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mestny [16]
3 years ago
15

South Celestial Pole is at the South Point

Mathematics
1 answer:
9966 [12]3 years ago
3 0

Answer:

<em>The south celestial pole is the point in the sky directly above Earth's southern axis. It's the point around which the entire southern sky appears to turn. The height of the south celestial pole in your sky depends on your latitude.</em>

Step-by-step explanation:                              

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What is 2 with an exponent of negative 2 equal to?
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An environmental group conducted a study to determine whether crows in a certain region were ingesting food containing unhealthy
nydimaria [60]

Part A

Given info:

  • xbar = sample mean = 4.90 ppm
  • s = sample standard deviation = 1.12 ppm
  • n = 23 = sample size

Because n > 30 is not true, and we don't know the population standard deviation (sigma), this means we must use a T distribution.

The degrees of freedom here are n-1 = 23-1 = 22.

At 90% confidence and the degrees of freedom mentioned, the t critical value is roughly t = 1.717

Use a T distribution table or calculator to determine this. If you don't have a calculator for the task, then you can search out "inverse T calculator" and there are tons of free options to pick from.

The margin of error E is

E = t*s/sqrt(n)

E = 1.717*1.12/sqrt(23)

E = 0.400982

This is approximate and accurate to 6 decimal places.

The confidence interval is going to be xbar plus or minus that E value

L = lower bound = xbar - E = 4.90 - 0.400982 = 4.499018 = 4.50

U = upper bound = xbar + E = 4.90 + 0.400982 = 5.300982 = 5.30

The confidence interval in the format of (L, U) is (4.50, 5.30)

You could also express it as the format L < mu < U and it would be 4.50 < mu < 5.30; however, I'll stick to the first method.

<h3>Answer:  (4.50, 5.30)</h3>

=====================================================

Part B

Since we know sigma = 2.6 is the population standard deviation, we can use a Z distribution now.

At 90% confidence, the z critical value is roughly 1.645; use a table or calculator to determine this.

n = \left(\frac{z*\sigma}{E}\right)^2\\\\n \approx \left(\frac{1.645*2.6}{0.03}\right)^2\\\\n \approx 20325.254444 \\\\

Round this up to the nearest integer to get 20326. For min sample size problems, <u>always</u> round up.

<h3>Answer: 20326</h3>
6 0
2 years ago
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