c=4

The discriminant is

b^2 -4ac

(-7)^2 - 4(3)(4)

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1

Since the discriminant is greater than zero, there are two real solutions

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3 years ago
The fraction 5/8 is located between what two fractions?
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zloy xaker [14]
Diagonal of a rectangle formula:

Diagonal = √l² + w²
Diagonal = √10² + 8²
Diagonal = √100 + 64
Diagonal = √164
Diagonal = 12.806 feet.

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3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

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P(at most 2 defective) = 0.99

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Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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Answer:

216

Step-by-step explanation:

5 0
3 years ago
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