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4 years ago

(-2g+7)-(g+11) Find the difference! Thanks!

Mathematics

2 answers:

maksim [4K]4 years ago

5 0

Rewritten: -2g + 7 - g - 11

Grouping like terms: -2g - g + 7 -11

That equals to: -3g - 4

konstantin123 [22]4 years ago

4 0

-2g + 7 - g -11
-3g - 4

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part b is: if the energy release of one earthquake is 10,000 times that of another, how much larger is the Richter scale reading

kotegsom [21]

You could use the information from part A to get B. I'm not sure if you want A or not, so I'll do it as well.

Eo = 10^4.4 Joules

E = 2 * 10^15

Formula

M = (2/3) log (E/Eo)

M = 2/3 * log (2 * 10^15/10^(4.4) )

M = 2/3 * log( 7.9621* 10^10)

M = 2/3 * 10.901

M = 7.26735 on the Richter scale. That is a huge amount of energy.

Part B

Suppose that you use Eo and your base. Eo is 10^4.4

Now the new earthquake is E = 10000 * Eo

So what you get now is M = (2/3)* Log(10000 * Eo / Eo )

The Eo's cancel out.

M = 2/3 * log(10000)

M = 2/3 of 4

M = 8/3

M = 2.6667 difference in the Richter Scale Reading. It is still an awful lot of energy.

What this tells you is that if the original reading was (say) 6 then the 10000 times bigger reading would 8.266667

Answer: M = 2.6667

8 0

3 years ago

Read 2 more answers

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

3 0

4 years ago

Eight less than five times a number is fewer than twenty-four

Hoochie [10]

The clear answer according to me would be

5a-8<24

8 0

3 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

3 years ago

1.) 2x +y = 3

Vika [28.1K]

Okay I don’t think I have to do it anymore I

3 0

3 years ago