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9514 1404 393

  3 < x < 9.09593

Step-by-step explanation:

In the attached, we defined TU=TW=y and TV=z. Then we used the Law of Cosines to write equations involving x, y, and z:

  42² = y² +z² -2yz·cos((6x-18)°)

  44² = y² +z² -2yz·cos((4x+2)°)

Solving each of these quadratics for z, and setting those solutions equal, gives a relation between x and y. We impose the constraints that x > 3, y > 0, and z > 0 so that angles and segment lengths are non-negative.

The attached graphs show the relation has solutions in the range of ...

  3 < x < 9.09593 (approximately)

_____

Comment on the question

Perhaps there is a simple way to solve this, but I haven't seen it yet. There is a set of constraints on x imposed by ∠UVW ≤ 180° and by ∠UTV ≥ 0°, but it turns out the geometry is actually more restrictive than that. The hard part of the solution is maintaining the required relation between angles VTU and VTW.

Added comment

The problem is symmetrical in that y and z are interchangeable. As the graph in the attachment shows, they converge to a single value at the extreme value of x. That is, the maximum x can be found by solving the equation ...

  (1 -cos((6x-18)°))/42² = (1 -cos((4x+2)°)/44²

That solution is x ≈ 9.09592829977.

3,000,540

step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.