**Thanksgiving Math Quiz**

#1 There are \(\frac{6!}{4}=180\) distinct arrangements.

#1 bonus. Hopefully the probability is 0 since you enjoy the company of everyone with you! Otherwise, if there are \(n\) people you don’t want to sit by, the probability is \(1-\frac{\binom{5-n}{2}}{\binom{5}{2}}\). This gives us the following table.

\(n=\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |

\(P(n)\) | \(0\) | \(\frac{2}{5}\) | \(\frac{7}{10}\) | \(\frac{9}{10}\) | \(1\) | \(1\) |

#2a There are \(\binom{10}{4}=210\) different ways to arrange people sitting at the kids or adult table.

#2b The probability you will be at the kids table is \(\frac{4}{10}\).

#3 Justin ate 1.75 pounds of turkey, Jaci ate .875 pounds of turkey and Joey ate 1.375 pounds of turkey.

#4 The total calories eaten will be \[ \begin{align*}\int_{0}^{12}\frac{1}{4}t^{4}-6t^{3}+36tdt& \\ &=\frac{1}{20}t^{5}-\frac{6}{4}t^{4}+18t^{2}|_{0}^{12} \\&=11,577.6\end{align*}\] Therefore, you will eat 11,577.6 calories over the course of the day.

#5 You get \(\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}\). Therefore, you will eat half the pie (assuming you could go up to get more infinitely many times).

#6 We assume the depth is constant, therefore the volume of pie is dependent only on the area. We therefore have that the area of the slice should be \(9 in^{2}\). We then have the area of a slice is \(\frac{\theta}{2}r^{2}\) where \(\theta\) is the angle of the slice and \(r\) is the radius, in this case 6 in. Then, solving \(\frac{\theta}{2}(6)^{2}=9\), we get \(\frac{1}{2}\). Therefore the angle if \(\frac{1}{2}\) radians or \(\frac{90}{\pi} \approx 28.65\) degrees.

### Thanks for reading and answering!

We hope you enjoyed the quiz. Counting #1 and #2 as two questions, if you got

- 8/8 you are indeed a genius,
- 6/8 you are extremely intelligent,
- 4/8 you are smarter than the average person,
- 2/8 you are average,
- 0/8 you should spend more time reading this blog so that you can better understand mathematics!

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**Geometry Quiz**

- \(BC=\sqrt{20}\) and \(BD=\sqrt{5}\).
- \(a=72, b=36,c=72\).
- \(\angle A +\angle C =180\).
- \(AF=4\).
- \(AP=4\).

### Thanks for reading and answering!

We hope you enjoyed the quiz. Counting #1 and #2 as two questions, if you got

- 5/5 you are indeed a geometry genius,
- 4/5 you are extremely intelligent at geometry,
- 2/5 you are smarter than the average person at geometry,
- 1/5 you are average at geometry,
- 0/5 you should spend more time reading this blog so that you can better understand geometry!

**Studying Exam**

- Note that \[\frac{dG}{dt}=12e^{-.2t}.\] Therefore, the rate of change of grade with respect to time after studying for 4 hours is \(12e^{-.2(4)}=5.4\). That is, your grade is going up by 5.4 points per hour of studying.
- Here, we have that \(s=12-t\), so \[\begin{align*}G(t)&=100-60e^{-.2t}-5(8-(12-t))^{2} \\ &=20+40t-5t^{2}-60e^{-.2t}.\end{align*}\] We then find that \[\frac{dG}{dt}=40-10t+12e^{-.2t}.\] This is undefined at the endpoints of the domain \(t=0,12\) and is \(0\) when \(t \approx 4.5\). Making a table we have

Note that while you can’t actually get a negative score on your exam, the values should at least indicate that it is not a good idea to sleep for the next twelve hours or study for the next twelve hours (however sleeping appears to be better). Your best solution would be to study for 4.5 hours and sleep for 7.5.t G(t) 0 -40 4.5 74.4 12 -225 - Note that the total scores on both problems would be \[P(t_{1},t_{2})=10-10e^{-.23t_{1}}+10-10e^{-.16t_{2}}.\] Since \(t_{2}=10-t_{1}\), we have that \[\begin{align*}P(t_{1})&=10-10e^{-.23t_{1}}+10-10e^{-.16(10-t_{1})} \\ &=10(2-e^{-.23t_{1}}-e^{.16t_{1}-1.6}. \end{align*}\] We then find the \[\frac{dP}{dt_{1}}=2.3e^{-.23t}-.32e^{-.16t}.\] We find this is undefined at the endpoints of the domain \(t=0,10\) and is 0 if \(t=5.03\). Making a table again, we have

Therefore, it would be best to spend 5 minutes and 2 seconds on the first problem and 4 minutes 58 seconds on the second problem. This would result in a combined score of 12.3 on both problems.\(t_{1}\) \(P(t_{1})\) 0 7.98 5.03 12.34 10 8.99 - Note that the roads form a right triangle where the radar will detect the change in the distance from your car to the cop with respect to time. If we call this distance \(c\) for \(c\), the distance east to the interestection \(e\), we then have \[e^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}=c^{2}.\] Implicitly differentiating we have, \[2e\frac{de}{dt}=2c\frac{dc}{dt}.\] Using that \(\frac{de}{dt}=-55\) (\(e\) is decreasing), and \(e=\frac{1}{2}\) and \(c=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=1\), we get that \(\frac{dc}{dt}=-27.5\).
That is, the you are getting closer to the cop car at a rate of 27.5mph. Therefore, the radar reads your speed as being under 45 and you will therefore not get a ticket.
- Bonus. If the police car is closer to the intersection, it will read your speed as higher, if it is further away it will read your speed lower. If it is moving toward you, the speed will be read higher, whereas if it is moving away from you, the speed will be lower. As an insight, this is the reason police cars are usually directly on the side of the road and not moving. This results in the most accurate reading of speed.