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SSSSS [86.1K]
2 years ago
15

Construct the confidence interval for the population mean. c=. 95 x=6. 4 o=. 3 and n=42.

Mathematics
1 answer:
Damm [24]2 years ago
3 0

The confidence interval is (63.909, 64.0907).

The confidence interval of a population mean is found using the formula:

CI = x ± z(σ/√n), where CI is the confidence interval, x is the mean, z is the z-score corresponding to the confidence level, σ is the standard deviation, and n is the population size.

In the question, we are given the confidence level, c = 95%.

Z-score corresponding to this, z = 1.96.

The mean of the population, x = 6.4.

The standard deviation of the population, σ = 0.3.

The population size, n = 42.

Thus, using the formula, the confidence interval is:

CI = 6.4 ± 1.96(0.3/√42),

or, CI = 6.4 ± 0.09073037.

Thus, the confidence interval is (64 - 0.09073037, 64 + 0.09073037), or, (63.909, 64.0907).

Learn more about the confidence interval at

brainly.com/question/17097944

#SPJ4

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Answer:

A straight line of approximately 75 meters, 1.4º north

Step-by-step explanation:

Hi, let's make it step by step to make it clearer

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in R^{2}. To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

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Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle

Now, let's calculate each vector length by calculating the norm.

\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.

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