The equation A = lw is an example of an

On April 11, 2012, two earthquakes were measured off the northwest coast of Sumatra. The first had a magnitude of 8.6. The secon

Hoochie [10]

Answer:

Option D. 2.51 is correct.

Step-by-step explanation:

Magnitude of first earthquake = 8.6

Let intensity of first earthquake be x

$Magnitude = \log(\frac{x}{S}),\text{ where S is seismographic factor}\\\\\implies 8.6=\log(\frac{x}{S}).......(1)$

Magnitude of second earthquake = 8.2

Let intensity of second earthquake be y

$Magnitude = \log(\frac{y}{S}),\text{ where S is seismographic factor}\\\\\implies 8.2=\log(\frac{y}{S}).......(2)$

Now, to find the factor by which intensity of first earthquake is greater than intensity of second : we divide equation (1) by equation (2)

$\implies\log(\frac{x}{y})=8.6 - 8.2\\\\\implies \log(\frac{x}{y})=0.4\\\\\implies\frac{x}{y}=10^{0.4}\\\\\implies\frac{x}{y}=2.51$

Hence the required factor is 2.51

Therefore, Option D. 2.51 is correct.

4 0

2 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

2 years ago

What is the slope of the line that goes through (-5,5) and (5,-7)

sladkih [1.3K]

Answer:

-1/5

Step-by-step explanation:

slope formula:

m = y₂ - y₁ / x₂ - x₁

Two points are given:

(-5, -5) (5, -7)

m = -7 - (-5) / 5 - (-5)

m = -2 / 10

m = -1/5

7 0

2 years ago

Read 2 more answers

Which is a reasonable unit to express a ropes length ? PLZ ANSWER FAST

Mrac [35]

Answer:

Depends on the type of rope but I would say feet

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The steps for adding integers are the same as for subtracting them. <br><br> True or False?

lisov135 [29]

Answer:

I'm really going to say false im not sure, I just started learning this.

Step-by-step explanation:

4 0

2 years ago

The equation A = lw is an example of an​

The equation A = lw is an example of an