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BlackZzzverrR [31]
3 years ago
15

Estimate the diameter of the supergiant star Betelgeuse, using its angular diameter of about 0.05 arcsecond and distance of abou

t 600 light years.
Express your answer to three significant figures and include the appropriate units.
Mathematics
2 answers:
8090 [49]3 years ago
7 0

<span>0.05 arc-second = 1 degree/72000 = (pi radians)/(180*72000) = 2.424 x 10^(-7) radians</span>

<span>The distance is roughly: </span>

<span>R*(theta) = (600 light-years)*2.424 x 10^(-7) = 0.00014544 light-years = 1.275 light-hours = (3600 seconds)*(3 x 10^8 m/s)*(1.275) = 1.38 x 10^12 meters.</span>

forsale [732]3 years ago
5 0

Answer:

Diameter of the star, d=1.38\times 10^{12}\ m

Step-by-step explanation:

Given that,

Angular diameter of the star, angle, \theta=0.05\ arcsecond=2.42\times 10^{-7}\ radian

Distance, D=600\ ly=5.67\times 10^{18}\ m

We need to find the diameter of the supergiant star Betelgeuse. The relationship between the diameter of star and angle subtended is given by :

\theta=\dfrac{d}{D}, d = diameter

d=\theta\times D

d=2.42\times 10^{-7}\times 5.67\times 10^{18}

d=1.37214\times 10^{12}\ m

or

d=1.38\times 10^{12}\ m

Hence, this is the required solution.

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|x-99|<3

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|x-99|<3

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4 years ago
If the area of a square is 100 cm2, what is the perimeter?<br>​
Vitek1552 [10]

Answer:

40

Step-by-step explanation:

To find the perimeter, first we need to find the side lengths.

If 100 cm is the area then we need to find what number squared is 100

10 x 10 = 100

So 10 is the side length. P= s + s + s +s

10 + 10 + 10 +10 = 40

6 0
3 years ago
Use the graph below to find the coordinates of J'after J(-5, 1) was reflected over the linex=-3.
k0ka [10]

Answer:

J's would be (-5,-4) if J was reflected over the line x=3.

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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
3 years ago
Solve the equation e/3 = 7 e= (answer)
Bond [772]

Answer:

21

Step-by-step explanation:

e/3 = 7 you can rewrite as e/3 = 7/1 and cross multiply e*1 = 3*7 so e= 21

5 0
3 years ago
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