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KiRa [710]
3 years ago
7

The regular price of a grill is $599.99 if there is a 20% discount and the sales tax is 6%, what is the final price of a grill​

Mathematics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

$508.79 .............................................................

       

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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.

A

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
Write an inequality for the situation below a number is no more than -4?
Anarel [89]
X is less than or equal to -4 (couldn’t find the symbol)
7 0
2 years ago
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A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe
PSYCHO15rus [73]

Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2  bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

7 0
3 years ago
Which is the most accurate way to estimate 51% of 57?
kow [346]
Multiply 0.57 by 51
=29.07

----------------------------------------------------------------------

8 0
2 years ago
Read 2 more answers
100% of 2.34567 is what
worty [1.4K]

2.34567

Happy valentines day!

5 0
3 years ago
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