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Jobisdone [24]
2 years ago
14

I don't remember sector area, and this is on my review for my geometry final

Mathematics
1 answer:
Fantom [35]2 years ago
8 0
\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad 
\begin{cases}
\theta=\textit{angle in degrees}\\
r=radius\\
---------\\
\theta=45\\
r=8
\end{cases}\implies A=\cfrac{45\cdot \pi \cdot 8^2}{360}
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6x – 2х + 8у – 6y simplified expression
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4x+2y

Step-by-step explanation:

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I dont know the answer to this question I dont know what I can put for the x and the y
harina [27]

Answer:

  (x, y) = (-4, 4)

Step-by-step explanation:

You find the values of x and y by working the problem using any of the methods of solving simultaneous equations that you have been taught.

Here, you observe that the y-coefficients are the same in each equation. That means you can cancel the y-terms by subtracting one equation from the other.

  (7x +6y) -(4x +6y) = (-4) -(8)

  3x = -12 . . . . simplify

  x = -4 . . . . . . divide by 3

Now, you can substitute this into either equation to find y.

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The solution to this system of equation is ...

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5 0
2 years ago
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In a survey, the planning value for the population proportion is . How large a sample should be taken to provide a confidence in
tatuchka [14]

Answer:

n=350

Step-by-step explanation:

Notation and definitions

n random sample taken  (variable of interest)

\hat p=0.35 estimated proportion  (value assumed)

p true population proportion

Confidence =0.95 or 95% (value assumed)

Me=0.05 (value assumed)

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.586  

And rounded up we have that n=350

4 0
2 years ago
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