3x^2+7x+2 how to factor
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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.
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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 . So we have T = 1.9905.
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The margin of error is:
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
There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/-
x = +/- i