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3 years ago

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In 2011, Japan experienced an intense earthquake with a magnitude of 9.1 on the Richter scale. In 2003, Japan experienced anothe

r intense earthquake that measured 8.3 on the Richter scale. Compare the intensities of the two earthquakes. Use a logarithmic model to solve. Round to the nearest whole number.

2 answers:

zepelin [54]3 years ago

6 0

Answer:

The intensity of the 2011 earthquake was about 6 times the intensity of the 2003 earthquake.

Step-by-step explanation:

To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities.

Convert the magnitudes to intensities and write them in exponential form.

R=logI

2011 earthquake:

9.1I=logI=109.1

2003 earthquake:

8.3I=logI=108.3

Form a ratio of the intensities.

intensity for 2011intensity for 2003

Substitute in the values and divide by subtracting the exponents to find

109.1108.3100.8≈6.

The intensity of the 2011 earthquake was about 6 times the intensity of the 2003 earthquake.

Your answer:

The intensity of the 2011 earthquake was about 15 times the intensity of the 2003 earthquake.

Reika [66]3 years ago

3 0

Answer:

The intensity of the 2011 earthquake was about 6 times the intensity of the 2003 earthquake.

Step-by-step explanation:

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3 years ago

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$x^4-81=0\\
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4 years ago

A number is greater than 50 and less than 85 the number has 13 and 3 as factors

alina1380 [7]

Answer:

The Answer is: 78.

Step-by-step explanation:

This one makes you think!

Let n = the number.

The number is greater than 50 and less than 85:

50 < n < 85

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So, multiply 13 * 3 = 39.

Multiply again by 2:

39 * 2 = 78.

78 can be factored as:

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The number is 78.

Hope this helps! Have an Awesome Day!! :-)

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4 years ago

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Answer:

Step-by-step explanation:

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3 years ago

A researcher collected data on the hours of TV watched per day from a sample of five people of different ages. Here are the resu

frozen [14]

Answer:

1. The least squares regression is y = -0.1015·x + 6.51

2. The independent variable is b) age

Please see attached table

Step-by-step explanation:

The least squares regression formula is given as follows;

$\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}}$

We have;

$\bar x$ = 24

$\bar y$ = 4

$\Sigma (x_i - \bar x) (y_i - \bar y)$ = -79

$\Sigma (x_i - \bar x)^2$ = 778

$\therefore \hat \beta =\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y} \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}} = \frac{-79}{778} = -0.1015$

The least squares regression is y = -0.1015·x + α

∴ α = y -0.1015·x = 6 - (-0.1015 × 5) = 6.51

The least squares regression is thus;

y = -0.1015·x + 6.51

2. The independent variable is the age b)

3. Steps to create an ANOVA table with α = 0.05

The overall mean = (43 + 30 + 22 + 20 + 5 + 1 + 6 + 4 + 3 + 6 )/10 = 14

There are 2 different treatment = $df_{treat} = 2 - 1 = 1$

There are 10 different treatment measurement = $df_{tot} = 10 - 1 = 9$

$df_{res} = 9 - 1 = 8$

$df_{treat} + df_{res} = df_{tot}$

The estimated effects are;

$\hat A_1 = 24 - 14 = 10$

$\hat A_2 = 4 - 14 = -10$

$SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000$

$\sum_{i}\SS_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y)= [(1 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (3 - 4)^2 + (6 - 4)^2] = 18$

$\sum_{i} S S_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y) ^2= [(43 - 24)^2 + (30 - 24)^2 + (22 - 24)^2 + (20 - 24)^2 + (5 - 24)^2] = 778$

$S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796$

$SS_{tot}$ = (43 - 14)² + (30 - 14)² + (22 - 14)² + (20 - 14)² + (5 - 14)² + (1 - 14)1² + (6 - 4 )² + (3 - 14)² + (6 - 14)² = 1796

$MS_{treat} = \dfrac{SS_{treat} }{df_{treat} } = \dfrac{1000}{1} = 1000$

$MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5$

F- value is given by the relation;

$F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05$

We then look for the critical values at degrees of freedom 1 and 8 at α = 0.05 on the F-distribution tables 5.3177

Hence; $F = 10.05 > F_{1,8}^{Krit}(5\%) = 5.3177$ , we reject the null hypothesis.

7 0

3 years ago