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pychu [463]
3 years ago
9

JUST NEED ANSWER THANKS people never answer my questions :(

Mathematics
2 answers:
miss Akunina [59]3 years ago
7 0
The answer is 10 because m^-2 equals -2*-2.Which equals 4 and then 7*-2 equals-14.So 4- -14 equals 18.Then 18-8 equals 10.
Stolb23 [73]3 years ago
5 0
(-2)^{2} is 4.

Then, 7(-2) is -14.

Then, 4 - (-14) is (-18).

Finnaly, (-18) - 8 is [-26]
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Given f(x) = 9x - 5 and g(x) = x², choose<br> the expression for (fºg)(x).
svetlana [45]

Answer:

  9x² -5

Step-by-step explanation:

  (f○g)(x) means f(g(x))

Put the definition of g(x) in place of x in the definition of f(x):

  f(g(x)) = f(x²) = 9(x²) - 5

  (f○g)(x) = 9x² - 5

5 0
3 years ago
Consider the Inequality <br> —2n +7 &gt; -11<br> What is the solution of the inequality
Marta_Voda [28]

Answer:

  • n < 9 or n = (-oo, 9)

Step-by-step explanation:

  • -2n +7 > -11
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5 0
3 years ago
A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the pop
pantera1 [17]

Answer:

The initial population at the beginning of the 10 years was 129.

Step-by-step explanation:

The population of the village may be modeled by the following function.

P(t) = P_{0}e^{rt}

In which P is the population after t hours, P_{0} is the initial population and r is the growth rate, in decimal.

In this problem, we have that:

P(10) = 158, r = 0.02.

So

158 = P_{0}e^{0.02*10}

P_{0} = 158*e^{-0.2}

P_{0} = 129

The initial population at the beginning of the 10 years was 129.

6 0
4 years ago
Read 2 more answers
Please answer its urgent‼️‼️‼️
loris [4]

Answer:

-30 m

Step-by-step explanation:

lmk if its right

5 0
3 years ago
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
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