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inysia [295]
3 years ago
8

a company gives raises once-a-year amir's present salary of $16,000 his manager would like to raise amir's salary to 35,000. Wha

t percent raise would Amir be receiving?
Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

Subtract 16,000 from 35,000 to find the quantity increase.

35,000 - 16,000 = 19,000

Now divide 19,000 by 16,000 to find the percent increase.

19,000 ÷ 16,000 = ‭1.1875‬

Multiply this by 100 to change it from a decimal to a percentage.

‭1.1875‬ × 100 = 118.75

<h2>Answer:</h2>

<u>Amir received a </u><u>118.75%</u><u> raise.</u>

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Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat
anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

 M be the event of failure due to mechanical defects, P(M) = 0.88

 LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

 IA be the event of mechanical defect due to improper assembly, P(IA/M)   =0.73

 DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

 IC be the event of electrical connects due to improper connections,

  P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

  P(PWW/F) = 0.52

(b)                                     <u> </u><u>Keyboard failures</u>

<h2>                              /               \</h2>

           <u> </u><u> Faulty electrical connects   </u>            <u>Mechanical Defects </u>          

                      P(F) = 0.12                                             P(M) = 0.88

<h2>        /            |             \                  /            \</h2>

<u><em>Defective wires</em></u>  <u><em>Improper</em></u>        <u><em>Poorly</em></u>                  <u><em>Loose Keys</em></u>      <u><em>Improper</em></u><em> </em>

P(DW/F)=0.35   <u><em>Connections</em></u>   <u><em>Welded wires</em></u>      P(LK/M)=0.27   <em> </em><u><em>Assembly</em></u>

                           P(IC/F)=0.13     P(PWW/F)=0.52                            P(IA/M)=0.73              

This is the required tree diagram.

(c) Probability that a failure is due to loose keys is given by:

  P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose  

                                               keys}

    P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

     wires is given by P(IC \bigcup PWW) ;

 P(IC \bigcup PWW) = P(IC) + P(PWW) - P(IC \bigcap PWW) { Here P(IC \bigcap PWW) = 0 }

 P(IC) = P(IC/F) * P(F)  = 0.13 * 0.12 = 0.0156

 P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC \bigcup PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

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Suzy has a collection of quarters and dimes. Together she has 34 coins worth $6.40. How many quarters and how many dimes does Su
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Answer:

Quarters = 20

Dimes = 14

Step-by-step explanation:

Dime = x = 10 cent = $0.1

Quarter = y = 25 cent = $0.25

x + y = 34 - - - (1)

0.1x + 0.25y = 6.40 - - - - (2)

From (1)

x = 34 - y

0.1(34 - y) + 0.25y = 6.40

3.4 - 0.1y + 0.25y = 6.40

3.4 + 0.15y = 6.40

0.15y = 6.40 - 3.40

0.15y = 3

y = 3 / 0.15

y = 20

x = 34 - y

x = 34 - 20

x = 14

Quarters = 20

Dimes = 14

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If you round to the nearest thousands then it would be 4988000.
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2 years ago
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