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Ksenya-84 [330]
3 years ago
14

In a right triangle, the acute angles measure 2x+50 and 3x degrees

Mathematics
2 answers:
Alika [10]3 years ago
8 0
Triangle int. angles add up to 180 degrees.
In a right triangle, there are 2 acute angles and 1 right angle.

Here are the angles given:
a1) 90°
a2) (2x + 50)°
a3) (3x)°

Use the equation of the sum of interior angles on a triangle and solve for x.

180 = a1 + a2 + a3
180 = 90 + (3x) + (2x + 50)
180 = 90 + 5x + 50
180 = 140 + 5x
  40 = 5x
  /5     /5
    8 = x

Therefore x = 8. Now we can put these values into the acute angles to get the measurement of all angles.

a1 = 90

a2 = 2x + 50
     = 2(8) + 50
     = 16 + 50
     = 66

a3 = 3x
     = 3(8)
     = 24

Therefore, the measurement of all the angles in the triangle is 90°, 66°, and 24°.

Proof:

180 = a1 + a2 + a3
180 = 90 + 66 + 24
180 = 156 + 24
180 = 180

So it is true. Hope I helped.
xenn [34]3 years ago
6 0
2x+50 + 3x= 90 degrees
5x +50 = 90
5x =40
x=8

Angle1= 2(8) +50= 66
Angle2= 3(8)=24

66+24+90= 180
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a) 48.80% probability that his travel time to work is less than 30 minutes

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 30.7, \sigma = 23

a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?

This is the pvlaue of Z when X = 30. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{30 - 30.7}{23}

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Z = -0.03 has a pvalue of 0.4880.

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b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.

n = 36

Applying the Central Limit Theorem, the mean is 30.7 minutes and the standard deviation is s = \frac{23}{\sqrt{36}} = 3.83

c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?

This is 1 subtracted by the pvalue of Z when X = 35. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

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

Z = \frac{35 - 30.7}{3.83}

Z = 1.12

Z = 1.12 has a pvalue of 0.8687

1 - 0.8687 = 0.1313

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