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Simora [160]
3 years ago
9

Question 30 The Royal Fruit Company produces two types of fruit drinks. The first type is pure fruit juice, and the second type

is pure fruit juice. The company is attempting to produce a fruit drink that contains pure fruit juice. How many pints of each of the two existing types of drink must be used to make pints of a mixture that is pure fruit juice
Mathematics
1 answer:
Flura [38]3 years ago
4 0

Answer:

The answer is below

Step-by-step explanation:

The Royal Fruit Company produces two types of fruit drinks. The first type is 65% pure fruit juice, and the second type is 90% pure fruit juice. The company is attempting to produce a fruit drink that contains 85% pure fruit juice. How many pints of each of the two existing types of drink must be used to make 80 pints of a mixture that is 85% pure fruit juice?

Answer: Let x be the number of pints of the first fruit juice (i.e 65%) and y be the number of pints of the second fruit juice (i.e 90%).

Since the total number of pints to make the 85% pure fruit juice is 80, it can be represented using the equation:

x + y = 80                              .                  .                 .    1)

Also, x pints of the first juice = 0.65x, y pints of the second juice = 0.9y and 80 pints of the mixture to be produced = 80(0.85) = 68. Therefore:

0.65x + 0.9y = 68               .                 .                  .        2)

We have to solve equation 1 and 2 simultaneously, first multiply equation 1 by 0.65 to get equation 3:

0.65x + 0.65y = 52            .                  .               .        3)

Subtract equation 3 from 2 and solve for y:

0.25y = 16

y = 16/0.25 = 64

y = 64 pints

Put y = 64 in equation 1:

x + 64 = 80

x = 80 - 64 = 16

x = 16 pints

Therefore 16 pints of the 65% pure fruit juice, and 64 pints of the 90% pure fruit juice is required to make 80 pints of 80% fruit juice.

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Step-by-step explanation:

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8 0
2 years ago
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
3 years ago
Someone help me pls!!!!
Alecsey [184]

Answer:

the answer is c

Step-by-step explanation:

3 0
2 years ago
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