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sladkih [1.3K]
3 years ago
5

The length of the hypotenuse of a right triangle is 34 inches and the length is 34 inches and the length of one of its legs is 1

6 inches. What is the length, in inches, of the other leg of this right triangle?
Mathematics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

The length of the other leg of this right angled triangle is 30 inches.

Step-by-step explanation:

We are given the following in the question:

A right angles triangle with the hypotenuse of 34 inches and the length of one of its legs is 16 inches.

Pythagoras theorem:

  • The sum of of square of two sides of a triangle is equal to the square of the hypotenuse.

Let x inches be the length of the other leg of this right triangle.

Thus, we can write the equation:

x^2 + (16)^2 = (34)^2\\x^2 + 256 = 1156\\x^2 = 900\\x = 30

Thus, the length of the other leg of this right angled triangle is 30 inches.

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20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes tha
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10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

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Normal probability distribution

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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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 300, p = 0.2

So

\mu = E(X) = np = 300*0.2 = 60

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.2*0.8} = 6.9282

What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Using continuity corrections, this is P(X \leq 51 + 0.5) = P(X \leq 51.5), which is the pvalue of Z when X = 51.5 So

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

Z = \frac{51.5 - 60}{6.9282}

Z = -1.23

Z = -1.23 has a pvalue of 0.1093.

10.93% probability of observing 51 or fewer vapers in a random sample of 300

4 0
3 years ago
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