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mestny [16]
2 years ago
14

They are arranged randomly because manual typewriters tended to jam if the user typed too fast - therefore the arrangement was i

ntended to slow early typists down.
Mathematics
2 answers:
solmaris [256]2 years ago
8 0

Answer:

Huh

Step-by-step explanation:

Huh......................

andre [41]2 years ago
5 0

Answer:

They are arranged randomly because manual typewriters tended to jam if the user typed too fast - therefore the arrangement was intended to slow early typists down.

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A bag contains 5 white, 3 black, and 2 green balls. Balls are picked at random.
Soloha48 [4]

Answer:

look down there

Step-by-step explanation:

First ball:

Probability of drawing a white ball is 5/8

Probability of drawing a black ball is 3/8

Second ball:

This depends on the first ball drawn, lets say you drew a white ball initially, 4 white balls are left out of 7 balls in total. The probability of a white ball in the second pick is 4/7.

Total probability of drawing two white balls is 5/8*4/7 (since they are independent events).

If you picked a black ball initially, picking another black ball would have a probability of 2/7, on similar grounds , total prob for 2 blacks would be 3/8*2/7.

The probability that you pick 2 balls of same color is (5/14 + 3/28) = 13/28. (Since they are mutually exclusive events)

6 0
3 years ago
Plz help me plz f(87) = -590.6 and f(132) = -901.1
quester [9]

Answer:

  ididk

Step-by-step explanation:

  ddkkdk

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 .

8 0
3 years ago
How do you solve the problem for this equation? 3+5x=3
Mazyrski [523]

Answer:

x=0

Step-by-step explanation:

3+5(0)=3

4 0
3 years ago
Read 2 more answers
Solve the compound inequality.
andreev551 [17]

Answer: 2u+4 <8= u<2

2u+4<8

Step 1: Subtract 4 from both sides.

2u+4−4<8−4

2u<4

Step 2: Divide both sides by 2.

2u/2<4/2

u<2

<u>4u-53 15</u>

=4u−795

Step-by-step explanation:

5 0
2 years ago
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