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bazaltina [42]
1 year ago
11

Now let's assume that you have a newer toilet that uses 6 L per flush, but you are considering upgrading to a dual-flush toilet.

If you flush the toilet four times per day, but three of those times you could use the 3 L flush, how many liters of water would you save in 1 year
Mathematics
1 answer:
nydimaria [60]1 year ago
8 0

Answer: 3,285 Liters

Step-by-step explanation:

In the actual case, all your flushes consume 6 L of water.

You flush 4 times per day, then you consume 4 times 6 L per day, this is:

4*6L = 24 L per day.

And in one year there are 365 days, then in one year you consume:

365*24 L = 8,760 L

Now, if you upgrade the toilet, now per day you use:

one time 6 L and 3 times 3 L

Then in day, you use:

1*6 L + 3*3 L = 6L + 9L = 15L

And in the complete year, you will use 365 times that, which is:

365*15 L = 5,475 L

The amount of water that you would save is equal to the difference between the amount that you use in one year with no upgrade (8,760 L) and the amount that you would use in a year if you upgraded your toilet ( 5,475 L)

The difference is:

D = 8,760 L - 5,475 L = 3,285 L

This means that, in this situation, you would save 3,285 liters of water per year if you upgrade your toilet.

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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

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:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

Z = \frac{46 - 40}{5.66}

Z = 1.06

Z = 1.06 has a pvalue of 0.8554

X = 34

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

Z = \frac{34 - 40}{5.66}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

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Answer:

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Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

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Answer:

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Answer:

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