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4 years ago

Which of the following represents 5x 4/9 in radical formj

Mathematics

1 answer:

solong [7]4 years ago

3 0

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
5x^{\frac{4}{9}}\implies 5\sqrt[9]{x^4}$

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A suit is on sale for 27% off. The sale price is $438. What is the regular price?

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9. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of

netineya [11]

Answer:

9.18% probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 12, \sigma = 3$

What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

$Z = \frac{16 - 12}{3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

1 - 0.9082 = 0.0918

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3 years ago

Suppose that salaries for recent graduates of one university have a mean of $25,700 with a standard deviation of $1350. Using Ch

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Answer:

75% of the data will reside in the range 23000 to 28400.

Step-by-step explanation :

To find the range of values :

We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,

$1 - \frac{1}{k^2} =\frac{75}{100}$

Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.

Range is given by : mean +/- (z score) × (value of a standard deviation)

⇒ Range : 25700 +/- 2 × 1350

⇒ Range : (25700 - 2700) to (25700 + 2700)

Hence, 75% of the data will reside in the range 23000 to 28400.

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3 years ago