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3 years ago

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A 25% vinegar solution is combined with triple the amount of a 45% vinegar solution and a 5% vinegar solution resulting in 20 mi

lliliters of a 30% vinegar solution.

1 answer:

Murrr4er [49]3 years ago

7 0

Complete Question: A 25% vinegar solution is combined with triple the amount of a 45% vinegar solution and a 5% vinegar solution resulting in 20 milliliters of a 30% vinegar solution.

Write an equation that models this situation and explain what each part represents in the situation. Then solve the equation.

Step-by-step explanation:

let the volume of the 45% vinegar solution be x litres ,

the volume of the 25% vinegar solution be 3x

and the volume of the 5% vinegar solution be y

summation of all gives,

x+3x+y = 20

4x + y = 20

subtracting 4x from both sides to make y the subject of the equation, we have;

y = 20 - 4x

To the concentration:

substituting for the value of x and y into equation x + 3x + y = 20, we have,

.45x + .25(3x) + .05(20-4x) = .3(20)

multiplying through by 100 to make whole figures we have;

45x + 25(3x) + 5(20-4x) = 600

45x + 75x + 100 - 20x = 600

120x - 20x + 100 = 600

100x + 100 = 600

subtracting 100 from both sides, we have;

100x = 600 - 100

100x = 500

dividing both sides by 100,

x = 5.

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NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400

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Answer:

The test statistics is $t = -1.727$

Step-by-step explanation:

From the question we are told that

The data given is

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The sample size is n = 16

The null hypothesis is $\mu \le 14400$

The alternative hypothesis is $H_a : \mu > 14400$

The sample mean is mathematically evaluated as

$\= x = \frac{\sum x_i}{n}$

So

$\= x = \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}$

=> $\= x = 10799.9$

The standard deviation is mathematically represented as

$\sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}$

So

$\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}} \ ..$

$..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2 }{16}} \ ...$

$\ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}$

=> $\sigma = 8340$

Generally the test statistic is mathematically represented as

$t = \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }$

$t = -1.727$

From the z-table the p-value is

$p-value = P(Z > t) = P(Z > -1.727) = 0.95792$

From the values obtained we see that

$p-value > \alpha$ so we fail to reject the null hypothesis

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Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63$

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

$Z = \frac{120.1 - 119}{1.63}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 117.9

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

$Z = \frac{117.9 - 119}{1.63}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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