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2 years ago

The sail of a boat is in the shape of a right triangle, as shown below:

Mathematics

2 answers:

dolphi86 [110]2 years ago

6 0

Answer: Choice C
7 divided by cosec 50 degrees

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Explanation:

Abbreviations:
sec = secant
cot = cotangent
cosec = cosecant (csc is also a widely used abbreviation)

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Let the horizontal leg be x

This leg is opposite the angle 50 degrees. The hypotenuse is 7

Opposite = x
Hypotenuse = 7

We'll either use the sine rule or the cosecant (csc) rule
sine = opposite/hypotenuse
cosecant = hypotenuse/opposite

Since "sine" is nowhere to be found in the answer choices, we'll use the cosecant rule

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cosecant = hypotenuse/opposite
csc(angle) = hypotenuse/opposite
csc(50) = 7/x
x*csc(50) = 7
x = 7/csc(50)

This is why the answer is choice C

Vladimir79 [104]2 years ago

5 0

Answer:

C. 7/cosec50

I did the test. This is correct.

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A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi

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Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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.

For the sum of n values of a distribution, the mean is $\mu \times n$ and the standard deviation is $\sigma\sqrt{n}$

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that $\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300$

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

$Z = \frac{20000 - 20000}{300}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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