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ser-zykov [4K]
2 years ago
13

9 is 50% of what number?

Mathematics
2 answers:
Nadusha1986 [10]2 years ago
8 0

Answer:

18

Step-by-step explanation:

18*0.50=9

hope this helped!!

Mashcka [7]2 years ago
6 0

its 18 bc half of 18 is 9 so ye

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Translate the sentence into an equation. Five more than the product of a number and 9 is equal to 6
kap26 [50]
The first thing we will do for this case is to define the following variable:
 x = real number.
 We then have the following sentence:
 "Five more than the product of a number and 9 is equal to 6"
 The equation sought is:
 9x + 5 = 6
 Answer:
 
An equation for the sentence is:
 
9x + 5 = 6
6 0
3 years ago
If you were 2 years old in 2011 how old would you be in 2020?
Roman55 [17]

Answer:

11 years

Step-by-step explanation:

2 years that you were in 2011

2020-2011= 9

2+9=11

also meaning they were born in 2009

2020-2009 also equals 11

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2 years ago
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4 mi = ___ ft <br> answer please!!
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Answer:

21120

Step-by-step explanation:

1 mile = 5280 feet

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4 0
2 years ago
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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levacccp [35]
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