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Dahasolnce [82]
2 years ago
14

find the distance between point A(-3,5) and point B(4,-6) in the coordinate plane. 1- What is the distance formula? 2- plug in t

he correct values from the problem, write the diatance formula with substituted values 3- simplify the expression, what is the distance between the two points?
Mathematics
2 answers:
IrinaK [193]2 years ago
4 0
<span>1- What is the distance formula?
distance = </span>√(x2-x1)² + (y2-y1)²<span>

2- plug in the correct values from the problem, write the diatance formula with substituted values
</span>distance = √(4-(-3))² + (-6-5)²
<span>
3- simplify the expression, what is the distance between the two points?
distance = </span>√170 = 13.04
zhuklara [117]2 years ago
3 0

Answer:

1- What is the distance formula?

distance = √(x2-x1)² + (y2-y1)²

2- plug in the correct values from the problem, write the diatance formula with substituted values

distance = √(4-(-3))² + (-6-5)²

3- simplify the expression, what is the distance between the two points?

distance = √170 = 13.04

Step-by-step explanation:

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Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

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The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

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