#2**+3 **

Well, I guess I committed a cardinal sin: I did not factor completely. I thought there was no way that either the numerator or denominator could have a common factor. How wrong I was!

\(-3x^7-3x^6-3x^5+3x^4+3x^2+3\) | First, factor out a common factor of -3 from every term. |

\(-3(x^7+x^6+x^5-x^4-x^2-1)\) | Well, I can use the rational root theorem to determine that 1 is a root of the seven-degree polynomial, so \(x-1\) must also be a factor. This results in the following result. After using synthetic division, I got the following! |

\(-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)\) | Let's worry about the denominator. |

\(x^3(x^2+x-2)\) | I was able to use factoring here, too, by finding the products of -2 that equal 1: 2 and -1. Let's complete the factoring, then! |

\(x^3(x-1)(x+2)\) | |

\(\frac{-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x-1)(x+2)}\) | Look at that! \(x-1\) is a common factor. That's crazy! This leaves us with the following! |

\(g(x)=\frac{-3\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x+2)}, x(x+2)\neq0\text{ and }x\neq 1\) | I guess this is better! |

TheXSquaredFactor
10.04.2018

#1**+3 **

Asymptotes are always expressed as equations because they are lines, so I would assume that the horizontal asymptote lies at y=0.

Of course, horizontal asymptotes have many rules associated with them. In this case, we are concerned with the following rule: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes exists at the x-axis (also known as y=0).

If we take \(f(x) = \frac{3(x^4+x^3+x^2+1)}{x^2+x-2}\) and multiply it by \(\frac{1}{x^3}\), then we would have finished this problem. Therefore, let's find g(x) that does this.

^{\(\frac{3(x^4+x^3+x^2+1)}{x^2+x-2}+g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}\)} | Let's do some subtraction here. |

\(g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3(x^4+x^3+x^2+1)}{(x^2+x-2)}\) | Let's convert the rightmost fraction into one with a common denominator so that further simplification is possible. |

\(g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3x^3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}\) | Let's do some distributing. |

\(g(x)=\frac{3x^4+3x^3+3x^2+3}{x^3(x^2+x-2)}-\frac{(3x^7+3x^6+3x^5+3x^3)}{x^3(x^2+x-2)}\) | Now, subtract! |

\(g(x)=\frac{-3x^7-3x^6-3x^5+3x^4+3x^2+3}{x^3(x^2+x-2)}\\ \) | Not a lot of simplification happens, though, but let's do it anyway! |

You should see that, when adding g(x) to f(x), this results in a horizontal asymptote of y=0.

TheXSquaredFactor
10.04.2018

#1**+1 **

We have two pieces of data from a study from separate groups.

In order to create the baseline, it is necessary to find the difference of the individual means of both data.

\(\text{Sedentary: }\overline{x}_1=56.5, SD_1=14.1,s_1=55\\ \text{Manual: }\overline{x}_2=51.3,SD_2=13.5,s_2=50\)

SD = sample standard deviation

s_{1} and s_{2} are the individual values for sampling size.

Since the question asks for the **difference between the population means**, one should do that in order to create a baseline.

\(\mu_{\overline{x}_1-\overline{x}_2}=\overline{x}_1-\overline{x}_2=56.5-51.3=5.2\)

This is the exact middle of the data. In order to establish a 95% confidence level of the difference, we will have to find the number of standard deviations from the current \(\mu_{\overline{x}_1-\overline{x}_2}\) . We can utilize the empirical rule (sometimes referred to the 68-95-99.7% rule) for this. An image will probably do the best explaining.

Source: http://simulation-math.com/_Statistics/EmpiricalRule.png

This image indicates to me that 95% of the data falls within 2 standard deviations from the mean. Because a bell-shaped distribution is symmetric in nature, we know that, from the mean, 95% will fall between \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) . By calculating this information, there should be a 95% chance that 5.2 (the \(\mu_{\overline{x}_1-\overline{x}_2}\)) lies within \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) of the mean. Now, we must find that standard deviation. Let's use some algebra manipulation to find a formula for this.

\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\sigma_{\overline{x}_1}^2+\sigma_{\overline{x}_2}^2\) | In other words, the variance of the means of the distribution equals the sum of the variance of each sampling distribution. We can rewrite this equation further because the variance of each sampling distribution also equals the variance of the population distribution divided by the sampling size. |

\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\frac{\sigma_{x_1}^2}{s_1}+\frac{\sigma_{x_2}^2}{s_2}\) | There is only one problem: We do not know the sample variances. However, we can approximate by using the values from the sample standard deviations instead. |

\(\sigma_{\overline{x}_1-\overline{x}_2}^2=\frac{SD_{1}^2}{s_1}+\frac{SD_{2}^2}{s_2}\) | Now, solve for sigma by taking the square root of both sides. |

\(\sigma_{\overline{x}_1-\overline{x}_2}=\sqrt{\frac{SD_{1}^2}{s_1}+\frac{SD_{2}^2}{s_2}}\) | Now, plug in the values for variables in the equation. Normally, I would rationalize the denominators, but this is a calculator's job anyway, so it does not really matter. |

\(\sigma_{\overline{x}_1-\overline{x}_2}=\sqrt{\frac{14.1^2}{55}+\frac{13.5^2}{50}}\) | |

\(\sigma_{\overline{x}_1-\overline{x}_2}\approx 2.6944\) | This is the variance calculated to the nearest ten-thousandth place. Since 95% is two standard deviations away, we need to double this answer. |

\(2\sigma_{\overline{x}_1-\overline{x}_2}\approx 2*2.6944\approx5.3888\) |

Remember this statement from earlier? I said there should be a 95% chance that 5.2 (the \(\mu_{\overline{x}_1-\overline{x}_2}\)) lies within \(\pm2\sigma_{\overline{x}_1-\overline{x}_2}\) of the mean. We can derive the interval with this. Therefore, 95% of the data lies within \(5.2\pm5.3888\) .

We are, therefore, 95% confident that the true mean lies between \(-0.1888\) and \(10.588\) .

I encourage you to try the next one! See how you do!

TheXSquaredFactor
31.03.2018

#1**+1 **

The definitions are given clearly to us; now, we must interpret the information.

"The function c(x)=50+5x represents price charged per customer where x is the number of $5 increases they charge at a rate of $50 per person. "

This segment states that a particular function \(c(x)=50+5x\) determines the cost for one attendee for this horse-drawn carriage tour company.

"The function p(x)=80−2x represents the number of customers expected for the day, where x is the number of $5 increases they charge at a rate of $50 per person."

\(p(x)=80−2x\) is a function that determines the number of attendees for this event.

\((p*c)(2)=p(2)*c(2)\), therefore, determines the total cost of the fee for all attendees when two five-dollar increments are placed on the cost.

\(\textcolor{red}{p(2)}*\textcolor{blue}{c(2)}\) | I decided to use colors to help differentiate the two quantities. |

\(\textcolor{red}{p(2)=80-2*2}\) | Determine the number of attendees when two five-dollar increments are placed on the cost. |

\(\textcolor{red}{p(2)=76}\) | |

\(\textcolor{blue}{c(2)=50+5*2}\) | Evaluate. |

\(\textcolor{blue}{c(2)=60}\) | |

\(\textcolor{red}{p(2)}*\textcolor{blue}{c(2)}\\ \textcolor{red}{76}\hspace{4mm}*\textcolor{blue}{60}\) | |

\($4560\) | |

Knowing this information, I would agree with you! The answer is C!

TheXSquaredFactor
22.03.2018

#1**0 **

1. \((h-f)(4)=h(4)-f(4)\)

The left hand side and the right hand of the equation are equivalent. The notation on the right might be more intuitive, though.

\(h(4)-f(4)\) | Evaluation both functions when x equals 4. |

\(h(4)=\frac{1}{4}*4+6\) | Here, I have replaced all instances of an "x" with a 4. Now, let's evaluate h(4). |

\(h(4)=1+6=7\) | Now, let's find f(4). |

\(f(4)=5*4-9\) | Yet again, every appearance of "x" is replaced with the input, 4. |

\(f(4)=11\) | The original question wants you to subtract the two functions, so let's do that. |

\(h(4)-f(4)\\ 7\hspace{6mm}-\hspace{3mm}11\) | |

\(-4\) | |

2) If h(x)=9, then we can use substitution to find the value of x:

\(h(x)=\frac{1}{4}x+6\) | Replace h(x) with 9 since they are equal. |

\(9=\frac{1}{4}x+6\) | Subtract 6 from both sides of the equation. |

\(3=\frac{1}{4}x\) | Multiply by 4 on both sides to isolate the variable. |

\(x=12\) | |

3) If f(n)=f(3n+1), then we can evaluate both functions for the given input and set them equal to each other.

\(f(x)=5x-9\) | \(f(x)=5x-9\) | |

\(f(n)=5n-9\) | \(f(3n+1)=5(2n+1)-9\) | |

\(f(3n+1)=10n+5-9\) | ||

\(f(n)=5n-9\) | \(f(3n+1)=10n-4\) | |

As aforementioned, these values are equal, so let's set them equal

\(\hspace{4mm}f(n)=f(3n+1)\\ 5n-9=10n-4\) | Now, solve for n. Move the constants and linear terms over to one side of the equation. |

\(-5=5n\) | Finally, divide by 5. |

\(-1=n\) | |

TheXSquaredFactor
22.03.2018

#1**+1 **

A 30-60-90 triangle is a special type of right triangle where the ratio of the sides are already known. The sides have a constant ratio of \(1:\sqrt{3}:2\). Using this information, it is possible to find the hypotenuse by setting up a proportion.

\(\frac{\text{long}}{\sqrt{3}}=\frac{\text{hypotenuse}}{2}\)

Of course, we already know the length of the "long" side, so it is possible to solve for the length of the hypotenuse.

\(\frac{\text{long}}{\sqrt{3}}=\frac{\text{hypotenuse}}{2}\) | Replace "long" with 3 since that is its side length. |

\(\frac{3}{\sqrt{3}}=\frac{\text{hypotenuse}}{2}\) | Multiply by 2 on both sides to get rid of the fraction on the right hand side. |

\(\text{hypotenuse}=\frac{6}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}\) | Multiplying by the square root of 3 like this will rationalize the denominator. |

\(\text{hypotenuse}=\frac{6\sqrt{3}}{3}=2\sqrt{3}\approx 3.464\) | The decimal approximation seems reasonable because it should be the longest side of the triangle. |

TheXSquaredFactor
22.03.2018