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nordsb [41]
2 years ago
13

Given that x = 5.4 m and 0 = 26°, work out BC rounded to 3 SF.

Mathematics
1 answer:
bagirrra123 [75]2 years ago
4 0

Answer:

BC ≈ 4.85 m

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos26° = \frac{adjacent}{hypotenuse} = \frac{BC}{AC} = \frac{BC}{5.4} ( cross- multiply )

5.4 × cos26° = BC , then

BC ≈ 4.85 m ( to 3 s f )

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Terri Vogel, an amateur motorcycle racer, averages 129.01 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o
Vlad1618 [11]

Answer:

a. X is the time, in seconds, of a randomly selected lap of Vogel.

b. X~N(129.01,2.26)

c. 67% of her laps that are completed in less than 130 seconds.

d. So the answer is 124.8 seconds.

e. The middle 80% of her laps are from 126.1 seconds to 131.9 seconds.

Step-by-step explanation:

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.

Terri Vogel, an amateur motorcycle racer, averages 129.01 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.26 seconds

This means that \mu = 129.01, \sigma = 2.26

a. In words, define the random variable X.

The problem states that X is the time, in seconds, of a randomly selected lap of Vogel.

b. X ~ ________(_____ , _____)

Normal with mean and standard deviation. So

X~N(129.01,2.26)

c. Find the percent of her laps that are completed in less than 130 seconds.

The proportion is the pvalue of Z when X = 130. So

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

Z = \frac{130 - 129.01}{2.26}

Z = 0.44

Z = 0.44 has a pvalue of 0.67.

0.67*100% = 67%

67% of her laps that are completed in less than 130 seconds.

d. The fastest 3% of her laps are under

Under the 3rd percentile, which is X when Z has a pvalue of 0.03. So X when Z = -1.88.

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

-1.88 = \frac{X - 129.01}{2.26}

X - 129.01 = -1.88*2.26

X = 124.8

So the answer is 124.8 seconds.

e. The middle 80% of her laps are from ___________seconds to ________seconds.

The 50 - (80/2) = 10th percentile to the 50 + (80/2) = 90th percentile.

10th percentile:

X when Z has a pvalue of 0.1, so X when Z = -1.28.

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

-1.28 = \frac{X - 129.01}{2.26}

X - 129.01 = -1.28*2.26

X = 126.1

90th percentile:

X when Z has a pvalue of 0.9, so X when Z = 1.28.

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

1.28 = \frac{X - 129.01}{2.26}

X - 129.01 = 1.28*2.26

X = 131.9

The middle 80% of her laps are from 126.1 seconds to 131.9 seconds.

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