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3 years ago

Given that a:b = 4:5 and b:c=3.2 find the ratio a:b:c in its simplest form

2 answers:

5 0

A:b = 4:5
b:c=3:2
b was 5 then it was 3, so you have to find the common multiple which is 15.
you’d have to times the 4 (a) by 3 since that 5x3 is 15
you’d have to times the 2 (c) by 5 since 5x3 is 15 again.
therefore the answer would be
a:b:c
12:15:10
I hope this makes sense it was kinda hard to explain sorry :)

Mkey [24]3 years ago

5 0

Hello,

Step-by-step explanation:

$\frac{a}{4} = \frac{b}{5}$ and $\frac{b}{3} = \frac{c}{2}$

or 5a = 4b and 2b = 3c ⇔ 15a = 12b and 10b = 15c

In fractions :

$\frac{a}{12} = \frac{b}{15}$ and $\frac{b}{15} = \frac{c}{10}$

We have :

a:b:c = 12:15:10

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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

Ms. Burkhard wants to trim the border below to get rid of the part that is torn. After

motikmotik

What is there to zoom inn?

3 0

3 years ago

Given the graph y = f(x), explain and contrast the effect of the constant c on the graphs y = f(cx) and y = cf(x).

Igoryamba

Given the graph y = f(x)

The graph y = f(cx), where c is a constant is refered to as horizontal stretch/compression

A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression is the squeezing of the graph towards the y-axis. A compression is a stretch by a factor less than 1.

If | c | < 1 (a fraction between 0 and 1), then the graph is stretched horizontally by a factor of c units.
If | c | > 1, then the graph is compressed horizontally by a factor of c units.

For values of c that are negative, then the horizontal compression or horizontal stretching of the graph is followed by a reflection across the y-axis.

The graph y = cf(x), where c is a constant is referred to as a vertical stretching/compression.

A vertical streching is the stretching of the graph away from the x-axis. A vertical compression is the squeezing of the graph towards the x-axis. A compression is a stretch by a factor less than 1.

If | c | < 1 (a fraction between 0 and 1), then the graph is compressed vertically by a factor of c units.
If | c | > 1, then the graph is stretched vertically by a factor of c units.

For values of c that are negative, then the vertical compression or vertical stretching of the graph is followed by a reflection across the x-axis.

3 0

3 years ago

Find the distance between the points (-21, 17) and (33, 17)

Norma-Jean [14]

Plug it into the distance formula.

d= (sqrt) (x2-x1)^2 +(y2-y1)^2

d= (sqrt) (17-17)^2 + (33-(-21)^2
d=(sqrt) (0)^2 + (33 +21)^2
d= (sqrt) (54)^2
d= (sqrt) 2916
d=54

The distance is 54 units.

I hope this helps!
~kaikers

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3 years ago

Question 1 of 10<br>Which of the following functions is graphed below?

Sedaia [141]

Attach picture, where is below

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3 years ago