Question

which choice shows the x-values of all the discontinuities of the function?
$f(x) = \frac{2x + 3}{2x^3 + 7x^2 - 7x - 12}$
\bigcirc -4, -1, 1.5
\bigcirc -1.5, 1, 4
\bigcirc -4, -1
\bigcirc 1, 4

in a poll taken in the summer of 2013, a national polling firm asked 1,025 american adults whether they were hockey fans, and 27% of respondents said that they were. which of these represents the 95% confidence interval for the percentage of american adults who are hockey fans?
\bigcirc 24.28% to 29.72%
\bigcirc 24.72% to 29.28%
\bigcirc 25.00% to 29.00%
\bigcirc 25.61% to 28.39%

Explanation:

First Question (Discontinuities of the Function)

Step1: Find Denominator Roots

To find discontinuities, we solve \(2x^3 + 7x^2 - 7x - 12 = 0\). Factor the cubic:
Group terms: \((2x^3 + 7x^2) + (-7x - 12) = x^2(2x + 7) - 1(7x + 12)\). Wait, better to use rational root theorem. Possible roots: \(\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{3}{2}\).
Test \(x = -4\): \(2(-4)^3 + 7(-4)^2 - 7(-4) - 12 = -128 + 112 + 28 - 12 = 0\). So \((x + 4)\) is a factor.
Divide cubic by \((x + 4)\): Using polynomial division or synthetic division, we get \(2x^2 - x - 3\).
Factor \(2x^2 - x - 3 = (2x - 3)(x + 1)\). So roots are \(x = -4\), \(x = -1\), \(x = \frac{3}{2}=1.5\).

Step2: Check Numerator

Numerator \(2x + 3\) at these roots: \(2(-4)+3=-5
eq0\), \(2(-1)+3=1
eq0\), \(2(1.5)+3=6
eq0\). So all are vertical asymptotes (discontinuities).

Second Question (Confidence Interval)

Step1: Formula for CI

For a proportion, 95% CI is \(\hat{p} \pm z^*\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\). Here, \(\hat{p} = 0.27\), \(n = 1025\), \(z^* = 1.96\) (for 95% CI).

Step2: Calculate Margin of Error (ME)

\(ME = 1.96\sqrt{\frac{0.27(0.73)}{1025}} \approx 1.96\sqrt{\frac{0.1971}{1025}} \approx 1.96\sqrt{0.0001923} \approx 1.96 \times 0.01387 \approx 0.0272\) (or 2.72%).

Step3: Compute CI

Lower bound: \(0.27 - 0.0272 = 0.2428\) (24.28%), Upper bound: \(0.27 + 0.0272 = 0.2972\) (29.72%).

Answer:

-4, -1, 1.5