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2 years ago

Simplify 6r · s · 4rt.

Mathematics

2 answers:

yanalaym [24]2 years ago

5 0

(6r)(s)(4rt)
(6r)(s)(4rt) = 24r²st

Sindrei [870]2 years ago

3 0

The answer would be ~ $24r^2st
$

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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

sweet [91]

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

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

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

$X - 32 = -1.645*1.5$

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2 years ago

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1

BigorU [14]

Answer:

a: Sample mean: 20, Sample standard deviation: 1.732

b: 19.13 <µ < 20.87

c: 18.84< µ < 21.16

d: as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a: Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

The t-value is 1.415

See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population. Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

The t-value is 1.895

See the third attached photo for the construction of the confidence interval