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strojnjashka [21]
3 years ago
13

How much of a 60% acid solution must be mixed with 50 gallons of a 24% acid solution to obtain a solution that is 50% acid?

Mathematics
2 answers:
icang [17]3 years ago
4 0
X gallons of a 60% acid solution contains 60%×x=0.6x gallons of acid.
50 gallons of a 24% acid solution contains 24%×50=0.24×50=12 gallons of acid.

After mixing, there will be x+50 gallons of a solution, and 0.6x+12 gallons of acid. The concentration must be 50%, so the volume of acid divided by the volume of the solution is 50%, or 0.5.

\frac{0.6x+12}{x+50}=0.5 \\
0.6x+12=0.5(x+50) \\
0.6x+12=0.5x+25 \\
0.6x-0.5x=25-12 \\
0.1x=13 \\
0.1x \times 10=13 \times 10 \\
x=130

The answer is A. 130 gallons.
Alex787 [66]3 years ago
3 0

Answer:

a. 130

Step-by-step explanation:

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The mean per capita consumption of milk per year is 131 liters with a variance of 841. If a sample of 132 people is randomly sel
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Answer:

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

The mean per capita consumption of milk per year is 131 liters with a variance of 841.

This means that \mu = 131, \sigma = \sqrt{841} = 29

Sample of 132 people

This means that n = 132, s = \frac{29}{\sqrt{132}}

What is the probability that the sample mean would be less than 133.5 liters?

This is the p-value of Z when X = 133.5. So

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

By the Central Limit Theorem

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

Z = \frac{133.5 - 131}{\frac{29}{\sqrt{132}}}

Z = 0.99

Z = 0.99 has a p-value of 0.8389

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

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