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larisa86 [58]
3 years ago
5

One eighth of the class graduated with an A average, 1/2 got a B, and the rest of the class had a C average. In a class of 800,

how many had a c average
Mathematics
2 answers:
pogonyaev3 years ago
8 0
I believe the answer would be 300. One eighth of the class would be 100 and 1/2 of the class would be 400. 800-(400+100)=300. Hope this helped!
EleoNora [17]3 years ago
8 0
1/8 <span> of the class graduated with an A average
</span><span>1/2= 4/8 got a B
</span><span>the rest of the class had a C average = 3/8

In a class of 800

800/8 = 100

so 

</span>1/8  of the class graduated with an A average = 100
1/2 got a B = 400
the rest of the class had a C average = 300

answer
300 had a C average.

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Step-by-step explanation:

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6 five dollar bill is $30.00

so 5 two dollar bill is $10.00

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5 0
2 years ago
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3 years ago
The amount of coffee that a filling machine puts into an 8 dash ounce 8-ounce jar is normally distributed with a mean of 8.2 oun
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Answer:

73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100​ jars, will be at most 0.02​ ounce

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theore.

Normal probability distribution

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

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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.

In this question, we have that:

\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100​ jars, will be at most 0.02​ ounce

That is, probability of the sample mean between 8.2-0.02 = 8.18 and 8.2 + 0.02 = 8.22, which is the pvalue of Z when X = 8.22 subtracted by the pvalue of Z when X = 8.18.

X = 8.22

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

By the Central Limit Theorem

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

Z = \frac{8.22 - 8.2}{0.018}

Z = 1.11

Z = 1.11 has a pvalue of 0.8665.

X = 8.18

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

Z = \frac{8.18 - 8.2}{0.018}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335.

0.8665 - 0.1335 = 0.7330

73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100​ jars, will be at most 0.02​ ounce

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