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makvit [3.9K]
3 years ago
5

What is the value of x ?enter your answer, as a decimal, in the box

Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

Final answer is x=125.45

Step-by-step explanation:

given that AB is parallel to NP

then ratio of corresponding segments must be equal

\frac{MA}{AN}=\frac{MB}{BP}

\frac{MN-AN}{AN}=\frac{MB}{BP}

\frac{80.5-35}{35}=\frac{x}{96.5}

\frac{45.5}{35}=\frac{x}{96.5}

Cross multiply

35x=45.5\cdot96.5

35x=4390.75

x=\frac{4390.75}{35}

x=125.45

Hence final answer is x=125.45

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Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A samp
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Answer:

0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 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.

Mean of 20, standard deviation of 5:

This means that \mu = 20, \sigma = 5

Sample of 170:

This means that n = 170, s = \frac{5}{\sqrt{170}}

What is the probability that a sample of 170 steady smokers spend between $19 and $21?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.

X = 21

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{\frac{5}{\sqrt{170}}}

Z = 2.61

Z = 2.61 has a p-value of 0.9955

X = 19

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

Z = \frac{19 - 20}{\frac{5}{\sqrt{170}}}

Z = -2.61

Z = -2.61 has a p-value of 0.0045

0.9955 - 0.0045 = 0.9910

0.9910 = 99.10% probability that a sample of 170 steady smokers spend between $19 and $21

3 0
3 years ago
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