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Doss [256]
3 years ago
8

What is the length of the altitude to the short side of an isosceles triangle with side

Mathematics
1 answer:
solniwko [45]3 years ago
8 0

Answer:

Step-by-step explanation:

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A video store charges $12 for used games and $25 for new games. Jonathan has saved $300 to spend on video games. This scenario i
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Suppose that in a certain population, individuals’ heights are approximately nor-mally distributed with parametersμ=70 andσ=3 in
ruslelena [56]

Answer:

a) 25.14% of the population is over 6ft tall.

b) Both are going to be normally distributed.

Centimeters: \mu = 177.8, \sigma = 7.62cm

Meters: \mu = 1.778, \sigma = 0.0762cm

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

Normally distributed with parameters \mu = 70, \sigma = 3.

a.What proportion of the population is over 6 ft. tall?

We have to convert from ft to inches.

Each feet has 12 inches, so 6ft = 72in.

This proportion is 1 subtracted by the pvalue of Z when X = 72. So

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

Z = \frac{72 - 70}{3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

This means that 1-0.7486 = 0.2514 = 25.14% of the population is over 6ft tall.

b.What is the distribution of heights if they are expressed in centimeters? Inmeters?

It is going to still be normally distributed. We just have to convert the mean and the standard deviation from inches to centimeters and meters.

In centimeters.

Each inch has 2.54cm. So it is going to be normally distributed with \mu = 70*2.54 = 177.8cm and \sigma = 3*2.54 = 7.62xm

In meters

Each m has 100cm. So

\mu = 1.778m, \sigma = 0.0762m.

7 0
4 years ago
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