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leva [86]
3 years ago
8

A satellite in a circular orbit 1250 kilometers above Earth makes one complete revolution every 110 minutes. What is its linear

speed? Assume that Earth is a sphere of radius 6378 kilometers.
Mathematics
1 answer:
Ivenika [448]3 years ago
7 0

Answer:

\large \boxed{\text{approximately 435.7 km/min}}  

Step-by-step explanation:

1. Angular speed

The angular speed ω is the angle θ swept out by the satellite in a given time t.

\omega = \dfrac{\theta}{t} = \dfrac{2\pi}{\text{110 min}}

2. Linear speed

The formula for the linear speed v is

v = rω, where

r = the distance from the centre of the Earth = 6378 km + 1250 km = 7628 km

\begin{array}{rcl}v & = & r\omega\\& = & \text{7268 km} \times \dfrac{2\pi}{\text{110 min}}\\\\& \approx & \textbf{435.7 km/min}\\\end{array}\\\text{The linear speed of the satellite is $\large \boxed{\textbf{approximately 435.7 km/min}}$}

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Answer:

30

Step-by-step explanation:

He is walking 3 dogs an hour

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I need help with this I been trying to find the answer but I can’t
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The correct answer is 42

Step-by-step explanation:

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The health of the bear population in a park is monitored by periodic measurements taken from anesthetized bears. A sample of the
Vikki [24]

Answer:

182.167-2.03\frac{114.05}{\sqrt{36}}=143.580  

182.167+2.03\frac{114.05}{\sqrt{36}}=220.754  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

Step-by-step explanation:

Assuming the following dataset:

77, 349,417,349, 167 , 225, 265, 360,205

145,335,40,139, 177,108, 163, 202, 22

123,439, 125,135, 86,43, 217,49, 156

119,178, 151, 61, 350, 312, 91, 89,89

We can calculate the sample mean with the followinf formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}= 182.167

And the sample deviation with:

s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}=114.05

The sample size on this case is n =36.

Previous concepts  

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".  

\bar X=182.167 represent the sample mean  

\mu population mean (variable of interest)  

s=114.05 represent the sample standard deviation  

n=36 represent the sample size    

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

The point estimate of the population mean is \hat \mu = \bar X =182.167

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=36-1=35  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,35)".And we see that t_{\alpha/2}=2.03  

Now we have everything in order to replace into formula (1):  

182.167-2.03\frac{114.05}{\sqrt{36}}=143.580  

182.167+2.03\frac{114.05}{\sqrt{36}}=220.754  

So on this case the 95% confidence interval would be given by (143.580;220.754)  

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A movie theater offers 30% off the price of a movie ticket to students from your school. The regular price of a movie ticket is
Snowcat [4.5K]

Answer:

$5.95

Step-by-step explanation:

To get the discounted price, we need to find how much is taken off from the regular price.

x/8.50=30/100

30×8.50=255

255/100=2.55

$2.55

Next, we need to subtract $2.55 from $8.50 to get the discounted price.

8.50-2.55=5.95

$5.95

$5.95 is the discounted price.

<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u>!</u></em><em><u> </u></em><em><u>:</u></em><em><u>)</u></em>

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Step-by-step explanation:

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