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4 years ago

Solve for n: 21k − 3n + 9p > 3p + 12

Mathematics

2 answers:

soldier1979 [14.2K]4 years ago

6 0

Answer: n<7k+2p-4

Step-by-step explanation:

saveliy_v [14]4 years ago

3 0

$21k-3n+9p\ \textgreater \ 3p+12
\\21k-3n+9p+(-21k)\ \textgreater \ 3p+12+(-21k)
\\-3n+9p\ \textgreater \ -21k+3p+12
\\-3n+9p+(-9p)\ \textgreater \ -21k+3p+12+(-9p)
\\-3n\ \textgreater \ -21k-6p+12
\\\frac{-3n}{-3}\ \textgreater \ \frac{-21k-6p+12}{-3}
\\n\ \textless \ 7k+2p-4$

Your answer is $n\ \textless \ 7k+2p-4$ .

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podryga [215]

Answer:

Step-by-step explanation:

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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 .

8 0

4 years ago

The average price of a movie ticket in the United States from 1980 to 2000 can be modeled by the expression, 2.75+0.10x, where x

MissTica

Answer:

Step-by-step explanation:

step one:

Let the cost be y

to solve for the cost of a movie ticket, we need to solve the expression

y=2.75+0.10x

step two:

given that x is the number of years between 1980-2000

the number of years is 20years

put x= 20 in the expression for the cost we have

y=2.75+0.10(20)

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4 years ago

Which equation is represented by the graph below?

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Where is the graph?

7 0

4 years ago

The sum of the squares of two consecutive positive integers is 41. Find the two

Komok [63]

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

16 + 25 = 41

<h3>Answer: 4 and 5</h3>

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41 use (a + b)² = a² + 2ab + b²

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41 subtract 41 from both sides

2n² + 2n - 40 = 0 divide both sides by 2

n² + n - 20 = 0

n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

n + 1 = 4 + 1 = 5

<h3>Answer: 4 and 5.</h3>

7 0

3 years ago