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solong [7]
2 years ago
12

In a​ triangle, the measure of the first angle is three timesthree times the measure of the second angle. the measure of the thi

rd angle is 65 degrees65° more than the measure of the second angle. use the fact that the sum of the measures of the three angles of a triangle is 180degrees° to find the measure of each angle.
Mathematics
1 answer:
polet [3.4K]2 years ago
7 0

1st angle = 3x

2nd angle = x

3rd angle = 65 +x

3x + x +65+x = 180

5x+65 = 180

5x=115

x = 115/5 = 23

 2nd angle = 23 degrees

 1st angle = 23 x 3 = 69 degrees

3rd angle = 65 +23 = 88 degrees


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A) 1.15
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E) 5.74
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I need this the answer to 2/7 × 5/6
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Answer:

5/21

Step-by-step explanation:

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2 years ago
Anyone good at trigonometry please help
s344n2d4d5 [400]

to find the vertical height

 multiply the tangent of the angle by the horizontal distance

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6 0
2 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard
avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

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

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.

Central Limit Theorem

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

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that \mu = 33208, \sigma = 2503.

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when X = 33208+633 = 33841 subtracted by the pvalue of Z when X = 33208 - 633 = 32575

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57

So

X = 33841

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

Z = \frac{33841 - 33208}{357.57}

Z = 1.77

Z = 1.77 has a pvalue of 0.9616

X = 32575

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

Z = \frac{32575- 33208}{357.57}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0
3 years ago
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Diano4ka-milaya [45]

Answer:

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Step-by-step explanation:

3s + 7a = 1102

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a = 88

Now plug this into the second equation

s + a = 250

s + 88 = 250

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s = 162

If this answer is correct, please make me Brainliest!

7 0
2 years ago
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