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3 years ago

3n^2-5n-6=9 help please!

Mathematics

2 answers:

Luden [163]3 years ago

6 0

Answer:

3n^2 -5n -6 = 9 This is a quadratic Equation.

3n^2 -5n -15 = 0

It is solved with the Quadratic Formula:

n = [-b +-sqr root (b^2 - 4ac)] / 2a

a = 3

b = -5

c= -15

n = [--5 +- sqr root(25 -4*3*-15)] / 6

n = [5 +- sqr root (25 + 180) / 6

n = [5 +-sqr root (205) ] / 6

n = 5 + 14.3178210633 / 6

n1 = 19.3178210633 / 6

n1= 3.2196368439

n2= 5 - 14.3178210633 / 6

n2 = -9.3178210633 / 6

n2 = -1.5529701772

Step-by-step explanation:

velikii [3]3 years ago

5 0

Answer:

n = 5/6 ±√(205/36)

n ≈ -1.55297, 3.21964

Step-by-step explanation:

3n² - 5n - 6 = 9

3(n² - (5/3)n - 2) = 9

(n² - (5/3)n - 2) = 9/3

(n² - (5/3)n ) = 3 + 2

(n² - (5/3)n + (5/6)²) = 5 + (5/6)²

(the added term will always be the square of b/2)

(n - (5/6))² = 5 + 25/36

n - 5/6 = ±√(205/36)

n = 5/6 ±√(205/36)

n ≈ -1.55297, 3.21964

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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

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

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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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3 years ago

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Answer:

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