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miskamm [114]
3 years ago
8

Write the equation in logarithmic form. (1/3)^3=1/27

Mathematics
2 answers:
Andru [333]3 years ago
8 0

Answer:

\log_{\frac{1}{3}} (\frac{1}{27})=3

Step-by-step explanation:

The given equation is

(\frac{1}{3})^3=\frac{1}{27}

We need to write the equation in logarithmic form.

Taking log on both sides.

\log (\frac{1}{3})^3=\log (\frac{1}{27})

Using the property of logarithm we get

3\log (\frac{1}{3})=\log (\frac{1}{27})           [\because \log a^b=b\log a]

Divide both sides by \log (\frac{1}{3}).

3=\dfrac{\log (\frac{1}{27})}{\log (\frac{1}{3})}

Using the property of logarithm we get

3=\log_{\frac{1}{3}} (\frac{1}{27})          [\because \log_x y =\dfrac{\log_ay}{\log_ax}]

Therefore, the logarithmic form of given equation is \log_{\frac{1}{3}} (\frac{1}{27})=3.

juin [17]3 years ago
6 0

Answer: Log1/3(1/27)=3

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a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation  

n=2341 represent the random sample taken    

X represent the people that they have watched digitally streamed TV programming on some type of device

\hat p=0.55 estimated proportion of people that they have watched digitally streamed TV programming on some type of device  

\alpha=0.01 represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval  

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

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

Part a) Confidence interval

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523

0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51  

And rounded up we have that n=1831

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