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jarptica [38.1K]
3 years ago
9

On May 22, 1960, an earthquake in Chile measured 9.5 on the Richter scale. On February 27, 2010, another earthquake in Chile mea

sured 8.8 on the Richter scale. Compare the intensities of the two earthquakes. Round to the nearest whole number.
Mathematics
2 answers:
Talja [164]3 years ago
3 0
The answer would be 5
Zielflug [23.3K]3 years ago
3 0

Answer:

\frac{I_1}{I_2} =5

Step-by-step explanation:

We can use Richter scale formula

R=log(I)

where

I is intensity

R is Richter scale value

First case:

On May 22, 1960, an earthquake in Chile measured 9.5 on the Richter scale

so, we get

9.5=log(I_1)

Second case:

On February 27, 2010, another earthquake in Chile measured 8.8 on the Richter scale

so, we get

8.8=log(I_2)

we can subtract both equations

9.5-8.8=log(I_1)-log(I_2)

log(\frac{I_1}{I_2} )=0.7

\frac{I_1}{I_2} =10^{0.7}

\frac{I_1}{I_2} =5

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Read 2 more answers
Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A samp
shutvik [7]

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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