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3 years ago

Find the area of the sector.

Mathematics

2 answers:

Rasek [7]3 years ago

7 0

Answer:

104.72 m^2.

Step-by-step explanation

I'm taking the angle to be 30 degrees.

Area of sector = area of the circle * 30/360

= pi*20^2 * 30/360

= 400pi * 1/12

= 104.72 m^2.

Korvikt [17]3 years ago

3 0

Step-by-step explanation:

A=30/360*22/7*20

A=110/21m^2

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3 years ago

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3 years ago

Find the slope of the line passing through the points (-2,5) and (3/2, 2).

sveticcg [70]

Answer:

The answer is

Step-by-step explanation:

The slope of the line passing through two points can be found by using the formula

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are (-2,5) and (3/2, 2)

The slope is

<h3> $m = \frac{2 - 5}{ \frac{3}{2} + 2} \\ = - \frac{3}{ \frac{7}{2} } \\ = - 3 \div \frac{7}{2} \\ = - 3 \times \frac{2}{7}$ </h3>

We have the final answer as

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3 years ago

Which of the following tables represent functions? If the relation is not a function, indicate why it is not.

Varvara68 [4.7K]

Step-by-step explanation:

1. the table on the left is a function, with the formula : y = x²

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3 years ago

(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the

goblinko [34]

Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

$98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\ \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\ \\ =10.0023\pm2.33(0.000089) \\ \\ =10.0023\pm0.00021 \\ \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\ \\ =(10.0021, \ 10.0025)$

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.

Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

$\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\ \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\ \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\ \\ \Rightarrow n=4.66^2=21.7156\approx22$

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>

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4 years ago