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forsale [732]
3 years ago
13

Can someone help please? Easy math for others but I just can’t understand! Pic included!

Mathematics
1 answer:
____ [38]3 years ago
4 0
<h3>Answer:  Choice A) -16, 0, 0, 144</h3>

================================================

Work Shown:

plug in x = 1

y = x^2 - 8x - 9

y = (1)^2 - 8(1) - 9

y = 1 - 8(1) - 9

y = 1 - 8 - 9

y = -16

So x = 1 pairs up with y = -16

We can stop here since choice A is the only one that has -16.

So choice A must be the answer.

For the sake of completeness, let's plug in the other x values.

---------------------

plug in x = -1

y = x^2 - 8x - 9

y = (-1)^2 - 8(-1) - 9

y = 1 - 8(-1) - 9

y = 1 + 8 - 9

y = 0

x = -1 pairs up with y = 0

---------------------

plug in x = 9

y = x^2 - 8x - 9

y = (9)^2 - 8(9) - 9

y = 81 - 8(9) - 9

y = 81 - 72 - 9

y = 0

x = 9 pairs up with y = 0

---------------------

plug in x = -9

y = x^2 - 8x - 9

y = (-9)^2 - 8(-9) - 9

y = 81 - 8(-9) - 9

y = 81 + 72 - 9

y = 144

x = -9 pairs up with y = 144

---------------------

When you plug in x = 1, -1, 9 and -9, then the y values are y = -16, 0, 0 and 144 in that exact order.

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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)

c)

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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