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3 years ago

Solve the system of equations algebraically using the substitution and elimination 2x+y=5 and x-3y=6

Mathematics

1 answer:

algol133 years ago

8 0

The answer is x=3 and y=-1.

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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

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

Factor tree for 30 and 54

elena-14-01-66 [18.8K]

Answer:

30|2

15|3

5|5

54|2

27|3

9|3

3|3

5 0

4 years ago

Read 2 more answers

A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th

Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

$\tan\theta=\frac{Perpendicular}{base}$

Here, $\theta=11.1^{\circ}$ ,Perpendicular AB = 4000

$\tan\theta=\frac{AB}{BC}$

$\tan 11.1^{\circ} =\frac{4000}{BC}$

Solving for BC, we get,

$BC=\frac{4000}{\tan 11.1^{\circ} }$

$BC=\frac{4000}{0.196}$ (approx)

$BC=20408.16$ (approx)

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

8 0

3 years ago

There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit

DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = $nCr * p^{r} * q^{n-r}$

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = $15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}$

p(x=3) = 455 * $0.4^{3} * 0.6^{12}$

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337

3 0

3 years ago

The perimeter of the trapezoid-shaped window frame is 23.59 feet. Write and solve an equation to find the unknown side length x

krok68 [10]

Answer:

$5.62 + 3.65 + 5.62 + x = 23.59$

$x = 8.7 ft$

Step-by-step explanation:

Perimeter of the trapezoid-shaped window = sum of all the sides of the window frame

Thus, the equation to find the unknown side length, x, would be:

✔️ $5.62 + 3.65 + 5.62 + x = 23.59$

Solve for x

$14.89 + x = 23.59$

Subtract 14.89 from each side

$14.89 + x - 14.89 = 23.59 - 14.89$

✔️ $x = 8.7 ft$

4 0

3 years ago