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andriy [413]
3 years ago
9

In ΔSTU, s = 6.1 cm, t = 3.2 cm and u=7.4 cm. Find the measure of ∠S to the nearest 10th of a degree.

Mathematics
2 answers:
GuDViN [60]3 years ago
8 0

Answer:

m∠S = 54.1°

Step-by-step explanation:

Use the law of cosines

s^{2}  = t^{2} + u^{2}  - 2utcos S\\6.1^{2} = 3.2^{2}  + 7.4^{2}  - 2(3.2)(7.4) cos S

37.21 = 10.24 + 54.76 - 47.36 cos S

-27.79 = -47.36 cos S

.586782 = cos S

arccos .586782 = S

m∠S = 54.1°

MA_775_DIABLO [31]3 years ago
7 0

Answer:

54.1°

Step-by-step explanation:

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The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.972 g and a standard deviation
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Answer:

P( ¯ x < 0.929 g) = 0.1867

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 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 = 0.972, \sigma = 0.289, n = 36, s = \frac{0.289}{\sqrt{6}} = 0.0482

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 36 cigarettes with a mean of 0.929 g or less. P( ¯ x < 0.929 g) =

This is the pvalue of Z when X = 0.929. So

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

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

Z = \frac{0.929 - 0.972}{0.0482}

Z = -0.89

Z = -0.89 has a pvalue of 0.1867. So

P( ¯ x < 0.929 g) = 0.1867

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For the given data set

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<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

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Minimum = 48

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Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

Learn more on Measures of a Data here: brainly.com/question/15097997

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