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d1i1m1o1n [39]
2 years ago
10

A vector in standard position has its terminal point at (–10, –4). What is the approximate direction angle of the vector?

Mathematics
2 answers:
Vlad [161]2 years ago
7 0

Answer:

202 degrees

Step-by-step explanation:

Nina [5.8K]2 years ago
6 0

Answer: D- 202° on edg

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The perimeter of a rectangle is 286 meters. Find the length and width if the length is an integer and the width is 5 times the n
Andreyy89

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See below

Step-by-step explanation:

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Determine whether y varies directly with x. If so, find the constant of variation k.
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Given the function f(x)=1/2x+8 find x so that, f(x)=10<br><br> please help im stumped
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2 years ago
The statistics of nequals22 and sequals14.3 result in this​ 95% confidence interval estimate of sigma​: 11.0less thansigmaless t
dimulka [17.4K]
Answer: no, the confidence interval for the standard deviation σ cannot be expressed as 15.7 \pm 4.7

There are three ways in which you can possibly express a confidence interval:

1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means <span>that the standard deviation can vary between 11.0 and 20.4

2) interval
</span>The two extremities will be inside a couple of round parenthesis, separated by a comma, always <span>the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.

3) point estimate </span><span>\pm margin of error</span>
This is the most common way to write a confidence interval because it shows straightforwardly some important information. 
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.

This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
\sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{\alpha / 2} } } \leq \sigma \leq \sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{1 - \alpha / 2} } }

which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.
4 0
3 years ago
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