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2 years ago

Add (a+b-3)+(b+2a-1)+(4-6b+2a)

Mathematics

2 answers:

igor_vitrenko [27]2 years ago

7 0

5a-4b is the answer

Mariulka [41]2 years ago

4 0

Answer:

$(a + b - 3) + (b + 2a - 1) + (4 - 6b + 2a)$

$a + 2a + 2a + b + b - 6b - 3 - 1 + 4$

$5a - 4b - 4 + 4$

$= 5a - 4b + 0$

$= 5a - 4b$

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Answer:

No, to be a function a relation must fulfill two requirements: existence and unicity.

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2 years ago

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Step-by-step explanation:

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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

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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) Keyboard failures

Faulty electrical connects Mechanical Defects

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

Defective wires Improper Poorly Loose Keys Improper

P(DW/F)=0.35 Connections Welded wires P(LK/M)=0.27 Assembly

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

This is the required tree diagram.

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 }

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Step-by-step explanation:

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2 years ago

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