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aniked [119]
2 years ago
11

Pls,pls,pls,help me I will give lots of pound 100+ points

Mathematics
1 answer:
Bond [772]2 years ago
4 0

Answer:

  1. 3log(10) -2log(5) ≈ 1.60206
  2. no; rules of logs apply to any base. ln(x) ≈ 2.302585×log(x)
  3. no; the given "property" is nonsense

Step-by-step explanation:

<h3>1.</h3>

The given expression expression can be simplified to ...

  3log(10) -2log(5) = log(10^3) -log(5^2) = log(1000) -log(25)

  = log(1000/25) = log(40) . . . . ≠ log(5)

  ≈ 1.60206

Or, it can be evaluated directly:

  = 3(1) -2(0.69897) = 3 -1.39794

  = 1.60206

__

<h3>2.</h3>

The properties of logarithms apply to logarithms of any base. Natural logs and common logs are related by the change of base formula ...

  ln(x) = log(x)/log(e) ≈ 2.302585·log(x)

__

<h3>3.</h3>

The given "property" is nonsense. There is no simplification for the product of logs of the same base. There is no expansion for the log of a sum. The formula for the log of a power does apply:

  \log(a)\log(b)=\log(a^{\log(b)})=\log(b^{\log(a)})

Numerical evaluation of Mr. Kim's expression would prove him wrong.

  log(3)log(4) = (0.47712)(0.60206) = 0.28726

  log(7) = 0.84510

  0.28726 ≠ 0.84510

  log(3)log(4) ≠ log(7)

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The median, upper and lower quartiles of the data set are:

Lower Quartile = 46

Median = 49

Upper Quartile = 90

<h3>What is the Median?</h3>

The center value or middle value of a data set is the median.

<h3>What are the Upper and Lower Quartiles?</h3>

Upper quartile (Q3) is the center or middle data point of the second half of a data set.

Upper quartile (Q3) is the center or middle data point of the second half of a data set.

Order the data set given as:

32, 46, 49, 49, 77, 90, 96

Lower Quartile = 32, (46), 49, 49, 77, 90, 96 = 46

Median = 32, 46, 49, (49,) 77, 90, 96 = 49

Upper Quartile = 32, 46, 49, 49, 77, (90,) 96 = 90

Learn more about the Median, Upper and Lower Quartiles on:

brainly.com/question/15572643

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Final Answer: x = -27

Steps/Equations:

Question: \frac{x}{-9} = 3

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With outlier (102)

t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415      

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Without outlier (102)

t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285      

p_v =P(t_{8}>3.285)=0.0056  

And we conclude that we reject the null hypothesis since p_v. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Step-by-step explanation:

Previous concepts  and data given  

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".  

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=126.2 represent the sample mean    

s=9.138 represent the sample standard deviation  

n=10 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:    

Null hypothesis:\mu \leq 125    

Alternative hypothesis:\mu > 125    

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415      

P-value  

First we need to calculate the degrees of freedom given by:

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p_v =P(t_{9}>0.415)=0.344    

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Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

\bar X=128.89 represent the sample mean    

s=3.551 represent the sample standard deviation

t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285      

First we need to calculate the degrees of freedom given by:

df=n-1=9-1= 8

Then since is a right tailed sided test the p value would be:    

p_v =P(t_{8}>3.285)=0.0056  

And we conclude that we reject the null hypothesis since p_v. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

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