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barxatty [35]
3 years ago
6

at the pet fair,darlene's dog weighed 5 times as much as leah's dog.together the dog weighed 84 pounds. how much did each dog we

igh?
Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0
Darlene's dog weighed 70 pounds. And leah's dog weighed 14 pounds. 
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superheros

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Mary predicts that the store will see 4,238 customers next week. Approximately how many high school students should the
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3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 139 millimeters,
FinnZ [79.3K]

Answer:

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters

Step-by-step explanation:

To solve this question, we need to understan 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, the sample means with size n of at least 30 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 = 139, \sigma = 6, n = 40, s = \frac{6}{\sqrt{40}} = 0.9487

Either the sample mean differs by 0.8 mm or less from the population mean, or it differs by more. The sum of these probabilities is decimal 1.

Probability it differs by less than 0.8mm

pvalue of Z when X = 139 + 0.8 = 139.8mm subtracted by the pvalue of Z when X = 139 - 0.8 = 138.2 mm

X = 139.8

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

By the Central Limit Theorem

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

Z = \frac{139.8 - 139}{0.9487}

Z = 0.84

Z = 0.84 has a pvalue of 0.7995

X = 138.2

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

Z = \frac{138.2 - 139}{0.9487}

Z = -0.84

Z = -0.84 has a pvalue of 0.2005

0.7995 - 0.2005 = 0.5990

What is the probability that the sample mean would differ from the population mean by more than 0.8 millimeters?

p + 0.5990 = 0.4010

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters

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Use sod six itch yah onto
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