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EastWind [94]
3 years ago
15

Thomas calculates 9×7 by thinking avout it as 70-7=63. Explain thomas' strategy.

Mathematics
2 answers:
asambeis [7]3 years ago
6 0
Thomas uses the distributive property 

9 X 7 = (10 X 7) - (1 X 7) 
          = 70 - 7
          = 63

9 groups of 7 is equal to 10 groups of 7 subtract 1 group of 7.
madam [21]3 years ago
4 0
10 x 7 = 70
1 x 7 = 7
9 x 7 = 70 - 7
9 x 7 = 63
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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
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Answer:

a) P[C]=p^n

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

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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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Answer:

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Step-by-step explanation:

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The triangles share a side so we cant use anything involving SSS SAS.

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