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Lady_Fox [76]
2 years ago
13

A tree casts a shadow that is 150 feet long. If the angle of elevation from the tip of the shadow to the top of the tree is 30°,

how tall is the tree to the nearest foot?
A) 75 feet

B) 87 feet

C) 106 feet

D) 212 feet

Mathematics
1 answer:
Eduardwww [97]2 years ago
7 0
<h3>Answer:</h3>

B) 87 feet

<h3>Step-by-step explanation:</h3>

The mnemonic SOH CAH TOA reminds you the relationship between angle, adjacent, and opposite sides is ...

... Tan = Opposite/Adjacent

In this geometry, the side adjacent to the angle is marked 150 ft, and the side opposite the angle is the height we want to find. This means ...

... tan(30°) = height/(150 ft)

Multiplying by 150 ft, we get ...

... height = (150 ft)·tan(30°) ≈ 87 ft

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Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o
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Answer:

0% probability that the mean of the sample taken is less than 2.2 feet.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 2.5 feet and a standard deviation of 0.2 feet.

This means that \mu = 2.5, \sigma = 0.2

Sample of 41

This means that n = 41, s = \frac{0.2}{\sqrt{41}}

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This is the p-value of Z when X = 2.2 So

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

By the Central Limit Theorem

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

Z = \frac{2.2 - 2.5}{\frac{0.2}{\sqrt{41}}}

Z = -9.6

Z = -9.6 has a p-value of 0.

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