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jasenka [17]
2 years ago
10

a solution of salt and water is 5 parts of salt to 25 parts of water . express the ratio of the amount of salt to the amount of

water. what percentage is the solution.​
Mathematics
1 answer:
labwork [276]2 years ago
4 0

Answer:

5 to 25 = 1 to 5 = 1/5 = 20%

Step-by-step explanation:

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Question number 1 You are comparing the costs of producing jewelry at two different manufacturers. Company 1 charges $3.25 per p
Assoli18 [71]

The value of x at equal total cost is 266.67

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\begin{gathered} 3.25x\text{ + 400 = 2.50x + 600} \\ 3.25x-2.50x\text{ = 600-400} \\  \\ 0.75x\text{ = 200} \\  \\ x\text{ = }\frac{200}{0.75} \\ x\text{ = 266.67} \end{gathered}

4 0
1 year ago
Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, 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, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

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3 years ago
What’s another fraction for 4/10
kaheart [24]

Answer:

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Step-by-step explanation:

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