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SSSSS [86.1K]
3 years ago
14

Find the value of cos P rounded to the nearest hundredth, if necessary.

Mathematics
2 answers:
Lemur [1.5K]3 years ago
7 0

Answer:

\cos P\approx 0.26

Step-by-step explanation:

In any right triangle, the cosine of an angle is equal to its adjacent side divided by the hypotenuse, or longest side, of the triangle.

In the given triangle, the adjacent side to angle P is marked as \sqrt{7} and the hypotenuse of the triangle is 10. Therefore, we have:

\cos P=\frac{\sqrt{7}}{10},\\\cos P=0.2645751311,\\\cos P \approx \boxed{0.26}

Leni [432]3 years ago
3 0

Answer:

\boxed {\boxed {\sf cos \  P \approx 0.26}}

Step-by-step explanation:

There are three main trigonometric functions: sine, cosine, and tangent.

We are asked to find the cosine of angle P. The cosine is the ratio of the adjacent side to the hypotenuse.

  • cos \theta = \frac{adjacent}{hypotenuse}

In this triangle, the side measuring √7 is the adjacent side because it is next to angle P. The side measuring 10 is the hypotenuse because it is opposite the right angle.

  • adjacent = √7
  • hypotenuse= 10

cos P= \frac{ \sqrt{7}}{10}

cos P=0.264575131106

Round to the nearest hundredth. The 4 in the thousandths place to the right ( 0.26<u>4</u>575131106) tells us to leave the 6 in the hundredths place.

cos P \approx 0.26

The cosine of angle P is approximately <u>0.26</u>

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As I understand, we need to evaluate the given rational expression, we will get that the evaluated expression is equal to - 1/10

<h3>Evaluating expressions:</h3>

Evaluating an expression just means that we need to replace the variable in the expression for a given value, here we have the expression:

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Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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 normal distributions 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 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.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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