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Wewaii [24]
3 years ago
7

HELP PLEASE A satellite rotates 50 miles above Earth's atmosphere. An astronaut works on the satellite and sees the sun rise ove

r Earth. To the nearest mile, what is the distance from the astronaut to the horizon? (HINT: Earth's radius is about 4,000 miles.) PLEASE HELP ME PLEASE AND THANK YOU, SHOW ALL WORK.
Mathematics
1 answer:
dusya [7]3 years ago
8 0
I must warn you I am not certain about this answer but this is the best I can come up with. Let's assume that the astronaut is 50 miles above the surface of the earth, due east of the centre of the Earth. He will therefore see the satellite when it is 50 miles above the surface of the earth, due north of the centre of the Earth. You can draw a triangle ABC where AB is the distance from the earth's centre to the astronaut, AC is the distance from the centre to the horizon, and BC is the distance from the astronaut to the horizon. The radius of the earth is 4000 miles, so the distance from the centre to the astronaut is 4050miles. This is the same as the distance from the centre to the horizon. This means AB and AC are 4050 miles. You can then use pythagoras' theorem to find out the length from the astronaut to the horizon:
<span>√(4050^2+4050^2)=5727.56
</span>So the distance is 5728 miles from the astronaut to the horizon.
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For the hypotheses below test alpha = 0.025 with n = 100 and bar p = 0.67. State
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Answer:

Part a : Critical value

For this case we want a value of \alpha = 0.025 we need to find on the normal standard distribution a z score that accumulates 0.025 of th area on the left and this value is z_{crit}= -1.96

Part b: Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.67 -0.76}{\sqrt{\frac{0.76(1-0.76)}{100}}}=-2.107  

Part c: Statistical decision  

n \hat p= 100*0.67= 67 >5

n (1-\hat p)= 100*(1-0.67)= 33 >5

So then the normal approximation makes sense.

Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided and we can conclude that the true proportion is lower than 0.76

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

\hat p=0.67 estimated proportion of interest

p_o=0.76 is the value that we want to test

\alpha=0.025 represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.76:  

Null hypothesis:p\geq 0.76  

Alternative hypothesis:p < 0.76  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Part a : Critical value

For this case we want a value of \alpha = 0.025 we need to find on the normal standard distribution a z score that accumulates 0.025 of th area on the left and this value is z_{crit}= -1.96

Part b: Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.67 -0.76}{\sqrt{\frac{0.76(1-0.76)}{100}}}=-2.107  

Part c: Statistical decision  

n \hat p= 100*0.67= 67 >5

n (1-\hat p)= 100*(1-0.67)= 33 >5

So then the normal approximation makes sense.

Since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level provided and we can conclude that the true proportion is lower than 0.76

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Answer:

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