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2 years ago

1)log x 2+log x 4x^ 2 =1

Mathematics

1 answer:

BigorU [14]2 years ago

3 0

Step-by-step explanation:

5x is the answers

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There were 7,500 tickets sold for a concert, 20% of which were general admission. How many general tickets were sold?

konstantin123 [22]

7,500*20%=g

7,500*.20=g

1500=g

1,500 general admission tickets were sold.

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3 years ago

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2 years ago

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Find gcd for 5767 and 4453

Dima020 [189]

The gcd of 5767 and 4453 is 73.
Hope this helps. :)

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3 years ago

Given m||n, find the value of x.<br> +<br> (5x+2)<br> (4x+6)°

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2 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

Step-by-step explanation:

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

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2 years ago