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Pavlova-9 [17]
3 years ago
14

Which equation can be used to solve for b? Triangle A B C is shown. Angle A C B is 90 degrees and angle C B A is 30 degrees. The

length of side B C is 8 feet, the length of B A is c, and the length of C A is b. b = (8)tan(30o) b = StartFraction 8 Over tangent (30 degrees) EndFraction b = (8)sin(30o) b = StartFraction 8 Over sine (30 degrees) EndFraction

Mathematics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer:

b=(8)tan(30\°)

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In the right triangle ABC, the tangent of angle of 30 degrees is equal to divide the opposite side to the angle of 30 degrees (AC) by the adjacent side to the angle of 30 degrees (BC)

tan(30\°)=AC/BC

substitute the given values

tan(30\°)=b/8

Solve for b

b=(8)tan(30\°)

galina1969 [7]3 years ago
8 0

Answer:

b=(8)tan(30\°)

Step-by-step explanation:

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8.52 The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches
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Answer:

Heights of 29.5 and below could be a problem.

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.

The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches.

This means that \mu = 32, \sigma = 1.5

There may be a problem when a child is in the top or bottom 5% of heights. Determine the heights of 2-year-old children that could be a problem.

Heights at the 5th percentile and below. The 5th percentile is X when Z has a p-value of 0.05, so X when Z = -1.645. Thus

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

-1.645 = \frac{X - 32}{1.5}

X - 32 = -1.645*1.5

X = 29.5

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