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Law Incorporation [45]
3 years ago
15

Find the arc length of the semicircle. 8

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

π×8² =

201.0619298297

arc length formular = π ×r²

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Find the ordered pair whose x-coordinate is 5. 4x=6y+8
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(6, 2 )

substitute x = 5 into the equation and solve for y

6y + 8 = 20 ( subtract 8 from both sides )

6y = 12 ( divide both sides by 6 )

y = \frac{12}{6} = 2

the ordered pair is (5, 2 )


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How to solve -27/x^15
kompoz [17]

Answer:

\frac{-27}{x^{15}}=-\frac{27}{x^{15}}

Step-by-step explanation:

<em>If you just want to simplify -27/x^15, it is not very difficult. Here is the simple way.</em>

<em />

Given the expression

\frac{-27}{x^{15}}

solving

\frac{-27}{x^{15}}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

=-\frac{27}{x^{15}}

Therefore,

\frac{-27}{x^{15}}=-\frac{27}{x^{15}}

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3 years ago
A researcher has two percentages and wants to know if the percentages are statistically different. The researcher calculates the
love history [14]

Answer:

p_v =2*P(Z>4.21) =2.55x10^{-5}

And we can use the following excel code to find it:"=2*(1-NORM.DIST(4.21,0,1,TRUE)) "

With the p value obtained and using the significance level assumed for example\alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the percentage 1 is significantly different from the percentage 2.

D) Are statistically different.

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Where \mu_1 and \mu_2 represent the percentages that we want to test on this case.

The statistic calculated is on this case was Z=4.21. Since we are conducting a two tailed test the p value can be founded on this way.

p_v =2*P(Z>4.21) =2.55x10^{-5}

And we can use the following excel code to find it:"=2*(1-NORM.DIST(4.21,0,1,TRUE)) "

With the p value obtained and using the significance level assumed for example\alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the percentage 1 is significantly different from the percentage 2.

And the best option on this case would be:

 D) Are statistically different.

4 0
3 years ago
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