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RoseWind [281]
3 years ago
10

What is the radical form of each of the given expressions?

Mathematics
2 answers:
ehidna [41]3 years ago
7 0
ANSWER

The radical form of the expression
{a}^{ \frac{m}{n} } = \sqrt[n]{ {a}^{m} }
Note that
n = 2 \: or \: m = 1
are not written explicitly. However, greater values are written explicitly.

So, for {2}^{ \frac{1}{2} }

n=2, m=1



This implies that, we will just write
{2}^{ \frac{1}{2} } = \sqrt{2}



for {2}^{ \frac{2}{3} }

n=3, m=2

This implies that,

{2}^{ \frac{2}{3} } = \sqrt[3]{ {2}^{2} }


Similarly,

{3}^{ \frac{3}{2} } = \sqrt{ {3}^{3} }

{3}^{ \frac{1}{3} } = \sqrt[3]{ {3}}
Lelu [443]3 years ago
4 0
2^1/2     =           √2
2^2/3     =           ∛2^2
3^3/2     =            √3^3
3^1/3     =            ∛3

hope it helps  
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Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

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 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

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As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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