Answer:
a. μ
3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ
±(t*s)/sqrt(n)
where:
μ = is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.
When talking about orbits, it would have to be a mixture of both A. and B. since Newton's first law, gravity plays a huge part in an orbit. However, the universal gravitation law also tells us the relationship between two massive objects in orbit. But to choose only one, it would have to be B. Newton's first law
satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radius 4R.
(i) What does the force exerted on the satellite then become?
eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(ii) What happens to the satellite's speed?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(iii) What happens to its period?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large</span></span>
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static electricity and friction
This would be typical of an elastic collision.
