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k0ka [10]
2 years ago
13

Help needed ASAP it’s geometry

Mathematics
1 answer:
ivanzaharov [21]2 years ago
5 0

Answer:

<h2>           32</h2>

Step-by-step explanation:

\tan(29.4^o)=\dfrac{?}{56.8}\\\\\tan(29.4^o)\times56.8=?\\\\?\approx0,563471\times56.8\approx32

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The difference of two numbers is 8. When twice the first number is added to three times the second number, the result is 51. Wha
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The answer is b. 15 & 7
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Whats the square root of 10?? For each to the nearest 0.05.​
Vesnalui [34]

Answer:

3.16(i think)

Step-by-step explanation:

4 0
3 years ago
How do you do expressions in maths
daser333 [38]
1. Know order of operations (bedmas, Canada), (pemdas, USA)
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8 0
3 years ago
A random sample of n = 100 observations is selected from a population with mean 20 and standard deviation 15. What is the probab
hjlf

Answer:

25.14% probability of observing a mean greater than 21

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 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 = 20, \sigma = 15, n = 100, s = \frac{15}{\sqrt{100}} = 1.5

What is the probability of observing a mean greater than 21?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{1.5}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

1 - 0.7486 = 0.2514

25.14% probability of observing a mean greater than 21

6 0
3 years ago
Please help me on this I-Ready task, I have no clue
Alex787 [66]

Answer:

4.2 miles

Step-by-step explanation:

Using the pythagorean theorem, a^2 + b^2 = c^2

youll get 3.6^2 + 2.2^2 = 17.8

sqrt of 17.8 = 4.2

6 0
3 years ago
Read 2 more answers
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