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3 years ago

What is the expression in radical form (2m^2n)^3/2

Mathematics

1 answer:

Marysya12 [62]3 years ago

3 0

Answer:

sqrt[(2m^2n)^3]

Step-by-step explanation:

So let's break down the exponent. The top number represents the number of times the term is repeated. The bottom number represents the root to be taken of the final product. With this in mind, let's rewrite this expression.

(2m^2n)^3/2

= [(2m^2n)^3]^1/2

Notice we have 3 of the (2m^2n) terms, but they are all under the 2nd root (aka a square root).

So now, we'll rewrite this into the radical form.

sqrt[(2m^2n)^3]

I hope this helps.

Cheers.

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TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

yarga [219]

Answer:

The 99% confidence interval for the proportion of all adult Americans who watched streamed programming up to that point in time is (0.5336, 0.5864). This means that we are 99% sure that the true proportion of all adult Americans who watched streamed programming up to that point in time is between these two values.

A sample size of 664 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A poll reported that 56% of 2348 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

This means that $\pi = 0.56, n = 2348$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 2.575$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 - 2.575\sqrt{\frac{0.56*0.44}{2348}} = 0.5336$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 + 2.575\sqrt{\frac{0.56*0.44}{2348}} = 0.5864$

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value P?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

This sample size is given by n, with $M = 0.05$ and $\pi = 0.5$ .

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.05 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.05\sqrt{n} = 2.575*0.5$

Dividing both sides by 0.05.

$\sqrt{n} = 2.575*10$

$(\sqrt{n})^2 = (2.575*10)^2$

$n = 663.1$

Rounding up:

A sample size of 664 is required.

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Answer:

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Answer:

Step-by-step explanation:

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