All categories

3 years ago

After you analyzed demand, you took steps to make sure your business made sense financially.

Mathematics

1 answer:

Stells [14]3 years ago

3 0

Answer:

A store ships cans by weight. A small box can hold 3 to 5 pounds. A medium box can hold 5 to 8 pounds. A large box can hold 8 to 10 pounds. The weights of the cans are given below.

Drag cans into each box to show what the box could contain.

kfeffse

Step-by-step explanation:

A store ships cans by weight. A small box can hold 3 to 5 pounds. A medium box can hold 5 to 8 pounds. A large box can hold 8 to 10 pounds. The weights of the cans are given below.

Drag cans into each box to show what the box could contain.

You might be interested in

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

Drag and drop the correct answer into each box to complete the proof.

Rudiy27

1. algebra x
2. _______ a -----------b _______c _________d

3. right angle
4. quadrilateral
5. algebra project a
opposites
lines
angles
right angles
defines

4 0

3 years ago

A n =−1+3(n−1) an=−1+3(n−1) what is the 55 term in the sequence

Sindrei [870]

ANSWER

$a_{55}=161$

EXPLANATION

The general term for the sequence is

$a_n=-1+3(n-1)$

To find the 55th term, we have to substitute
$n = 55$
in to the general term and simplify.

This implies that,

$a_{55}=-1+3(55-1)$

$a_{55}=-1+3(54)$

$a_{55}=-1+162$

$a_{55}=161$

Therefore the 55th term is 161.

8 0

3 years ago

you increase your walking speed from 1 m/s to 3 m/s in a period of 1 s. What is your acceleration? A) 2 m/s ^2 B) 3 m/s ^2 C) 5

arsen [322]

Acceleration is to change in speed divided by the time for the change. Change in speed is end speed minus started speed.

3m/s-1m/s and equals to 2m/s.

The answer should be A. (2m/s^2.)

Hope it helped you!

Have a great day!

8 0

3 years ago

Read 2 more answers

Evaluate the expression:<br> 2 (m + 1) – n3 when m = -2 and n = 3

QveST [7]

Answer:

-15

Step-by-step explanation:

2 * (-4+2) = -6

-6 - 9 = -15

8 0

3 years ago