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3 years ago

Need help urgent ... I will brainlist

Mathematics

1 answer:

victus00 [196]3 years ago

4 0

The new set of data would be
69, 83, 89, 94, 100 (from least to greatest)

Mean= 87 (You add all of the numbers in the set together, and divide by how many numbers there are)

Median= 89 (the middle number)

Mode= none (There is no number that repeats its self)

Range= 31 (Subtract the greatest number by the smallest number)

Hope this helps!

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3 years ago

Jack has 120 songs on his music player. Some are rock, some are jazz, and the rest of classical pieces. If his music player is o

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2/5 and 1/3 have a common denominator of 15. 6/15, 5/15, and 4/15 add up to 15/15. So the chances of rock are 4/15. 120 divided by 15 is 8. Six times eight equals fourty eight, five times eight equals forty, and four times eight equals thirty two. So there are 48 classical songs, 40 jazz songs, and 32 rock songs.

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3 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) 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 }

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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3 years ago