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dolphi86 [110]
3 years ago
6

Determine the quadrant in which the terminal side of the given angle lies. -300°

Mathematics
1 answer:
Lilit [14]3 years ago
3 0

Answer:

-270 to -360 degrees I quadrant

And for this case since -300° is between -270 to -360 degrees we can conclude that this angle is in the I quadrant

Step-by-step explanation:

For this case we can solve this problem taking in count the following rule when we have negative angles:

0 to -90 degrees IV quadrant

-90 to -180 degrees III quadrant

-180 to -270 degrees II quadrant

-270 to -360 degrees I quadrant

And for this case since -300° is between -270 to -360 degrees we can conclude that this angle is in the I quadrant

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Answer:

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Step-by-step explanation:

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A kite is a <em>quadrilateral</em> which has <u>two</u> equal adjacent sides. Therefore <em>measuring</em> the lengths a<u>ccurately</u> would make the pieces of wood <em>perpendicular</em>. Since the diagonals of a kite are at <em>right angle</em> to each other.

<u>Quadrilaterals</u> are shapes which has four straight sides. Examples are; square, trapezium, kite, rectangle etc.

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Visit: brainly.com/question/21979359

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Answer:What’s the question-

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A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are
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Answer:

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

Step-by-step explanation:

Using a .10 significance level, can he conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree

The null hypothesis is that they are equal(that is, subtraction between the means is 0), while the alternate hypothesis is that they are greater(that is, subtraction between the means is larger than 0). So

H_0: \mu_1 - \mu_2 = 0

H_a: \mu_1 - \mu_2 > 0

In which \mu_1 is the mean pine trees height while \mu_2 is the mean spruce trees height.

The test statistic is:

z = \frac{X - \mu}{s}

In which X is the sample mean, \mu is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that \mu = 0

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Spruce trees - 30

The sample mean is given by the subtraction of the means. So

X = 35 - 30 = 5

Sample Size 40 80

Population variance 160 160

This means that the standard error for each sample is given by:

s_1 = \frac{\sqrt{160}}{\sqrt{40}} = 2

s_1 = \frac{\sqrt{160}}{\sqrt{80}} = \sqrt{2}

The standard error of the difference is the square root of the sum squared of the standard error of each sample.

s = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{6} = 2.4495

Test statistic:

z = \frac{X - \mu}{s}

z = \frac{5 - 0}{2.4495}

z = 2.04

Pvalue of the test:

Probability of a sample mean larger than 5, which is 1 subtracted by the pvalue of Z when X = 5.

Looking at the z-table, z = 2.04 has a pvalue of 0.9793.

1 - 0.9793 = 0.0207

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

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