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Ede4ka [16]
3 years ago
6

What is 25% of 24? What is 15% of 60? 9 is what percent of 72

Mathematics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer:

6

9

6.48

Step-by-step explanation:

this is the easiest way to solve for percentage:

\frac{part}{whole} =\frac{part}{whole}  Where all the percents (except 100) would go in part

and the whole number would go in whole

\frac{25}{100} =\frac{?}{24}  Then multiply 24 × 25 ÷ 100 = 6

\frac{15}{100} =\frac{?}{60}  Multiply 60 by 25 then divided by 100 = 9

Then for the last one 9 is what percent of 72

all we have to do is multiply 9 and then convert 72 into a decimal (0.72)

Then 9 × 0.72 = 6.48

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a canoe rental service charges a $20 transportation fee and $30 dollars an hour to rent a canoe. Write and graph an equation rep
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3 years ago
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ t
Fiesta28 [93]

Answer:

100% probability that the sample mean scores will be between 87 and 124 points

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 105, \sigma = 20, n = 20, s = \frac{20}{\sqrt{20}} = 4.47

What is the probability that the sample mean scores will be between 87 and 124 points

This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So

X = 124

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

By the Central Limit Theorem

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

Z = \frac{124 - 105}{4.47}

Z = 4.25

Z = 4.25 has a pvalue of 1

X = 87

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

Z = \frac{87 - 105}{4.47}

Z = -4.25

Z = -4.25 has a pvalue of 0

1 - 0 = 1

100% probability that the sample mean scores will be between 87 and 124 points

3 0
4 years ago
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