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solmaris [256]
3 years ago
5

What is the square root 6n carried to the power of 2/3 in radical form

Mathematics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer: \sqrt[3]{6n}

Step-by-step explanation:

We have the following expression:

(\sqrt{6n})^{\frac{2}{3}}

Which can be written as follows:

((6n)^{\frac{1}{2}})^{\frac{2}{3}}

Multiplying the exponents:

(6n)^{\frac{1}{3}}

Writing in radical form we finally have the result:

\sqrt[3]{6n}

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So fill in what you know:

120=.5b*15

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the next step after writing the equation would be to subtract 10 from both sides to get 0.05x=12. after this you would then have to divide 12 by 0.05 to get 240 minutes. you could check it by putting 240 in for x to get 10+0.05(240)=22. you would then multiply 0.05 by 240 to get 12 and then add 10 onto it to get 22 which was already given.

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You are ordering lunch for everyone in your office from a small restaurant nearby. The following shows the menu items available.
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Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

And we have :

\mu_p = 0.66

\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

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