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Maslowich
3 years ago
14

Can someone help me on this question please? Factorize x²-5x-6

Mathematics
2 answers:
Bond [772]3 years ago
8 0

Answer:

x²-5x-6

Factor the expression by grouping. First, the expression needs to be rewritten as x²+ax+bx-6 to find a and b, set up a system to be solved.

a+d=-5

a+d=-5ab=1(-6)=-6

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

1,-6

1,-62,-3

Calculate the sum for each pair.

1−6=−5

1−6=−52-3=-1

The solution is the pair that gives sum -5.

a=-6

b=1

Rewrite x²5x-6 as

(x²-6x)+(x-6)

Factor out x in x²-6x.

x(x-6)+x-6

Factor out common term x−6 by using distributive property.

(x-6)(x+1)

<em><u>D</u></em><em><u>O</u></em><em><u>N</u></em><em><u>E</u></em>

Phoenix [80]3 years ago
7 0

Answer: (x−6)(x+1)

Step-by-step explanation:

Ask: Which two numbers add up to −5 and multiply to −6?

Answer: −6 and 1

Rewrite the expression using the above.

You will get (x−6)(x+1)

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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

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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 quality control consultant is to select 4 of these workers for in-depth interviews:

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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)}

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(^{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)}

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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

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

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