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vodomira [7]
3 years ago
7

A tank is filled with 1000 liters of pure water. Brine containing 0.07 kg of salt per liter enters the tank at 9 liters per minu

te. Another brine solution containing 0.06 kg of salt per liter enters the tank at 8 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 17 liters per minute. A. Determine the differential equation which describes this system. Let S(t) denote the number of kg of salt in the tank after t minutes. Then dSdt= B. Solve the differential equation for S(t). S(t)=
Mathematics
1 answer:
DiKsa [7]3 years ago
7 0

Answer: A. S(t)= 1000 + 0.63t + 0.48t - 17t

              B. S(t)= 984.11

Step-by-step explanation:

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Rewrite (120 + 20) using factoring.
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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

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A ​$220 suit is marked down by 30​%. Find the sale price.
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KiRa [710]

Answer:

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Step-by-step explanation:

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