Which characteristic does a translation not preserve?

$4x^2+12x-7=0\\\\a=4;\ b=12;\ c=-7\\\Delta=b^2-4ac\\\\\Delta=12^2-4\cdot4\cdot(-7)=144+112=256 \ \textgreater \ 0\\\\therefore\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{256}=16\\\\x_1=\dfrac{-12-16}{2\cdot4}=\dfrac{-28}{8}=\boxed{-\dfrac{7}{2}}\\\\x_2=\dfrac{-12+16}{2\cdot4}=\dfrac{4}{8}=\boxed{\dfrac{1}{2}}$

3 0

3 years ago

In a study on the time that

krok68 [10]

The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

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To solve this question, we need to find the confidence interval for the amount of time it takes the students to get the degree.

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

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The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 80 - 1 = 79

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95% confidence interval

Standard level of confidence, we have to find a value of T, which is found looking at the t table, with 79 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9905.

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The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9905\frac{2.2}{\sqrt{80}} = 0.5$

In which s is the standard deviation of the sample and n is the size of the sample.

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The lower end of the interval is the sample mean subtracted by M. So it is 4.8 - 0.3 = 4.3 years.

The upper end of the interval is the sample mean added to M. So it is 4.8 + 0.3 = 5.3 years.

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The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

A similar question is given at brainly.com/question/24278748

6 0

3 years ago