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Setler [38]
3 years ago
5

In Triangle XYZ, measure of angle X = 49° , XY = 18°, and

Mathematics
1 answer:
marissa [1.9K]3 years ago
6 0

Answer:

There are two choices for angle Y: Y \approx 54.987^{\circ} for XZ \approx 15.193, Y \approx 27.008^{\circ} for XZ \approx 8.424.

Step-by-step explanation:

There are mistakes in the statement, correct form is now described:

<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>

The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:

YZ^{2} = XZ^{2} + XY^{2} -2\cdot XY\cdot XZ \cdot \cos X (1)

If we know that X = 49^{\circ}, XY = 18 and YZ = 14, then we have the following second order polynomial:

14^{2} = XZ^{2} + 18^{2} - 2\cdot (18)\cdot XZ\cdot \cos 49^{\circ}

XZ^{2}-23.618\cdot XZ +128 = 0 (2)

By the Quadratic Formula we have the following result:

XZ \approx 15.193\,\lor\,XZ \approx 8.424

There are two possible triangles, we can determine the value of angle Y for each by the Law of Cosine again:

XZ^{2} = XY^{2} + YZ^{2} - 2\cdot XY \cdot YZ \cdot \cos Y

\cos Y = \frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ}

Y = \cos ^{-1}\left(\frac{XY^{2}+YZ^{2}-XZ^{2}}{2\cdot XY\cdot YZ} \right)

1) XZ \approx 15.193

Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]

Y \approx 54.987^{\circ}

2) XZ \approx 8.424

Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]

Y \approx 27.008^{\circ}

There are two choices for angle Y: Y \approx 54.987^{\circ} for XZ \approx 15.193, Y \approx 27.008^{\circ} for XZ \approx 8.424.

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tatuchka [14]

Answer:

Slope of Parallel Line: -7

Step-by-step explanation:

A set of parallel lines will ALWAYS have the same slope. Perpendicular lines will have complete opposite slopes, for example, the slope of a line perpendicular to this one would be 1/7. If the slope is positive, the other slope will be negative and vice versa. For perpendicular lines, you also have to flip the fractions.

8 0
3 years ago
An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than
Sedbober [7]

Testing the hypothesis, it is found that:

a)

The null hypothesis is: H_0: \mu \leq 10

The alternative hypothesis is: H_1: \mu > 10

b)

The critical value is: t_c = 1.74

The decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

c)

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

H_0: \mu \leq 10

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

H_1: \mu > 10

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: t_c = 1.74.

Hence, the decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem, we have that:

\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18

Thus, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}

t = 1.41

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0
3 years ago
The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance wo
Keith_Richards [23]
This question has this set of answer choices:

a) when the plane cuts three faces of the cube, separating one corner from the others

b) when the plane passes through a pair of vertices that do not share a common face

c) when the plane is perpendicular to the base and intersects two adjacent vertical faces

d) when the plane makes an acute angle to the base and intersects three vertical faces

e) not enough information to answer the question

The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others

You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.

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lakkis [162]

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We round that to the nearest 100th to get our final answer.

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LekaFEV [45]

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In graph(below) given line is passing through (-2,-4) and (2,2) .

Slope of the given line passing through (-2,-4) and (2,2) =\dfrac{-4-2}{-2-2}=\dfrac{-6}{-4}=\dfrac{3}{2}

Since parallel lines have equal slope . That means slope of the required line would be .

Equation of a line passing through (a,b) and has slope m is given by :_

(y-b)=m(x-a)

Then, Equation of a line passing through(-3, 1) and has slope =  is given by

(y-1)=\dfrac32(x-(-3))\\\\\Rightarrow\ y-1=\dfrac32(x+3)

Required equation: y-1=\dfrac32(x+3)

8 0
3 years ago
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