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mina [271]
4 years ago
6

The magnitude, M, of an earthquake is defined to be M=log I/S, where I is the intensity of the earthquake (measured by the ampli

tude of the seismograph wave) and S is the intensity of a “standard” earthquake, which is barely detectable. What is the magnitude of an earthquake that is 35 times more intense than a standard earthquake? Use a calculator. Round your answer to the nearest tenth.
–1.5?
–0.5?
1.5?
3.6?
Mathematics
2 answers:
Nitella [24]4 years ago
7 0
The answer is 1.5 have a good day
Brilliant_brown [7]4 years ago
6 0
Given:
M = log I/S

M = magnitude
I = intensity of the earthquake
S = intensity of the standard earthquake

The minimum intensity of a standard earthquake is 10.
Intensity of the earthquake is 35 times the standard earthquake. So,
10 x 35 = 350

M = log 350/10
M = log 35
M = 1.544 or 1.5  Third option
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The American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day. Legacy
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We conclude that the null hypothesis is rejected which means residents of Legacy Ranch use less water on average.

Step-by-step explanation:

We are given that the American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day.

Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day.

<u><em>Let </em></u>\mu<u><em> = average water usage by residents of Legacy Ranch.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 67 gallons     {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}

Alternate Hypothesis, H_A : \mu < 67 gallons    {means that the residents of Legacy Ranch use less water on average}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                                 T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean water use per day = 63 gallons

             s = sample standard deviation = 8.1 gallons

             n = sample of responded = 23

So, <u><em>test statistics</em></u>  =  \frac{63-67}{\frac{8.1}{\sqrt{23} } }  ~ t_2_2

                               =  -2.368

The value of the test statistics is -2.368.

<em>Now at 0.025 significance level, the </em><u><em>t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test</em></u><em>. Since our test statistics is less than the critical values of t as -2.368 < -2.074, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to </em><u><em>which we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the residents of Legacy Ranch use less water on average.

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4 years ago
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