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Mama L [17]
3 years ago
10

Diana uses 30 grams of coffee beans to make 48 fluid ounces of coffee. When company comes she makes 100 fluid ounces of coffee.

How many grams of coffee beans does Diana use when company comes?
Mathematics
2 answers:
mestny [16]3 years ago
8 0
Assuming a direct proportionality between the grams of coffee and the ounces of fluid made,

30 g ------- 48 ounces
x   g ------- 100 Ounces (when company comes)

Then,
x = (30*100)/48 = 62.5 grams of coffee are used when company comes.
Firlakuza [10]3 years ago
6 0

Answer:

62.5

Step-by-step explanation:

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3 years ago
Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample
fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

Z = \frac{0.38 - 0.37}{0.0322}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

Z = \frac{0.26 - 0.37}{0.0322}

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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