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Ganezh [65]
3 years ago
7

A Factory processes 1,560 ounces of olive oil per hour. the oil is packaged into 24-ounces bottles. how many bottles does the fa

ctory fill in one hour
Mathematics
1 answer:
klasskru [66]3 years ago
3 0
1560 oz / 24 oz bottles = 65 bottles are filled per hr
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Given the following equation in point-slope form. Identify the slope and the point.
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Answer:

point = (-3, 5)

slope = 3/4

Step-by-step explanation:

The formula for point slope is:

y - y1 = m (x - x1)

point (x1, y1)

slope m

So with your equation:

y - 5 = 3/4 (x+ 3)

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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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Show all your work!!! <br><br> B is already wrong (TRUST ME)
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Answer:

yes

Step-by-step explanation:

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