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BaLLatris [955]
3 years ago
10

Need help!! Don't understand algebra. I need help figuring out 18x2 + 9x - 20

Mathematics
1 answer:
Murrr4er [49]3 years ago
8 0
The answer is (6x-5)(3x+4)

Write 9x as a sum or difference.
18x^2+24x-15x-20
Factor out 6x from the expression.
18x^2+24x
Factor out -5 from the expression.
-15x-20
Factor out 3x+4 from the expression. 
6x(3x+4)-5(3x+4). 

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According to the information given above, the answer is: You hear that a storm is moving at 15 km/h north. You have been given the storm's <span>velocity</span>. 
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A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A
Zigmanuir [339]

Answer:

The probability that all 4 selected workers will be from the day shift is, = 0.0198

The probability that all 4  selected workers will be from the same shift is = 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

Step-by-step explanation:

Given that:

A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 4 of these workers for in-depth interviews:

The number of selections result in all 4 workers coming from the day shift is :

(^n _r) = (^{10} _4)

=\dfrac{(10!)}{4!(10-4)!}

= 210

The probability that all 5 selected workers will be from the day shift is,

\begin{array}{c}\\P\left( {{\rm{all \ 4 \  selected   \ workers\  will \  be  \ from  \ the \  day \  shift}}} \right) = \frac{{\left( \begin{array}{l}\\10\\\\4\\\end{array} \right)}}{{\left( \begin{array}{l}\\24\\\\4\\\end{array} \right)}}\\\\ = \frac{{210}}{{10626}}\\\\ = 0.0198\\\end{array}

(b) The probability that all 4 selected workers will be from the same shift is calculated as follows:

P( all 4 selected workers will be) = \dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}

where;

(^{8}_4) } = \dfrac{8!}{4!(8-4)!} = 70

(^{6}_4) } = \dfrac{6!}{4!(6-4)!} = 15

∴ P( all 4 selected workers is ) =\dfrac{210+70+15}{10626}

The probability that all 4  selected workers will be from the same shift is = 0.0278

(c) What is the probability that at least two different shifts will be represented among the selected workers?

P ( at least two different shifts will be represented among the selected workers)  = 1-\dfrac{ (^{10}_4) }{(^{24}_4)}+\dfrac{ (^{8}_4) }{(^{24}_4)} + \dfrac{ (^{6}_4) }{(^{24}_4)}

=1 - \dfrac{210+70+15}{10626}

= 1 - 0.0278

The probability that at least two different shifts will be represented among the selected workers is = 0.9722

(d)What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

The probability that at least one of the shifts will be unrepresented in the sample of workers is:

P(AUBUC) = \dfrac{(^{6+8}_4)}{(^{24}_4)}+ \dfrac{(^{10+6}_4)}{(^{24}_4)}+ \dfrac{(^{10+8}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0

P(AUBUC) = \dfrac{(^{14}_4)}{(^{24}_4)}+ \dfrac{(^{16}_4)}{(^{24}_4)}+ \dfrac{(^{18}_4)}{(^{24}_4)}- \dfrac{(^{6}_4)}{(^{24}_4)}-\dfrac{(^{8}_4)}{(^{24}_4)}-\dfrac{(^{10}_4)}{(^{24}_4)}+0

P(AUBUC) = \dfrac{1001}{10626}+ \dfrac{1820}{10626}+ \dfrac{3060}{10626}-\dfrac{15}{10626}-\dfrac{70}{10626}-\dfrac{210}{10626} +0

The probability that at least one of the shifts will be unrepresented in the sample of workers is P(A∪B∪C) = 0.5256

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3 years ago
What is 14c12?<br> A. 78<br> B. 91<br> C. 105<br> D. 120
Delicious77 [7]

9514 1404 393

Answer:

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Step-by-step explanation:

14C12 = 14!/(12!(14-12)!) = 14·13/(2·1) = 7·13 = 91

5 0
3 years ago
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