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2 years ago

Given that 32 = 9, write this in logarithm form.

Mathematics

1 answer:

Gwar [14]2 years ago

6 0

It is

log 39=2

Explanation:

By definition the logarithmic form

log bx=y is equivalent to the exponential form

by=x

Hence from the given 32=9 you have b=3 , y=2 , x=9

hence the logarithmic form is

log 3 9=2

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Four time a number r plus half a number s equals 12. Twice the number r plus one fourth of the numbers s equals 8. What are the

Snezhnost [94]

The system of equations have no solution.

Explanation:

The equations are $4r+\frac{1}{2}s=12$ and $2r+\frac{1}{4} s=8$

To find the two numbers, let us solve the equations using substitution method.

From the equation $2r+\frac{1}{4} s=8$ , let us determine the value of r,

$\begin{aligned} 2 r+\frac{1}{4} s &=8 \\ 2 r &=8-\frac{1}{4} s \\ r &=\frac{1}{2}\left(8-\frac{1}{4} s\right) \\ r &=4-\frac{1}{8} s \end{aligned}$

Let us substitute the value of r in $4r+\frac{1}{2}s=12$ , we get,

$\begin{aligned} 4\left(4-\frac{1}{8} s\right)+\frac{1}{2} s &=12 \\ 16-\frac{1}{2} s+\frac{1}{2} s &=12 \\ 16 &=12 \end{aligned}$

which is not possible.

This means that the system has no solution.

6 0

3 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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Answer:

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Step-by-step explanation:

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I'm not 100% sure on this answer, but I don't think anyone else will be answering.

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