After trial and error, you discover that - 2 works.
-2 || 3 16 18 -4
-6 -20 4
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3 10 - 2 0
What you have left is a quadratic
3x^2 + 10x - 2 You can just use the quadratic formula to solve for the other 2 roots.
x1 = [ - 10 +/- sqrt(10^2 - 4*3*(-2) ) ] / 6
x1 = [ -10 +/- sqrt (100 + 24) ) /6
x1 = [ - 10 +/- sqrt(124) ) / 6
x1 = [ - 10 +/- 11.14 ] / 6 = 0.1893
x2 = [ - 21.14 ] / 6 = - 3.522
Answer
x1 = - 2
x2 = 0.1893
x3 = - 3.522
Answer:
The 95% confidence interval for the mean zinc concentration in the river is between 1.75 and 3.45 grams per milliliter.
The 99% confidence interval for the mean zinc concentration in the river is between 1.48 and 3.72 grams per milliliter.
Step-by-step explanation:
95% confidence interval:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.96.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.6 - 0.85 = 1.75 grams per milliliter.
The upper end of the interval is the sample mean added to M. So it is 2.6 + 0.85 = 3.45 grams per milliliter.
The 95% confidence interval for the mean zinc concentration in the river is between 1.75 and 3.45 grams per milliliter.
99% confidence level:
By the same logic as for the 95% confidence interval, we have that . So
The lower end of the interval is the sample mean subtracted by M. So it is 2.6 - 1.12 = 1.48 grams per milliliter.
The upper end of the interval is the sample mean added to M. So it is 2.6 + 1.12 = 3.72 grams per milliliter.
The 99% confidence interval for the mean zinc concentration in the river is between 1.48 and 3.72 grams per milliliter.
Answer:
x=0, x= - 3, x=9
Step-by-step explanation:
f(x)=7/2x(x+3)(x-9)
f(x) doesn't exist for x=0,-3,9
as
vertical asymptote:
x=0, x= - 3, x=9