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alexdok [17]
1 year ago
15

Un círculo con circunferencia igual a 20 tiene un arco con ángulo central igual a 72°.

Mathematics
1 answer:
Citrus2011 [14]1 year ago
6 0

Answer:

4

Step-by-step explanation:

\frac{20}{360}=\frac{x}{72}\\x = 4

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Three people become infected with a virus that spreads quickly. Each day that passes, the number of infected people doubles. How
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3x \times 2 = y
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M is directly proportional to the square of X
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Jonathon stated that 0.20 of his coins were quaters. What percentage of jonathons coins are quaters
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20%, because 20/100 is equal to 20&.
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Simplify the expression <br> 4(-8x+5)-(-33x-26)=
Pavel [41]

Answer:

x+46

Explanation:

4(−8x+5)−(−33x−26)

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=4(−8x+5)+−1(−33x−26)

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=4(−8x+5)+33x+26

Distribute:

=(4)(−8x)+(4)(5)+33x+26

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Combine Like Terms:

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3 0
3 years ago
A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

            \sigma = population standard deviation = 0.001 g

            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

                                             = [ 3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } } , 3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } } ]

                                             = [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
2 years ago
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