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OleMash [197]
4 years ago
14

The Schwartz family spent a total of $111.75 for Internet service for 3 months. Each month they received $5.50 as a credit on th

e bill. Which equation and solution shows the cost of their monthly Internet service before the credit?
Mathematics
2 answers:
Lesechka [4]4 years ago
4 0
3(x+5.50)=111.75 .       $31.75

kupik [55]4 years ago
3 0

Answer:

The cost of their monthly Internet service before the credit is $31.75

The required equation =  3(x+5.50)=111.75

Step-by-step explanation:

Let the monthly cost be x

Each month they received $5.50 as a credit on the bill.

The Schwartz family spent a total of $111.75 for Internet service for 3 months.

So, total cost of three moths = 3(x+5.50)

Since we are given that The Schwartz family spent a total of $111.75

So,  3(x+5.50)=111.75

x+5.50=\frac{111.75}{3}

x+5.50=37.25

x=37.25-5.50

x=31.75

Hence the cost of their monthly Internet service before the credit is $31.75

The required equation =  3(x+5.50)=111.75

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Read 2 more answers
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.

6 0
4 years ago
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