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givi [52]
3 years ago
5

PLEASE HELP ASAP!!!!!! WORTH 90 POINTS Find each angle measure

Mathematics
2 answers:
Ganezh [65]3 years ago
4 0

Answer:

a) UXW = 78 degrees

b) UWY = 48 degrees

c) WZX = 48 degrees

d) XYZ = 42 degrees

Step-by-step explanation:

Apply rules and theorems:

Angles in a triangle add up to 180 degrees.

Angles on a straight line add up to 180 degrees.

Right angles are 90 degrees.

Vertically opposite angles are equal.

With these we can solve for the angles.

kirill115 [55]3 years ago
3 0

Answer:

a) UXW = 78        b) UWY = 48         c)  WZX = 48          d) XYZ = 42

Step-by-step explanation:

There are many ways you can do this, but I'll just describe one for each.

a) Angle YWV is vertical to XWZ, so they are congruent. Angle XWZ is an alternate interior angle with angle UXW, so UXW = 78.

b) All the angles that revolve around point W must add up to 360. We already know 4 angles due to vertical angles, so 78 + 78 + 54 + 54 = 264.

360 - 264 = 96.   There are two angles left, and since they are vertical they must be congruent. 96/2 = 48, so angle UWY is 48 degrees.

c) We already know that the angle vertical to UWY is 48, so WZX is also 48 degrees because they are alternate interior angles.

d) Angle UWY is 48 degrees and angle WUY is 90 degrees because of the perpendicular line. Now we can find XYZ, the remaining angle.

90 + 48 = 138.           180 - 138 = 42.          So XYZ = 42 degrees.

I hope this helped. I'm open to any questions you may have. :)

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Ian wants to run 412 miles to get ready for his upcoming race. He has set up 3 markers the same distance apart to show how far h
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Answer:

Each marker will be exactly <u>103</u> miles apart.

Step-by-step explanation:

It is given that Ian wants to run 412 miles and has set up 3 markers the same distance apart.

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2 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
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