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netineya [11]
3 years ago
7

Sandra saves 10% of her salary for retirement. This year her salary was $2,000 more than in the previous year, and she saved $4,

200. What was her salary in the previous year?
Mathematics
2 answers:
Ede4ka [16]3 years ago
7 0

Answer:

2200................ here you go

telo118 [61]3 years ago
6 0

Answer:

2200 :)

Step-by-step explanation:

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Find the common factor and rewrite. Type your answer in the box. Do not use any spaces.
s2008m [1.1K]

Answer:

4X is the common factor

4X(y+2Z)

3 0
3 years ago
2. Find angle p.<br> A) 58<br> B) 67<br> C) 90<br> D)116
BARSIC [14]

Answer:

90°

Step-by-step explanation:

because 180 - 90

remember the quadrilateral must equal to 360 (all angles)

3 0
4 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

(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

5 0
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The red arrow is pointing to what part of the xy-plane?
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