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Amanda [17]
3 years ago
14

A crew consists of 1​ apprentice, 1​ journeyman, and 1 master carpenter. The crew receives a check for ​$4800 for a job they jus

t finished. A journeyman makes 200​% of what an apprentice​ makes, and a master makes 150​% of what a journeyman makes. How much does each person on the crew​ earn?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

Apprentice=$800

Journeyman=$1600

Master=$2400

Step-by-step explanation:

We have that all the crew won a total of $4800 then it is the salary of the apprentice, the journeyman and the master carpenter. It say that the journeyman makes 200% of the apprentice, then it means that the journeyman won the double of the apprentice. The master won 150% of what a journeyman, it means that the master won 1.5 of the salary of the journeyman.

With this information we can write the next set of equations.

A+J+M=\$4800\\J=2A\\M=1.5J

Using the second equation in the third equation we can know how much won the master in terms of the salary of the apprentice, then:

M=1.5J\\M=1.5(2A)\\M=3A

With this we could rewrite the first equation as:

A+J+M=\$4800\\A+2A+3A=\$4800\\6A=\$4800\\A=\dfrac{\$4800}{6}\\A=\$800

Then the aprentice earn $800, the journeyman $1600 and the master $2400

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Suppose babies born in a large hospital have a mean weight of 3242 grams, and a standard deviation of 446 grams. If 107 babies a
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Step-by-step explanation:

To solve this question, we have 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:

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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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 3242, \sigma = 446, n = 107, s = \frac{446}{\sqrt{107}} = 43.12

What is the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams

This is the pvalue of Z when X = 3242 + 40 = 3282 subtracted by the pvalue of Z when 3242 - 40 = 3202

X = 3282

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

By the Central Limit Theorem

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

Z = \frac{3282 - 3242}{43.12}

Z = 0.93

Z = 0.93 has a pvalue of 0.8238.

X = 3202

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

Z = \frac{3202 - 3242}{43.12}

Z = -0.93

Z = -0.93 has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams

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