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Igoryamba
3 years ago
7

Can someone help me i have no idea how to do this

Mathematics
2 answers:
rewona [7]3 years ago
5 0

Step-by-step explanation:

P=2+7sqrt2+6-2sqrt2+4+5sqrt2=2+6+4+sqrt2(7-2+5)=12+10sqrt2=2(6+5sqrt2)

Ratling [72]3 years ago
3 0

Answer:

Step-by-step explanation:

You have a right triangle,

, so basically, you add all the sides

2+7\sqrt{2}+6-2\sqrt{2}+4+5\sqrt{2}= 12+10\sqrt{2}

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What is the volume of a sphere with a radius of 6.4 in, rounded to the nearest tenth of
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3 years ago
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
777dan777 [17]

Answer:

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

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:

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 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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Answer:

The answer is c!

Step-by-step explanation:

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