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3 years ago

What is the solution to the system? [1] -3x+4y+2z=-3 [2] 2x-4y-z=0 [3] y=3x-13

Mathematics

1 answer:

Monica [59]3 years ago

8 0

Answer: x = 5

y = 2

z = 2

Step-by-step explanation:

-3x+4y+2z=-3 - ,- - - - - - 1

2x-4y-z=0 - - - - - - - - - - 2

y=3x-13 - - - - - - - - - - 3

We will use the method of substitution.

Substituting y = 3x -13 into equation 1 and equation 2, it becomes

-3x+4(3x-13)+2z=-3

-3x + 12x - 52 + 2z=-3+52

9x+2z = 49 - - - - - - -4

2x - 4(3x-13)-z= 0

2x - 12x + 52 - z = 0

-10x- z = -52 - - - - - - --5

Using elimination method to solve equation 4 and equation 5

Multiply equation 5 by 2 and equation 4 by 1

-20x - 2z = -104

9x+2z = 49

Adding both equations,

-11x = -55

x = -55/-11

x = 5

Substituting x = 5 into equation 3

y = 3x - 13

y = 3×5 -13 = 15-13

y = 2

Substituting x = 5 and y = 2 into equation 1,

-3x+4y+2z=-3

-3×5 + 4×2 +2z = -3

-15+8+2z = -3

-7+2z = -3

2z = -3+7

2z = 4

z = 4/2 = 2

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3 years ago

Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

$P[C]=p^n$

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.

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4 years ago