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Anna007 [38]
1 year ago
12

Find the value of each variable and the measure of each labeled angle.

Mathematics
2 answers:
labwork [276]1 year ago
7 0

Answer:

see explanation

Step-by-step explanation:

(8x - 75) and (5x) are vertically opposite angles and are congruent , then

8x - 75 = 5x ( subtract 5x from both sides )

3x - 75 = 0 ( add 75 to both sides )

3x = 75 ( divide both sides by 3 )

x = 25

Then

5x = 5(25) = 125°

8x - 75 = 8(25) - 75 = 200 - 75 = 125°

Sholpan [36]1 year ago
4 0

Answer: (5x)°=(8x-75)°=125°

Step-by-step explanation:

<em>Property of vertical angles: vertical angles are equal</em>

<em>Hence,</em>

<em></em>(8x-75)^0=(5x)^0\\(8x)^0-75^0=(5x)^0\\(8x)^0-75^0-(5x)^0=(5x)^0-(5x)^0\\(3x)^0-75^0=0\\(3x)^0-75^0+75^0=0+75^0\\(3x)^0=75^0\\

<em>Divide both parts of the equation by 3 :</em>

<em></em>x^0=25^0\\(5x)^0=(5*25)^0\\(5x)^0=125^0

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valentina_108 [34]
Radius = d/2 = 1.7 in

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3 years ago
The probability that a randomly selected 3-year-old male chipmunk will live to be 4 years old is 0.96516.
mezya [45]

Using the binomial distribution, it is found that there is a:

a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

-----------

For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.96516 probability of a chipmunk living through the year, thus p = 0.96516

Item a:

  • Two is P(X = 2) when n = 2, thus:

P(X = 2) = C_{2,2}(0.96516)^2(1-0.96516)^{0} = 0.9315

The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.

Item b:

  • Six is P(X = 6) when n = 6, then:

P(X = 6) = C_{6,6}(0.96516)^6(1-0.96516)^{0} = 0.80834

The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.

Item c:

  • At least one not living is:

P(X < 6) = 1 - P(X = 6) = 1 - 0.80834 = 0.19166

The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.

A similar problem is given at brainly.com/question/24756209

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Answer:

Step-by-step explanation:

57 divided by 6 = 7.5

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