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3 years ago

I NEED HELP PLEASE, THANKS! :)

Mathematics

1 answer:

Sav [38]3 years ago

3 0

Answer: $\bold{log_5\bigg(\dfrac{x^5}{(8-x)^{\frac{1}{4}}}\bigg)}$

<u>Step-by-step explanation:</u>

Notes: Coefficients condense to exponents.

Subtraction condenses to division

$5\log_5(x)-\dfrac{1}{4}\log_5(8-x)\\\\\\=\log_5(x)^5-\log_5(8-x)^\frac{1}{4}\\\\\\=\large\boxed{\log_5\bigg(\dfrac{x^5}{(8-x)^\frac{1}{4}}\bigg)}$

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In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

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3 years ago

A game has 50 cards with 10 each numbered 1 to 5,and a player must draw a 2 or a 3 to move out of the “start” position.

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Answer:

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Step-by-step explanation:

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4 years ago

What is the measure of each exterior angle of a regular triangle

Sophie [7]

Answer:Each pair of interior angle and the exterior angle

So the exterior angle = 360° - the interior angle.

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3 years ago

Read 2 more answers

7.5% of the grade has green eyes if there are 120 students in the grade how many dont have green eyes

GREYUIT [131]

Answer:

Step-by-step explanation:

If 7.5 % has green eyes, then 92.5% (the complement given by 100%-7.5%) dont have green eyes. 92.5% of 120 is given by

0.925*(120)=111

111 students in the grade dont have green eyes.

8 0

3 years ago

write the logarithm as a sum or difference of logarithms. simplify each term as much as possible. log4(6ab)

Dominik [7]

The logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

<h3>Simplifying Logarithms</h3>

From the question, we are to write the given logarithm expression as a sum or difference of logarithms

The given logarithm is

$log_{4 }6ab$

This can be written as

$log_{4 }6 \times a \times b$

From one of the rules of logarithm, we have that

$log_{x }yz= log_{x }y + log_{x }z$

Thus,

$log_{4 }6 \times a \times b$ becomes

$log_{4 }6 + log_{4 }a + log_{4 }b$

This can be further simplified into

$log_{4 }3 \times 2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + log_{4 }2 + log_{4 }a + log_{4 }b$

If desired, this can be further simplified into

$log_{4 }3 + log_{2^{2} }2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} log_{2}2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} (1)+ log_{4 }a + log_{4 }b$

$\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Hence, the logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Learn more on Simplifying logarithms here: brainly.com/question/17851187

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2 years ago