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shtirl [24]
3 years ago
15

Please help me with this

Mathematics
2 answers:
AfilCa [17]3 years ago
6 0

20,000 + 70 + 3 = 20,073

Troyanec [42]3 years ago
6 0

To solve this problem, you need to understand place value concepts. Each digit added makes a number greater, but when adding numbers you have to make sure that you add your other digits in the right places.

To solve 20,000 + 70 + 3, first we can add 70 + 3. This may seem like a simple problem, but it is a good example of place values. Because 3 is in the ones place, you add it to the ones place of the other number (currently a 0). 3 + 0 = 3, so you replace the digit 0 with a 3, giving you the answer 73.

Now, we must solve 20,000 + 73 in a similar manner. Because 73 takes up the tens and the ones places, we should add these numbers to the preexisting digits in those values. Because they are zeros, your answer will still be 73, making your final answer 20,073.

Hope this helps!

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Rashid [163]
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3 0
4 years ago
An 18 ounce box of cereal costs $2.76. How many ounces should a box priced at $2.07 contain?
Darya [45]
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5 0
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)

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
What’s<br> -2x+y= -14 <br> 6x + 5y = -6
Katarina [22]
Hey, here's a pic of it ! hope this helps !

p.s., photomath is a good app for algebraic answers with work, you should check it out (:

3 0
3 years ago
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