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hodyreva [135]
3 years ago
9

If one leg of a right triangle measures 3 centimeters and the other leg measures 6 centimeters, what is the length of the hypote

nuse?
A. 3√5 centimeters
B. 5√3 centimeters
C. 9 centimeters
D. √5 centimeters
Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
6 0
The answer is A. USE PYTHAGOREAN THEOREM 3^2+6^2=c^2 9+36=c^2 45=c^2 c=3 sqrt5
Salsk061 [2.6K]3 years ago
4 0

Answer:

The correct option is A. 3\sqrt{5}centimeters.

Step-by-step explanation:

Consider the provided information.

It is given that the leg of a right triangle measures 3 centimeters and the other leg measures 6 centimeters.

Use Pythagorean theorem:

(a)^2+(b)^2=(c)^2

Where <em>c</em> is hypotenuse, <em>a</em> is leg and <em>b</em> is other leg.

Now substitute the values of leg in the above formula as shown:

(3)^2+(6)^2=(c)^2

9+36=(c)^2

45=(c)^2

\sqrt{45}=c

3\sqrt{5}=c

Hence, the correct option is A. 3\sqrt{5}centimeters.

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Read 2 more answers
A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 33.9 wee
MArishka [77]

Answer:

P(35.3 < M < 35.4) = 0.0040.

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:

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, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 33.9, \sigma = 6.7, n = 119, s = \frac{6.7}{\sqrt{119}} = 0.6142

Find the probability that a single randomly selected value is between 35.3 and 35.4

This is the pvalue of Z when X = 35.4 subtracted by the pvalue of Z when X = 35.3. So

X = 35.4

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

By the Central Limit Theorem

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

Z = \frac{35.4 - 33.9}{0.6142}

Z = 2.44

Z = 2.44 has a pvalue of 0.9927

X = 35.3

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

Z = \frac{35.3 - 33.9}{0.6142}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

0.9927 - 0.9887 = 0.0040

So the answer is:

P(35.3 < M < 35.4) = 0.0040.

4 0
3 years ago
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