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2 years ago

Find the value of x. View picture for full problem. Please and thank you! m<2 = 6x + 12.

Mathematics

1 answer:

Karo-lina-s [1.5K]2 years ago

4 0

Answer:

x = 8

Step-by-step explanation:

All the sides are equal. So, it is an equilateral triangle. So, all angles are 60 each

6x + 12 = 60

6x = 60 - 12

6x = 48

x = 48/6

x = 8

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Write an equation of a line in slope intercept form that is parallel to the line x + 4y = 6 and passes through (-8, 5)

Annette [7]

If the line is parallel, then their slope would be equal.
4y = -x + 6
y = -1/4 x + 3/2

Compare it with, y = mx + b
m = -1/4 [ Slope ]

Now, y - y1 = m(x - x1)
y - 5 = -1/4 (x + 8)
y - 5 = -1/4x - 2
y = -1/4x + 3

In short, Your Answer would be: y = -1/4x + 3

Hope this helps!

3 0

3 years ago

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

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

5 0

3 years ago

Question 10 Multiple Choice Worth 1 points)

zheka24 [161]

Answer:

Linearly, because the table shows that the sunflowers increased by the same amount each month

Step-by-step explanation:

Given the table

$\begin{array}{lcccc}\text{Month}&1&2&3&4\\ \\\text{Sunflowers}&15&17.2&19.4&21.6\end{array}$

Note that months change one-by-one (21-1, 3-2=1, 4-3=1).

Also

$17.2-15=2.2\ [\text{from month 1 to month 2}]\\ \\19.4-17.2=2.2\ [\text{from month 2 to month 3}]\\ \\21.6-19.4=2.2\ [\text{from month 3 to month 4}]$