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MAXImum [283]
2 years ago
6

What is the concentration of a new solution obtained by mixing 25ml of water with 125ml of a 20% salt solution in water?

Mathematics
1 answer:
expeople1 [14]2 years ago
5 0
Given that the concentration of salt in a 125 ml of salt solution is 20%, this means that the ratio of salt to water in the solution is 20 : 80 = 1 : 4.

Thus, the amount of salt in the 125 ml 20% salt solution is given by 20% of 125 = 0.2 x 125 = 25ml and the amount of water in the 125ml 20% salt solution is 125 - 25 = 100 ml.

When 25 ml of water is added to the 125 ml 20% salt solution, the new amount of water is 100 + 25 =125 and the amount of salt remain 25 ml.

Thus, the new volume of the solution is 125 + 25 = 150 ml and the proprtion of salt in the new solution is given by 25 / 150 = 0.1667

Therefore, the concentration of a new solution <span>obtained by mixing 25ml of water with 125ml of a 20% salt solution in water is 16.67%</span>
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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
1 year ago
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